[R] Creating mean C columns out of As and Bs
I am working on a gene expression microarray dataset, each sample has been taking from animal A and animal B on the same type of microarray. I would like to take the mean of these columns and create a dataset with columns with data which are means of the corresponding rows of A and B For example: TissueExpressionA - HTissue[1:22284, c(29, 39, 57, 59, 63, 69, 71, 73, 77, 79, 83, 89, 99, 117, 119, 121, 123, 125, 137, 145, 159)] TissueExpressionB - HTissue[1:22284, c(30, 40, 58, 60, 64, 70, 72, 74, 78, 80, 84, 90, 100, 118, 120, 122, 124, 126, 138, 146, 160)] I would like TissueExpressionC - HTissue[1:22284, c[mean(29,30), mean(39,40), mean(57,58) ... mean(159,160)] I hope that makes sense. If you know of any other method of merging data sets like this any help would be appreciated. Michael Budnick Graduate Researcher St. John's University [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Conditional CCA and Monte Carlo - Help!
Hi All, I am using canonical correspondence analysis to compare a community composition matrix to a matrix of sample spatial relationships and environmental variables. In order to parse out how much variance is explained purely by space (S/E) or the environment (E/S) I am using a conditional (partial) CCA. I want to test significance via Monte Carlo but I can not find a way to do this with a conditional CCA. I have been using vegan for the CCA and attempting to use ade4 to run a Monte Carlo. However, these two packages conflict when it comes to CCA. If I use vegan I can run a conditional CCA, and if I use ade4 I can do a Monte Carlo - but I can't figure out how to do a conditional CCA with ade4 OR a Monte Carlo with vegan. If anyone has experience with this I would be truly grateful for your help! I am fairly new to R, and I have quickly found myself in a place where Google-ing has no longer proven useful. Below are my scripts and error messages. Using Vegan: vare.cca - cca(InvertR.csv ~ Space1 + Space2... + Condition(Env1) + Condition(Env2)..., HabitatSpaceR.csv) randtest(vare.cca, nrepet = 1000) Error in randtest.cca(vare.cca, nrepet = 1000) : Object of class dudi expected Using ade4: vare.cca - cca(InvertR.csv ~ Space1 + Space2... + Condition(Env1) + Condition(Env2)..., HabitatSpaceR.csv) Error in cca(InvertR.csv ~ Space1 + Space2... + Condition(Env1) + Condition(Env2)... + : data.frame expected -- View this message in context: http://r.789695.n4.nabble.com/Conditional-CCA-and-Monte-Carlo-Help-tp4662572.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating mean C columns out of As and Bs
On Mar 26, 2013, at 6:50 PM, Michael Budnick wrote: I am working on a gene expression microarray dataset, each sample has been taking from animal A and animal B on the same type of microarray. I would like to take the mean of these columns and create a dataset with columns with data which are means of the corresponding rows of A and B For example: TissueExpressionA - HTissue[1:22284, c(29, 39, 57, 59, 63, 69, 71, 73, 77, 79, 83, 89, 99, 117, 119, 121, 123, 125, 137, 145, 159)] TissueExpressionB - HTissue[1:22284, c(30, 40, 58, 60, 64, 70, 72, 74, 78, 80, 84, 90, 100, 118, 120, 122, 124, 126, 138, 146, 160)] I would like TissueExpressionC - HTissue[1:22284, c[mean(29,30), mean(39,40), mean(57,58) ... mean(159,160)] mean(29,30) will evaluate to 29.5 which would have picked out a single column allmeans - colMeans(HTissue[1:22284, sort( c(30, 40, 58, 60, 64, 70, 72, 74, 78, 80, 84, 90, 100, 118, 120, 122, 124, 126, 138, 146, 160), c(29, 39, 57, 59, 63, 69, 71, 73, 77, 79, 83, 89, 99, 117, 119, 121, 123, 125, 137, 145, 159)) ] # I do not see a pattern in this pairing. # Will return a vector rather than a data.frame. len - length( sort( c(30, 40, 58, 60, 64, 70, 72, 74, 78, 80, 84, 90, 100, 118, 120, 122, 124, 126, 138, 146, 160), c(29, 39, 57, 59, 63, 69, 71, 73, 77, 79, 83, 89, 99, 117, 119, 121, 123, 125, 137, 145, 159)) ) sapply(seq(1, len-1, by=2), function(colA) mean( c(allmeans[colA:(colA+1)]) ) ) #should return a vector that is half the length of len Testing not done in the absence of data. I hope that makes sense. Perhaps. If you know of any other method of merging data sets like this any help would be appreciated. Michael Budnick Graduate Researcher St. John's University [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] NaNS Error Message
Hi
If you can not share information you need to solve your problems yourself.
you can trace code by some print statements
sn-(S-n);snfact-gamma(sn); print(sn)
or use
?debug
as Sarah pointed out, n is probably greater than 100 and in that case gamma(sn)
results in NaN.
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Sahana Srinivasan
Sent: Tuesday, March 26, 2013 4:52 PM
To: Berend Hasselman
Cc: r-help@r-project.org
Subject: Re: [R] NaNS Error Message
Hi, sorry I can't provide the reproducible code - the formula is
currently being worked on for a research paper so I can't really give
out details in that regard.
That being said, k is initialized to 1 at the beginning of the loop. n
has various values (whole numbers, 0 or greater, no upper limit).
On Tue, Mar 26, 2013 at 3:41 PM, Berend Hasselman b...@xs4all.nl
wrote:
On 26-03-2013, at 16:25, Sahana Srinivasan
sahanasrinivasan...@gmail.com
wrote:
Hi,
I'm using R to do a series of calculation and I have gotten several
warnings that say NaNS produced. Whatever I could read on line
gives me an idea that this warning is produced when the number is
use is a
negative
log or otherwise mathematically problematic.
I'm getting this error while using factorial() and gamma () on
strictly positive numbers (always greater than zero).
Here is a snippet of my code:
k goes from 1 to a positive limit.
S is always 100.
Only positive values of n are allowed into the loop.
if(n0)
{
while(k=lim)
{
sn-(S-n);snfact-gamma(sn);
sn2-(2-n+S);sn2gam-gamma(sn2);
num-(ngam*sn2gam);
nk2-(2+k-(2*n)+S);
nk2gam-gamma(nk2);
den-(k*nk2gam);
prob-(num/den);
sum-(as.numeric((k*prob))+sum);
k-k+1;
}
}
The error message received for every instance of this loop is :
In gamma(sn) : NaNs produced
In gamma(sn2) : NaNs produced
In gamma(nk2) : NaNs produced
This not reproducible code.
Where is ngam?
Where is k set to 1?
What value does n have?
Berend
[[alternative HTML version deleted]]
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[R] Averaging Out many rows from a column AND funtion to string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
or anything similar in size as this is a matrix that needs plotting (so is ok
if it is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and averaging per
column (notice that the column number should stay the same)
b. I would like to be able to convert strings that are function names to real
function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction)
)
print(sprintf('Using the function %s',FunctionIndex[1,1]))
# call the FunctionIndex[1,1] somehow
I would like to thank you in advance for your help
Regards
Alex
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] conditional Dataframe filling
Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c d 0 0 0 0 1 2 3 0 1 0 1 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to use parentheses and degree symbol together?
On Tue, 26-Mar-2013 at 05:05PM +0900, Pascal Oettli wrote:
| Hi,
|
| You are right. The following should solve that problem:
|
| plot(0, 0, pch = )
| text(0, .5, expression(Temperature~(degree*C)))
It's not *exactly* the same. It uses a different font family for the
brackets, evidently from the Symbol family, not Helvetica (or
whatever). Probably more appropriate since it's more akin to the
other characters' features.
So, not only more elegant, but better looking.
|
| HTH,
| Pascal
|
|
| On 26/03/13 16:55, Patrick Connolly wrote:
| On Tue, 26-Mar-2013 at 04:20PM +0900, Pascal Oettli wrote:
|
| | Hi,
| |
| | Is it what you are looking for?
| |
| | plot(0, 0, pch = )
| | text(0, .5, expression(Temperature~(degree ~ C)))
|
| That produces an unwanted space between the degree symbol and the C.
| The search continues.
|
| Thanks
|
| | text(0, .4, substitute(paste(Temperature, B * degree, C)), list(B
| | = ()))
| |
| | Hope this help,
| | Pascal
| |
| |
| |
| | On 26/03/13 16:12, Patrick Connolly wrote:
| | I'm interested in using a regular bracket with the degree symbol as an
| | axis label but it's somewhat simpler to show what I mean in a text
| | statement.
| |
| | plot(0, 0, pch = )
| |
| | If I'm easy to please, this would suffice:
| | text(0, .5, expression(Temperature * degree ~ C))
| |
| | But I'm not that easily pleased. I prefer it to look like this:
| | text(0, .4, substitute(paste(Temperature, B * degree, C)), list(B =
()))
| |
| | It looks fine, but I'm sure there's a more elegant way to do it.
| |
| | Is there?
| |
| |
|
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_) . Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] conditional Dataframe filling
Here's a possible solution. dd - structure(list(a = c(TRUE, FALSE, FALSE), b = c(TRUE, FALSE, TRUE), c = c(TRUE, FALSE, FALSE), d = c(TRUE, TRUE, FALSE)), .Names = c(a, b, c, d), row.names = c(NA, -3L), class = data.frame) ds - as.data.frame(t(apply(!dd, 1, cumsum)-apply(dd, 1, cumsum))) ds[as.matrix(dd)] - 0 ds Best, Nello -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Camilo Mora Sent: Mittwoch, 27. März 2013 09:32 To: r-help@r-project.org Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c d 0 0 0 0 1 2 3 0 1 0 1 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction)
)
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
Regards
Alex
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction)
)
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
[[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newey West HAC for pooled cross-section data
On Tue, 26 Mar 2013, SHISHIR MATHUR wrote: Thanks for the reply Achim. The reason I suspect autocorrelation is because I think that within the same neighborhood, homes sold a few months back are likely to impact the price of homes sold subsequently. This may well be spatial (auto)correlation rather than temporal autocorrelation. In fact the DW test and Breusch-Pagan test come out to be significant. So even though the data is not time series (that is, I do not have repeated observations for the same house), however, the houses sold close in time to each other are in the data set. If there is a unique ordering of all observations by time, then you could in principle apply an autocorrelation correction for the data, e.g., via Newey-West. But from what you describe above, it seems to be more important to capture spatial effects in the data, e.g., by using a spatial lag model (see lagsarlm in spdep) or by using an additive spatial effect (see e.g. gam in mgcv). Thanks, Shish On Tue, Mar 26, 2013 at 3:51 PM, Achim Zeileis achim.zeil...@uibk.ac.at wrote: On Tue, 26 Mar 2013, SHISHIR MATHUR wrote: Hello: My dataset set contains several thousand rows of data, with each row containing information for a house. The variables include the sale price of the house, the quarter and year of sale, the attributes of the house, and the attributes of the neighborhood and the city in which the house is located. The data is for a 10-year period. No house is repeated in the dataset. In summary, the dataset can be termed pooled cross-section data. My question: Can I estimate Newey-West HAC standard errors for a model that estimates the effect of various independent variables on the sale price of the house? My understanding is that Newey-West can be used for time series and panel data. However, I am not sure whether it can be used for pooled cross-section data. If yes, can you refer me to a specific source, such as a paper or a book? The result of your aggregation is a cross-section data set. Thus, there should be no correlation between the different observations - or in other terms, the ordering of your observations is completely arbitrary. Consequently, there may be heteroskedasticity but not autocorrelation. So you may use HC standard errors but HAC should not be necessary. (Using HAC standard errors will still be consistent but less efficient.) -- Best, Shish [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Best, Shishir __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Automatic script for updating packages in R
Baer,Thank you for help.huu Date: Mon, 25 Mar 2013 08:20:28 -0500 From: rb...@atsu.edu To: petr.pi...@precheza.cz CC: uddessy2...@hotmail.com; r-h...@stat.math.ethz.ch Subject: Re: [R] Automatic script for updating packages in R try, update.packages(ask='graphics', checkBuilt=TRUE) ?update.packages ?download.file ?url # file:// URLs On 3/25/2013 6:09 AM, PIKAL Petr wrote: Hi Strange, I thought that only Windows users are sending Html mail and providing sparse info describing their problems. I presume, you use 2.15.3 R version. There is some readme in Ubuntu CRAN repository and if you followed that and fail it would be necessary to provide at least what you did (some code) and what was the error messages. http://cran.r-project.org/bin/linux/ubuntu/README Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Twaha Mlwilo Sent: Monday, March 25, 2013 11:21 AM To: r-h...@stat.math.ethz.ch Subject: [R] Automatic script for updating packages in R Hello all,Good day,Internet access have been a problem , and learning R forced to download packages manual.I have google for script for automatic R update didnt getPlease any one with help?am using ubuntu 12.04R-2.5.3Thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Robert W. Baer, Ph.D. Professor of Physiology Kirksille College of Osteopathic Medicine A. T. Still University of Health Sciences Kirksville, MO 63501 USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
Hi
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org
[mailto:r-help-bounces@r-
project.orghttp://project.org/] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call mean of what? I am sure I do not understand what is your intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Petr
PS. Do not use HTML mail messages.
Regards
Alex
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction)
)
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
[[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] conditional Dataframe filling
Hi, You could try: dat1- read.table(text= a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE ,sep=,header=TRUE) dat2-dat1 dat2[]-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: r-help@r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c d 0 0 0 0 1 2 3 0 1 0 1 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] conditional Dataframe filling
HI, Just a correction: : dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum #should also work A.K. - Original Message - From: arun smartpink...@yahoo.com To: Camilo Mora cm...@dal.ca Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1- read.table(text= a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE ,sep=,header=TRUE) dat2-dat1 dat2[]-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: r-help@r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c d 0 0 0 0 1 2 3 0 1 0 1 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] animated charts
Hello all! I want to create animated chart of temperature variation in last century. how can I do this with R? Thank you! -- --- Catalin-Constantin ROIBU Forestry engineer, PhD Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I show real values on a log10 histogram
Shane Carey careys...@gmail.com on Tue, 26 Mar 2013 11:03:20 + writes: Yup, Ive tried all these things and I think Johns might be the best approach, well, :-) you have not yet seen the following one : if(!require(sfsmisc)) install.packages(sfsmisc) require(sfsmisc) ## the data: set.seed(1); summary(x - rlnorm(100, m = 2, sdl = 3)) ## the plot (w/o x-axis) : r - hist(log10(x), xaxt = n, xlab = x [log scale]) ## the nice axis: axt - axTicks(1) eaxis(1, at = axt, labels = pretty10exp(10^axt, drop.1=TRUE)) Martin Maechler, ETH Zurich thanks On Tue, Mar 26, 2013 at 11:01 AM, PIKAL Petr petr.pi...@precheza.cz wrote: Hi maybe axis(3, (x)^10) Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Shane Carey Sent: Tuesday, March 26, 2013 11:19 AM To: r-help@r-project.org Subject: [R] How do I show real values on a log10 histogram Hi, I have a histogram with values logged to the base 10 hist(log10(x),breaks=60) How do I show the log values on the x-axis and a second x-axis showing the real values? Thanks -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
 num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
 num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx-cut(1:30, 10)
keep.ag-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
              c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Petr
PS. Do not use HTML mail messages.
Regards
Alex
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
            c(2,TakeFunction)
           )
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
   [[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I show real values on a log10 histogram
Well, I don't think that is what Shane wants either but my suggestion is clearly wrong. I was reading the question as a dual axis not just a suplimentary , equivalent value axis. DUH. What I think he wants is what IIRC Petr orginally suggested which is this.. Create the data to be graphed x-1:5 y1-x y2-x^2 # Set the par values # op - par(las=1,xaxs=r,mai=c(1,1,1,1)) # Draw the plot plot(x,y1,xlim=c(0,5),ylim=c(0,5), ylab=y1, las = 1) title(main=Equvalent Axes in R) axis(3, labels = c(10, 100, 1,000, 10,000, 100,000), at=1:5) text(12, 50, y2, srt = 270, xpd = TRUE) par(op) # reset par John Kane Kingston ON Canada -Original Message- From: maech...@stat.math.ethz.ch Sent: Wed, 27 Mar 2013 14:37:58 +0100 To: careys...@gmail.com Subject: Re: [R] How do I show real values on a log10 histogram Shane Carey careys...@gmail.com on Tue, 26 Mar 2013 11:03:20 + writes: Yup, Ive tried all these things and I think Johns might be the best approach, well, :-) you have not yet seen the following one : if(!require(sfsmisc)) install.packages(sfsmisc) require(sfsmisc) ## the data: set.seed(1); summary(x - rlnorm(100, m = 2, sdl = 3)) ## the plot (w/o x-axis) : r - hist(log10(x), xaxt = n, xlab = x [log scale]) ## the nice axis: axt - axTicks(1) eaxis(1, at = axt, labels = pretty10exp(10^axt, drop.1=TRUE)) Martin Maechler, ETH Zurich thanks On Tue, Mar 26, 2013 at 11:01 AM, PIKAL Petr petr.pi...@precheza.cz wrote: Hi maybe axis(3, (x)^10) Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Shane Carey Sent: Tuesday, March 26, 2013 11:19 AM To: r-help@r-project.org Subject: [R] How do I show real values on a log10 histogram Hi, I have a histogram with values logged to the base 10 hist(log10(x),breaks=60) How do I show the log values on the x-axis and a second x-axis showing the real values? Thanks -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Receive Notifications of Incoming Messages Easily monitor multiple email accounts access them with a click. Visit http://www.inbox.com/notifier and check it out! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] animated charts
maybe try the shiny examples that use googleviz. Have a look at the following link: https://code.google.com/p/google-motion-charts-with-r/ I would be interested to know what you decide. Thanks On Wed, Mar 27, 2013 at 1:20 PM, catalin roibu catalinro...@gmail.comwrote: Hello all! I want to create animated chart of temperature variation in last century. how can I do this with R? Thank you! -- --- Catalin-Constantin ROIBU Forestry engineer, PhD Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I show real values on a log10 histogram
Yes, what I wanted was the original suggestion and that is what I went with. John, I found a solution around par also. There is a function called split.screen that allows you split the screen up, and also use par at the same time. Thanks everyone for all your help. Cheers On Wed, Mar 27, 2013 at 2:02 PM, John Kane jrkrid...@inbox.com wrote: Well, I don't think that is what Shane wants either but my suggestion is clearly wrong. I was reading the question as a dual axis not just a suplimentary , equivalent value axis. DUH. What I think he wants is what IIRC Petr orginally suggested which is this.. Create the data to be graphed x-1:5 y1-x y2-x^2 # Set the par values # op - par(las=1,xaxs=r,mai=c(1,1,1,1)) # Draw the plot plot(x,y1,xlim=c(0,5),ylim=c(0,5), ylab=y1, las = 1) title(main=Equvalent Axes in R) axis(3, labels = c(10, 100, 1,000, 10,000, 100,000), at=1:5) text(12, 50, y2, srt = 270, xpd = TRUE) par(op) # reset par John Kane Kingston ON Canada -Original Message- From: maech...@stat.math.ethz.ch Sent: Wed, 27 Mar 2013 14:37:58 +0100 To: careys...@gmail.com Subject: Re: [R] How do I show real values on a log10 histogram Shane Carey careys...@gmail.com on Tue, 26 Mar 2013 11:03:20 + writes: Yup, Ive tried all these things and I think Johns might be the best approach, well, :-) you have not yet seen the following one : if(!require(sfsmisc)) install.packages(sfsmisc) require(sfsmisc) ## the data: set.seed(1); summary(x - rlnorm(100, m = 2, sdl = 3)) ## the plot (w/o x-axis) : r - hist(log10(x), xaxt = n, xlab = x [log scale]) ## the nice axis: axt - axTicks(1) eaxis(1, at = axt, labels = pretty10exp(10^axt, drop.1=TRUE)) Martin Maechler, ETH Zurich thanks On Tue, Mar 26, 2013 at 11:01 AM, PIKAL Petr petr.pi...@precheza.cz wrote: Hi maybe axis(3, (x)^10) Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Shane Carey Sent: Tuesday, March 26, 2013 11:19 AM To: r-help@r-project.org Subject: [R] How do I show real values on a log10 histogram Hi, I have a histogram with values logged to the base 10 hist(log10(x),breaks=60) How do I show the log values on the x-axis and a second x-axis showing the real values? Thanks -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Receive Notifications of Incoming Messages Easily monitor multiple email accounts access them with a click. Visit http://www.inbox.com/notifier and check it out! -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odd graphic device behavior
Any chance that you made an earlier call to par() resetting cex in your session? I just had that happen. John Kane Kingston ON Canada -Original Message- From: tea...@gmail.com Sent: Tue, 26 Mar 2013 10:15:33 -0400 To: r-help@r-project.org Subject: [R] Odd graphic device behavior I'm experiencing odd graphics device behavior running R 2.15.3 on Ubuntu. Regardless of what I try like: require(stats) plot(cars) lines(lowess(cars)) plot(sin, -pi, 2*pi) for example, the graphics device fills the entire screen with the graphic and a very large font. When I shrink the graphics device window, the lettering remains large and the line thicknesses stay quite thick. It may have been some time since I have done any R work on my Ubuntu computer, but clearly this did not happen previously. Does anyone have any thoughts? Thank you, Tom [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I show real values on a log10 histogram
Glad it worked. Sorry to mislead you so badly. John Kane Kingston ON Canada -Original Message- From: careys...@gmail.com Sent: Wed, 27 Mar 2013 14:09:58 + To: jrkrid...@inbox.com Subject: Re: [R] How do I show real values on a log10 histogram Yes, what I wanted was the original suggestion and that is what I went with. John, I found a solution around par also. There is a function called split.screen that allows you split the screen up, and also use par at the same time. Thanks everyone for all your help. Cheers On Wed, Mar 27, 2013 at 2:02 PM, John Kane [1]jrkrid...@inbox.com wrote: Well, I don't think that is what Shane wants either but my suggestion is clearly wrong. I was reading the question as a dual axis not just a suplimentary , equivalent value axis. DUH. What I think he wants is what IIRC Petr orginally suggested which is this.. Create the data to be graphed x-1:5 y1-x y2-x^2 # Set the par values # op - par(las=1,xaxs=r,mai=c(1,1,1,1)) # Draw the plot plot(x,y1,xlim=c(0,5),ylim=c(0,5), ylab=y1, las = 1) title(main=Equvalent Axes in R) axis(3, labels = c(10, 100, 1,000, 10,000, 100,000), at=1:5) text(12, 50, y2, srt = 270, xpd = TRUE) par(op) # reset par John Kane Kingston ON Canada -Original Message- From: [2]maech...@stat.math.ethz.ch Sent: Wed, 27 Mar 2013 14:37:58 +0100 To: [3]careys...@gmail.com Subject: Re: [R] How do I show real values on a log10 histogram Shane Carey [4]careys...@gmail.com on Tue, 26 Mar 2013 11:03:20 + writes: Yup, Ive tried all these things and I think Johns might be the best approach, well, :-) you have not yet seen the following one : if(!require(sfsmisc)) install.packages(sfsmisc) require(sfsmisc) ## the data: set.seed(1); summary(x - rlnorm(100, m = 2, sdl = 3)) ## the plot (w/o x-axis) : r - hist(log10(x), xaxt = n, xlab = x [log scale]) ## the nice axis: axt - axTicks(1) eaxis(1, at = axt, labels = pretty10exp(10^axt, drop.1=TRUE)) Martin Maechler, ETH Zurich thanks On Tue, Mar 26, 2013 at 11:01 AM, PIKAL Petr [5]petr.pi...@precheza.cz wrote: Hi maybe axis(3, (x)^10) Regards Petr -Original Message- From: [6]r-help-boun...@r-project.org [mailto:[7]r-help-bounces@r- [8]project.org] On Behalf Of Shane Carey Sent: Tuesday, March 26, 2013 11:19 AM To: [9]r-help@r-project.org Subject: [R] How do I show real values on a log10 histogram Hi, I have a histogram with values logged to the base 10 hist(log10(x),breaks=60) How do I show the log values on the x-axis and a second x-axis showing the real values? Thanks -- Shane [[alternative HTML version deleted]] __ [10]R-help@r-project.org mailing list [11]https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide [12]http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Shane [[alternative HTML version deleted]] __ [13]R-help@r-project.org mailing list [14]https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide [15]http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [16]R-help@r-project.org mailing list [17]https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide [18]http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Receive Notifications of Incoming Messages Easily monitor multiple email accounts access them with a click. Visit [19]http://www.inbox.com/notifier and check it out! -- Shane _ [20]3D Earth Screensaver Preview Free 3D Earth Screensaver Watch the Earth right on your desktop! Check it out at [21]www.inbox.com/earth References 1. mailto:jrkrid...@inbox.com 2. mailto:maech...@stat.math.ethz.ch 3. mailto:careys...@gmail.com
Re: [R] animated charts
Dear Catalin, If you have an R function that plots a fragment of your data and takes the specific fragment selection as parameter, you can readily turn this into a Shiny app. The slider control (see example in http://rstudio.github.com/shiny/tutorial/#sliders) in Shiny can be animated (you can have a play button) which seems to be just what you need. Kind regards, Andrius 2013/3/27 catalin roibu catalinro...@gmail.com Hello all! I want to create animated chart of temperature variation in last century. how can I do this with R? Thank you! -- --- Catalin-Constantin ROIBU Forestry engineer, PhD Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] FMOLS DOLS and ADL regression
Whether can any R package run Full modified OLS (Phillips and Hansen 1990 ), DOLS (Stock and Watson 1993) and ADL model (Pesaran and Shin 2001) for cointegrated VAR model? I cannot find any useful order in VAR and SVAR package. Thanks. Eric Wang [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Crrstep help
Hi, I'm using crrstep package to do stepwise covariate selection for the Fine Gray competing risks regression model. However, I keep getting an error (please see below). Please help!! PHstep - crrstep(years~1+var1+var2+var3+var4+var5,scope.min=~1,censorcmprsk, data=crisk, direction=c(forward), crr.object = FALSE, trace = TRUE, steps = 100) crrstep(formula = years ~ 1 + var1+var2+var3+var4+var5, scope.min = ~1, etype = censorcmprsk, data = crisk, direction = c(forward), crr.object = FALSE, trace = TRUE, steps = 100) *Error in crr(ftime, fstatus, cov1, variance = TRUE, ...) : unused argument(s) (variance = TRUE)* Thank you very much! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Distance calculation
HI,
Try this:
dat1- read.csv(DQP.csv,sep=\t)
res- do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x) {x1-rbind(dat1[i,],x);colnames(x1)-gsub([.],,colnames(x1));
if({indx- colSums(x1[,2:3]==0);indx[1]==0 indx[2]==1 }) #2 peaks 1 peak
comparison
{x2- x1[order(x1$Peak2t,x1$Npeak2t),];
with(x2,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))
})
}
else #cases where peaks are similar
{with(x1,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))
+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))})
}
}
res2-do.call(cbind,lapply(seq_len(ncol(res)),function(i)
c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
row.names(res2)-1:nrow(res2)
dim(res2)
#[1] 124 124
res2[1:5,1:5]
# [,1] [,2] [,3] [,4] [,5]
#1 0.000 1.512 7.031 3.662 13.030
#2 1.512 0.000 7.109 4.880 18.731
#3 7.031 7.109 0.000 0.056 1.280
#4 3.662 4.880 0.056 0.000 0.584
#5 13.030 18.731 1.280 0.584 0.000
A.K.
From: eliza botto eliza_bo...@hotmail.com
To: smartpink...@yahoo.com smartpink...@yahoo.com
Sent: Wednesday, March 27, 2013 8:58 AM
Subject: RE: Distance calculation
Dear Arun,
I would like to ask a small question.
In the distance calculation procedure, if there are only 2 peaks and 1 peaks
stations and there are no 3 and 4 peaks stations, like the file i attached. How
to modify the script below to eliminate 3 peaks and 4 peaks scripts??
I hope you mind my hasty questioning
Thanks in advance
Elisa
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
Hi
str(as.matrix(keep.ag[,-1]) ) # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:30] V1 V2 V3 V4 ...
Please read and follow what was recommended.
quote
use as.matrix(data.frame) on numeric part
^^^
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
And please can you explain how mean(rnorm(whatever)) shall be integer?
Regards
Petr
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org
[mailto:r-help-bounces@r-
project.orghttp://project.org/] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30. but
it looks at the end that I do not get the Data part of the data.fram correctly:
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx-cut(1:30, 10)
keep.ag-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Petr
PS. Do not use HTML mail messages.
Regards
Alex
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction)
)
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
[[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
I have fixed it like that:
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo-500
idx-cut(1:nrow(keep), ShrinkTo)
keep.ag-aggregate(keep, list(idx), mean)
PlotMe-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
str(as.matrix(keep.ag[,-1]) ) Â # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, dimnames)=List of 2
 ..$ : NULL
 ..$ : chr [1:30] V1 V2 V3 V4 ...
Â
Please read and follow what was recommended.
Â
quote
Â
use as.matrix(data.frame) on numeric part
                                       Â
                 ^^^
Â
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
Â
And please can you explain how mean(rnorm(whatever)) shall be integer?
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
 num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
 num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
Â
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx-cut(1:30, 10)
keep.ag-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
Â
Â
Â
Â
Â
Â
Â
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
              c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Petr
PS. Do not use HTML mail messages.
Regards
Alex
b. I would like to be able to convert strings that are function names
to real function calls. For example I have something like
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
            c(2,TakeFunction)
           )
print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf -function(fun=mean, arg) eval(call(fun, arg))
Regards
Petr
I would like to thank you in advance for your help
Regards
Alex
   [[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Controlling Raster plots
Hi,
I have a few raster layers and I would like to customized their x and y axis.
I have tried already something like:
require('raster')
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
xlab-seq(100e6,200e6,length.out=5)
test-raster(keep)
plot(test,ylab=,xaxt=n,yaxt=n)
axis(1, at=seq(0,1,length.out=5), xlab)
that normally would work in any other plot function.
Could you please help me add the xaxis ?
I would like to thank you in advance for your reply
Regards
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
Hi
system.time({
+ keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
+ ShrinkTo-500
+ idx-cut(1:nrow(keep), ShrinkTo)
+ keep.ag-aggregate(keep, list(idx), mean)
+ PlotMe-as.matrix(keep.ag[,-1])
+ })
user system elapsed
29.230.14 29.37
It takes 30 seconds when using 30 rows. It is not enough time to get a cup
of tee, which I do not consider ages. Maybe split lapply approach or
data.matrix or **ply could be quicker but I do not consider worth spending
hours to elaborate some solution which will spare 20 sec computing time.
Regards
Petr
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 4:12 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
I have fixed it like that:
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo-500
idx-cut(1:nrow(keep), ShrinkTo)
keep.ag-aggregate(keep, list(idx), mean)
PlotMe-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
str(as.matrix(keep.aghttp://keep.ag/[,-1]) ) # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:30] V1 V2 V3 V4 ...
Please read and follow what was recommended.
quote
use as.matrix(data.frame) on numeric part
^^^
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
And please can you explain how mean(rnorm(whatever)) shall be integer?
Regards
Petr
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org
[mailto:r-help-bounces@r-
project.orghttp://project.org/] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30. but
it looks at the end that I do not get the Data part of the data.fram correctly:
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx-cut(1:30, 10)
keep.ag-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Re: [R] Averaging Out many rows from a column AND funtion to string
Well my true matrix is
of
num [1:153899, 1:3415]
that I want to convert to something like
num [1:1000, 1:3415] (keeping column number the same).
I only give subsets here to allow others to run the code at their computer
Thanks a lot
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 4:45 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
system.time({
+ keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
+ ShrinkTo-500
+ idx-cut(1:nrow(keep), ShrinkTo)
+ keep.ag-aggregate(keep, list(idx), mean)
+ PlotMe-as.matrix(keep.ag[,-1])
+ })
  user system elapsed
  29.23   0.14  29.37
Â
Â
It takes 30 seconds when using 30 rows. It is not enough time to get a cup
of tee, which I do not consider ages. Maybe split lapply approach or
data.matrix or  **ply could be quicker but I do not consider worth spending
hours to elaborate some solution which will spare 20 sec computing time.
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 4:12 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
I have fixed it like that:
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo-500
idx-cut(1:nrow(keep), ShrinkTo)
keep.ag-aggregate(keep, list(idx), mean)
PlotMe-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
Â
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
str(as.matrix(keep.ag[,-1]) ) Â # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, dimnames)=List of 2
 ..$ : NULL
 ..$ : chr [1:30] V1 V2 V3 V4 ...
Â
Please read and follow what was recommended.
Â
quote
Â
use as.matrix(data.frame) on numeric part
                                       Â
                 ^^^
Â
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
Â
And please can you explain how mean(rnorm(whatever)) shall be integer?
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Â
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
 num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
 num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
Â
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx-cut(1:30, 10)
keep.ag-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
Â
Â
Â
Â
Â
Â
Â
2) I want to have a string that can be used also for calling a function with
the same name. Think of using mean to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction- function(){}
FunctionIndex- rbind (c(1,LogFunction),
              c(2,TakeFunction))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not
[R] find and replace characters in a string
Hi, I have a string of text as follows LOI . How do I replace the dot with (%) gsub(.,(%),LOI .) gives (%)(%)(%)(%)(%) Thanks -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odd graphic device behavior
John, Thanks for the suggestion, but no. I have even gone so far as to rebuild R from source, re-booted my computer, and tried the 'experiment': require(stats) plot(cars) immediately after starting R. Still the same result. I think it must be related to some default Ubuntu Unity window manager default I can't sort out. Regards, Tom On Wed, Mar 27, 2013 at 10:14 AM, John Kane jrkrid...@inbox.com wrote: Any chance that you made an earlier call to par() resetting cex in your session? I just had that happen. John Kane Kingston ON Canada -Original Message- From: tea...@gmail.com Sent: Tue, 26 Mar 2013 10:15:33 -0400 To: r-help@r-project.org Subject: [R] Odd graphic device behavior I'm experiencing odd graphics device behavior running R 2.15.3 on Ubuntu. Regardless of what I try like: require(stats) plot(cars) lines(lowess(cars)) plot(sin, -pi, 2*pi) for example, the graphics device fills the entire screen with the graphic and a very large font. When I shrink the graphics device window, the lettering remains large and the line thicknesses stay quite thick. It may have been some time since I have done any R work on my Ubuntu computer, but clearly this did not happen previously. Does anyone have any thoughts? Thank you, Tom [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! Check it out at http://www.inbox.com/marineaquarium [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] find and replace characters in a string
txt- LOI . gsub([.],%,txt) #[1] LOI % A.K. From: Shane Carey careys...@gmail.com To: r-help@r-project.org Sent: Wednesday, March 27, 2013 12:09 PM Subject: [R] find and replace characters in a string Hi, I have a string of text as follows LOI . How do I replace the dot with (%) gsub(.,(%),LOI .) gives (%)(%)(%)(%)(%) Thanks -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] find and replace characters in a string
Hello, The period is a metacharacter so you have to escape it. The period is escaped with a '\'. In it's turn, '\' is a metacharacter so it needs to be escaped. Hence the double'\\'. x - LOI . gsub(\\., (%), x) Hope this helps, Rui Barradas Em 27-03-2013 16:09, Shane Carey escreveu: Hi, I have a string of text as follows LOI . How do I replace the dot with (%) gsub(.,(%),LOI .) gives (%)(%)(%)(%)(%) Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
Hi
Such a big matrix is above capability of my computer as it requires about 4GB
(If I compute it correctly) and I have only 2GB.
Maybe you could try package data.table which is designed for fast data
manipulation.
Regards
Petr
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 5:05 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Well my true matrix is
of
num [1:153899, 1:3415]
that I want to convert to something like
num [1:1000, 1:3415] (keeping column number the same).
I only give subsets here to allow others to run the code at their computer
Thanks a lot
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 4:45 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
system.time({
+ keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
+ ShrinkTo-500
+ idx-cut(1:nrow(keep), ShrinkTo)
+ keep.ag-aggregate(keep, list(idx), mean)
+ PlotMe-as.matrix(keep.aghttp://keep.ag/[,-1])
+ })
user system elapsed
29.230.14 29.37
It takes 30 seconds when using 30 rows. It is not enough time to get a cup
of tee, which I do not consider ages. Maybe split lapply approach or
data.matrix or **ply could be quicker but I do not consider worth spending
hours to elaborate some solution which will spare 20 sec computing time.
Regards
Petr
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 4:12 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
I have fixed it like that:
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo-500
idx-cut(1:nrow(keep), ShrinkTo)
keep.ag-aggregate(keep, list(idx), mean)
PlotMe-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
str(as.matrix(keep.aghttp://keep.ag/[,-1]) ) # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:30] V1 V2 V3 V4 ...
Please read and follow what was recommended.
quote
use as.matrix(data.frame) on numeric part
^^^
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
And please can you explain how mean(rnorm(whatever)) shall be integer?
Regards
Petr
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
From: Alaios [mailto:ala...@yahoo.com]
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
To: Alaios ala...@yahoo.commailto:ala...@yahoo.com; R help
R-help@r-project.orgmailto:R-help@r-project.org
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org
[mailto:r-help-bounces@r-
project.orghttp://project.org/] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if it
is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values
Re: [R] find and replace characters in a string
Although I am not an expert, this is simple. txt- LOI . gsub(.,%,txt, fixed=TRUE) Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Rui Barradas Sent: Wednesday, March 27, 2013 5:17 PM To: Shane Carey Cc: r-help@r-project.org Subject: Re: [R] find and replace characters in a string Hello, The period is a metacharacter so you have to escape it. The period is escaped with a '\'. In it's turn, '\' is a metacharacter so it needs to be escaped. Hence the double'\\'. x - LOI . gsub(\\., (%), x) Hope this helps, Rui Barradas Em 27-03-2013 16:09, Shane Carey escreveu: Hi, I have a string of text as follows LOI . How do I replace the dot with (%) gsub(.,(%),LOI .) gives (%)(%)(%)(%)(%) Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] when to use which apply function?
Thanks for pointing that out.
Mike
On Tue, Mar 26, 2013 at 6:53 PM, David Winsemius dwinsem...@comcast.netwrote:
On Mar 26, 2013, at 2:51 PM, C W wrote:
Dear list,
I am a little confused as to when to use apply, sapply, tapply, vapply,
replicate. I've encountered this several times,
This is time, this is what I am working on,
mat - matrix(c(seq(from=1, to=10), rnorm(10)), ncol=2)
a=1; b=5
newfun - function(x, y, a, b)
{
x*y+a+b
}
sapply(i=1:10, newfun(x=mat[i, 1], y=mat[i, 2], a=a, b=b))
Error in match.fun(FUN) : argument FUN is missing, with no default
I want to use ith row of mat, evaluate newfun(). Am I making a parameter
mistake,
There is nothing that would naturally catch the values. After they are
evaluated sequentially by `sapply` the values 1:10 no longer are named i
and so don't naturally fall into the slots you thought you had constructed
for them. By constructing an anonymous function with any formal argument
name (as already demonstrated by another poster) you could get to use
positional matching.
or should I use a different apply function?
Yes, some other function since all those operations are vectorized, this
could be more easily done with:
mat[ ,1]*mat[ ,2]+a+b # and no need for a loop, right?
--
David Winsemius
Alameda, CA, USA
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odd graphic device behavior
On Mar 27, 2013, at 9:07 AM, Thomas Adams wrote: John, Thanks for the suggestion, but no. I have even gone so far as to rebuild R from source, re-booted my computer, and tried the 'experiment': require(stats) plot(cars) immediately after starting R. Still the same result. I think it must be related to some default Ubuntu Unity window manager default I can't sort out. Have your deleted the default workspace and history files? When they get corrupted, odd things can happen. Regards, Tom On Wed, Mar 27, 2013 at 10:14 AM, John Kane jrkrid...@inbox.com wrote: Any chance that you made an earlier call to par() resetting cex in your session? I just had that happen. John Kane Kingston ON Canada -Original Message- From: tea...@gmail.com Sent: Tue, 26 Mar 2013 10:15:33 -0400 To: r-help@r-project.org Subject: [R] Odd graphic device behavior I'm experiencing odd graphics device behavior running R 2.15.3 on Ubuntu. Regardless of what I try like: require(stats) plot(cars) lines(lowess(cars)) plot(sin, -pi, 2*pi) for example, the graphics device fills the entire screen with the graphic and a very large font. When I shrink the graphics device window, the lettering remains large and the line thicknesses stay quite thick. It may have been some time since I have done any R work on my Ubuntu computer, but clearly this did not happen previously. Does anyone have any thoughts? Thank you, Tom [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! Check it out at http://www.inbox.com/marineaquarium [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error while using register from fda package
Hi all, I think I've run into an error caused by my computer's having too little computational power, but I'm not sure. I used the following code to perform continuous registration on two functional objects: lambda1 - 10E-2 norder=8 samples-seq(0,780,length=781) nbasis - length(samples)+norder-2 mybasis-create.bspline.basis(c(0,780),nbasis,norder) myfdPar-fdPar(mybasis,4,lambda1) myfd.unemployment - smooth.basis(samples,unemployment.rate,myfdPar)$fd myfd.unemployment.squared - smooth.basis(samples,unemp.squared,myfdPar)$fd lambda2-10E-2 myfdPar-fdPar(mybasis,4,lambda2) myfd.inflation - smooth.basis(samples,inflation.rate,myfdPar,)$fd myfd.inflation.squared - smooth.basis(samples,inflation.squared,myfdPar)$fd plot(myfd.inflation) #register lambda=1 nbasis=myfd.unemployment$basis$nbasis ntrial=1 y0fd=mean(myfd.inflation) yfd=myfd.unemployment coef0-matrix(0,nbasis,ntrial) wfd-fd(coef0,mybasis) wfdPar=fdPar(coef0,2,lambda) reglist-register.fd(y0fd,yfd,wfdPar,iterlim=10,dbglev=1) I get the following error after the last command: Error in matrix(0, length(x), nbaspr) : too many elements specified Is this error associated with my computer's not having sufficient power to perform the registration? I can't think of what else that might be going on... Thanks, Zoe Richards [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Setting up a model in package dlm()
Hello, I apologize for such a basic question, but I have been trying to do this in multiple packages without much success. I am trying to set up a state space model for Kalman filtering. I am using package dlm. The DLM is specified by: observation: y(t) = F(t)*theta(t) + v(t) state: theta(t) = G(t)*theta(t-1) + w(t) I have no problem setting up a simple example where F is constant. I am trying to set up a model where F(t) (the output matrix) has time varying values (x1(t),x2(t)). I keep getting an incompatible dimension error despite trying a number of different permutations. I do have a copy of the Petris book, but I haven't had a chance to give it a thorough read at this point. Any assistance would be greatly appreciated: Example code: #n x p output matrix FMat-cbind(rnorm(123),rnorm(123)) #p x p state transition matrix GMat-matrix(c(1,0,0,1),ncol=2) #p x p system noise distribution WMat-matrix(c(0.02,0,0,0.02),ncol=2) # nxn measurement noise Vmat-matrix(0,ncol=123,nrow=123) diag(Vmat)-0.02 #initial state mean and variance m0Vec-c(1,1) c0Vec-c(0.05,0.05) ssMod-dlm(FF=FMat,V=Vmat,GG=GMat,W=WMat,m0=m0Vec,C0=c0Vec) Resulting error: Error in dlm(FF = FMat, V = Vmat, GG = GMat, W = WMat, m0 = m0Vec, C0 = c0Vec) : Incompatible dimensions of matrices [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Weighted Kaplan-Meier estimates with R
Thanks to all of you for your very helpful comments! As Terry suggested, svykm is what I was looking for. While testing that I get the same results with the package survey as with the package survival, I encountered the issue of how to draw survival curves. Apparently the implementations in the two packages differ, as I show below. I would very much welcome your views, since the tail of the survival curve has a major impact on the interpretation of my results. In my data, the last ‘death’ occurs at 2094 days, while the last censoring time is 3297 days. If possible, I would like to say something about the probability between 2100 days and 3300 days. So, my question is that after the last observed death, in the very simple example below at 883 days, how should one draw the survival curve? The graph produced by svykm (package survey) ends at 883 days, whereas survfit (package survival) continues the graph all the way to the last censoring time, which is at 1022 days. Please run the code below to see my point. There are no weights. With weights, I face the same issue. require(survival) lungSubSet - lung S - Surv(lung$time,lung$status) sKm - survfit(S~1) ## require(survey) svyDesignObject- svydesign(id=~1,weights=~1,data=lungSubSet) svyKm - svykm(S~1,design=svyDesignObject,se=T) ## plot(svyKm,xlim=c(0,1200)) lines(sKm,conf.int=T,mark.time=F,col='green') -- View this message in context: http://r.789695.n4.nabble.com/Re-Weighted-Kaplan-Meier-estimates-with-R-tp4662494p4662619.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odd graphic device behavior
On Mar 27, 2013, at 18:11 , David Winsemius wrote: On Mar 27, 2013, at 9:07 AM, Thomas Adams wrote: John, Thanks for the suggestion, but no. I have even gone so far as to rebuild R from source, re-booted my computer, and tried the 'experiment': require(stats) plot(cars) immediately after starting R. Still the same result. I think it must be related to some default Ubuntu Unity window manager default I can't sort out. Have your deleted the default workspace and history files? When they get corrupted, odd things can happen. ...or try R --vanilla The immediate suspicion is that something is tampering with your graphics device options, so - figure out what device you are using (dev.list()) - if it is X11, have a look at X11.options() - try running X11(width=7, height=7) to see if the automagic settings get it wrong. - try system(xdpyinfo); this may give a long list of gibberish, but look for dimensions and resolution and see if they look sane. E.g., I get screen #0: dimensions:2560x1418 pixels (677x375 millimeters) resolution:96x96 dots per inch Regards, Tom On Wed, Mar 27, 2013 at 10:14 AM, John Kane jrkrid...@inbox.com wrote: Any chance that you made an earlier call to par() resetting cex in your session? I just had that happen. John Kane Kingston ON Canada -Original Message- From: tea...@gmail.com Sent: Tue, 26 Mar 2013 10:15:33 -0400 To: r-help@r-project.org Subject: [R] Odd graphic device behavior I'm experiencing odd graphics device behavior running R 2.15.3 on Ubuntu. Regardless of what I try like: require(stats) plot(cars) lines(lowess(cars)) plot(sin, -pi, 2*pi) for example, the graphics device fills the entire screen with the graphic and a very large font. When I shrink the graphics device window, the lettering remains large and the line thicknesses stay quite thick. It may have been some time since I have done any R work on my Ubuntu computer, but clearly this did not happen previously. Does anyone have any thoughts? Thank you, Tom [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! Check it out at http://www.inbox.com/marineaquarium [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] a similar question
Hi Elisa,
Try this:
dat1- read.csv(DQV.csv,sep=\t)
res- do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),function(x)
{x1- rbind(dat1[i,],x); colnames(x1)- gsub([.],,colnames(x1));
if({indx- colSums(is.na(x1)); indx[2]==0 indx[3]==1}) # 2 peaks 1 peak
comparison
{x2- x1[rev(order(x1$Peak2v, x1$Npeak2v)),];
with(x2,{abs(Peak1v[1]-Peak1v[2])+abs(Peak1v[1]-Peak2v[2]) +
abs(Npeak1v[1]-Npeak1v[2])+abs(Npeak1v[1]-Npeak2v[2]) })
}
else {x1[is.na(x1)]-0; x1; #cases where peaks are similar
with(x1,{abs(Peak1v[1]-Peak1v[2])+ abs(Peak2v[1]-Peak2v[2])+
abs(Npeak1v[1]-Npeak1v[2])+ abs(Npeak2v[1]-Npeak2v[2])})
}
}
res2-do.call(cbind,lapply(seq_len(ncol(res)),function(i)
c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
row.names(res2)-1:nrow(res2)
dim(res2)
#[1] 124 124
res2[1:5,1:5]
# [,1] [,2] [,3] [,4] [,5]
#1 0.000 0.8024471 2.3537210 2.1929718 3.7746302
#2 0.8024471 0.000 3.1233237 2.9954189 4.5442329
#3 2.3537210 3.1233237 0.000 0.4094026 0.8108279
#4 2.1929718 2.9954189 0.4094026 0.000 0.8571442
#5 3.7746302 4.5442329 0.8108279 0.8571442 0.000
A.K.
From: eliza botto eliza_bo...@hotmail.com
To: smartpink...@yahoo.com smartpink...@yahoo.com
Sent: Wednesday, March 27, 2013 12:07 PM
Subject: RE: a similar question
Dear Arun,
The last question regarding this distance measurement is that how to modify the
following codes for calculation of distance if we have 2 peaks atmost. The data
file is attached
Thanks in advance
Elisa
Date: Tue, 26 Mar 2013 16:02:01 -0700
From: smartpink...@yahoo.com
Subject: Re: a similar question
To: eliza_bo...@hotmail.com
CC: r-help@r-project.org
HI Elisa,
You need to review the formulas. Some of them (especially between 84 85 or
4 84 is not making sense, possibly typos). I changed it according to the
pattern of the formulas.
source(ElisaNew.txt)
#mat1 # dataset
res-do.call(cbind,lapply(seq_len(nrow(mat1)),function(i)
do.call(rbind,lapply(split(rbind(mat1[i,],mat1[-i,]),1:nrow(rbind(mat1[i,],mat1[-i,]))),
function(x) {x1- rbind(mat1[i,],x);x1-as.data.frame(x1);
if({indx-colSums(is.na(x1));indx[1]==0 indx[2]==0 indx[3]==1
indx[4]==2}) #3 peaks 2 peaks comparison
{x2- x1[rev(order(x1$Peak3,x1$Npeak3)),];
with(x2,{abs(Peak1[1]-Peak1[2])+abs(Peak2[1]-Peak2[2])+abs(Peak1[1]-Peak3[2])
+ abs(Npeak1[1]-Npeak1[2])+abs(Npeak2[1]-Npeak2[2])+abs(Npeak1[1]-Npeak3[2])})
}
else if({indx[1]==0 indx[2]==1 indx[3]==2 indx[4]==2}) #2 peaks 1 peak
comparison
{x3- x1[rev(order(x1$Peak2,x1$Npeak2)),];
with(x3,{abs(Peak1[1]-Peak1[2])+ abs(Peak1[1]-Peak2[2])+
abs(Npeak1[1]-Npeak1[2])+ abs(Npeak1[1]-Npeak2[2])})
}
else if({indx[1]==0 indx[2]==0 indx[3]==1 indx[4]==1}) #4 peaks 2 peaks
comparison
{x4- x1[rev(order(x1$Peak3,x1$Peak4,x1$Npeak3,x1$Npeak4)),];
with(x4,{abs(Peak1[1]-Peak1[2])+abs(Peak2[1]-Peak2[2])+
abs(Peak1[1]-Peak3[2])+ abs(Peak2[1]-Peak4[2])+
abs(Npeak1[1]-Npeak1[2])+abs(Npeak2[1]-Npeak2[2])+ abs(Npeak1[1]-Npeak3[2])+
abs(Npeak2[1]-Npeak4[2])})
}
else if({indx[1]==0 indx[2]==0 indx[3]==0 indx[4]==1}) #4 peaks 3 peaks
comparison
{x5- x1[rev(order(x1$Peak4,x1$Npeak4)),];
with(x5,{abs(Peak1[1]-Peak1[2])+abs(Peak2[1]-Peak2[2])+
abs(Peak3[1]-Peak3[2])+ abs(Peak1[1]-Peak4[2])+
abs(Npeak1[1]-Npeak1[2])+abs(Npeak2[1]-Npeak2[2])+ abs(Npeak3[1]-Npeak3[2])+
abs(Npeak1[1]-Npeak4[2])})
}
else if({indx[1]==0 indx[2]==1 indx[3]==1 indx[4]==1}) #4 peak 1 peak
comparison
{x6-
x1[rev(order(x1$Peak2,x1$Peak3,x1$Peak4,x1$Npeak2,x1$Npeak3,x1$Npeak4)),];
with(x6,{abs(Peak1[1]-Peak1[2])+
abs(Peak1[1]-Peak2[2])+abs(Peak1[1]-Peak3[2])+ abs(Peak1[1]-Peak4[2])+
abs(Npeak1[1]-Npeak1[2])+ abs(Npeak1[1]-Npeak2[2])+abs(Npeak1[1]-Npeak3[2])+
abs(Npeak1[1]-Npeak4[2])})
}
else if({indx[1]==0 indx[2]==1 indx[3]==1 indx[4]==2}) # 3 peaks 1 peak
comparison
{x7- x1[rev(order(x1$Peak2,x1$Peak3,x1$Npeak2,x1$Npeak3)),];
with(x7,{abs(Peak1[1]-Peak1[2])+abs(Peak1[1]-Peak2[2])+abs(Peak1[1]-Peak3[2])+
abs(Npeak1[1]-Npeak1[2])+abs(Npeak1[1]-Npeak2[2])+abs(Npeak1[1]-Npeak3[2])})
}
else {x1[is.na(x1)]-0; x1; #cases where peaks are similar
with(x1,{abs(Peak1[1]-Peak1[2])+ abs(Peak2[1]-Peak2[2])+
abs(Peak3[1]-Peak3[2])+abs(Peak4[1]-Peak4[2]) + abs(Npeak1[1]-Npeak1[2])+
abs(Npeak2[1]-Npeak2[2])+ abs(Npeak3[1]-Npeak3[2])+abs(Npeak4[1]-Npeak4[2])})
}
}
res2-do.call(cbind,lapply(seq_len(ncol(res)),function(i)
c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
row.names(res2)-1:nrow(res2)
dim(res2)
#[1] 124 124
res2[1:5,1:5]
# [,1] [,2] [,3] [,4] [,5]
#1 0.0 65.59415 86.62556 407.9987 104.78294
#2 65.59415 0.0 42.14256 307.3830 39.18879
#3 86.62556 42.14256 0.0 314.4331 33.88839
#4 407.99871 307.38297 314.43309 0. 266.78887
#5 104.78294 39.18879 33.88839
Re: [R] Odd graphic device behavior
Peter, Thank you. When I run: X11(width=7, height=7), I get the same full-screen graphics device window. Running dev.list() gives me X11cairo Running system(xdpyinfo) looks reasonable I tried running options(device=x11) at the R prompt, but this did not seem to change anything. When I started R with R --vanilla and then did: require(stats) plot(cars) I got: Error in plot.new() : figure margins too large ... but the window was a reasonable size. However, after I resized the graphics device window modestly and re-ran plot(cars), the plot was generated in the resized window, but the lettering was large as before and the lines were thick as they were previously. Regards, Tom On Wed, Mar 27, 2013 at 1:43 PM, peter dalgaard pda...@gmail.com wrote: On Mar 27, 2013, at 18:11 , David Winsemius wrote: On Mar 27, 2013, at 9:07 AM, Thomas Adams wrote: John, Thanks for the suggestion, but no. I have even gone so far as to rebuild R from source, re-booted my computer, and tried the 'experiment': require(stats) plot(cars) immediately after starting R. Still the same result. I think it must be related to some default Ubuntu Unity window manager default I can't sort out. Have your deleted the default workspace and history files? When they get corrupted, odd things can happen. ...or try R --vanilla The immediate suspicion is that something is tampering with your graphics device options, so - figure out what device you are using (dev.list()) - if it is X11, have a look at X11.options() - try running X11(width=7, height=7) to see if the automagic settings get it wrong. - try system(xdpyinfo); this may give a long list of gibberish, but look for dimensions and resolution and see if they look sane. E.g., I get screen #0: dimensions:2560x1418 pixels (677x375 millimeters) resolution:96x96 dots per inch Regards, Tom On Wed, Mar 27, 2013 at 10:14 AM, John Kane jrkrid...@inbox.com wrote: Any chance that you made an earlier call to par() resetting cex in your session? I just had that happen. John Kane Kingston ON Canada -Original Message- From: tea...@gmail.com Sent: Tue, 26 Mar 2013 10:15:33 -0400 To: r-help@r-project.org Subject: [R] Odd graphic device behavior I'm experiencing odd graphics device behavior running R 2.15.3 on Ubuntu. Regardless of what I try like: require(stats) plot(cars) lines(lowess(cars)) plot(sin, -pi, 2*pi) for example, the graphics device fills the entire screen with the graphic and a very large font. When I shrink the graphics device window, the lettering remains large and the line thicknesses stay quite thick. It may have been some time since I have done any R work on my Ubuntu computer, but clearly this did not happen previously. Does anyone have any thoughts? Thank you, Tom [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! Check it out at http://www.inbox.com/marineaquarium [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Time trends with GAM
Thanks. On Tue, Mar 26, 2013 at 4:45 AM, Simon Wood s.w...@bath.ac.uk wrote: For 1) see reply just sent to R help... Re: [R] Use pcls in mgcv package to achieve constrained cubic spline (I'm assuming you mean to fix the value of the smooth at a particular knot location - i.e. to impose a constraint like f(t_0) = b). For 2) try 'model.matrix', or 'predict.gam' with argument 'type=lpmatrix' (see ?predict.gam) On 23/03/13 01:05, Antonio P. Ramos wrote: Hi all, I am using GAM to model time trends in a logistic regression. Yet I would like to extract the the fitted spline from it to add it to another model, that cannot be fitted in GAM or GAMM. Thus I have 2 questions: 1) How can I fit a smoother over time so that I force one knot to be at a particular location while letting the model to find the other knots? 2) how can I extract the matrix from the fitted GAM so that I can use it in as an impute for a different model. The types of models I am running are to the following form: gam - gam(mortality.under.2~ maternal_age_c+ I(maternal_age_c^2)+ s(birth_year,by=wealth2) + + wealth2 + sex + residence+ maternal_educ + birth_order, ,data=colombia2,family=**binomial) I've read the extensive documentation for the GAM but I am not sure still. Any suggestion is really appreciated. Thanks, Antonio Pedro [[alternative HTML version deleted]] __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Simon Wood, Mathematical Science, University of Bath BA2 7AY UK +44 (0)1225 386603 http://people.bath.ac.uk/sw283 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Archieve of mails from R forum
Dear R helpers, Everyday I do receive many many mails from R forum and after some period of times, INBOX is filled with numerous mails. At times if for some period of time, I haven't accessed mails, it becomes difficult to keep track of mails and many times simply due to the volume (and owing to the lack of time due to office constraints), I have to simply delete the mails without opening them and I understand this is a huge loss. If in case I wish to refer to all the old emails that have been appeared in the R forum, where do I get these? Is there any list where I will get subject-wise of thread-wise archive of old emails? I understand that will be an ocean of quality information and one can learn a lot from these old mails and I don't need to keep track of my emails all the time. Kindly guide. Regards Katherine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Archieve of mails from R forum
On Mar 27, 2013, at 1:58 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear R helpers, Everyday I do receive many many mails from R forum and after some period of times, INBOX is filled with numerous mails. At times if for some period of time, I haven't accessed mails, it becomes difficult to keep track of mails and many times simply due to the volume (and owing to the lack of time due to office constraints), I have to simply delete the mails without opening them and I understand this is a huge loss. If in case I wish to refer to all the old emails that have been appeared in the R forum, where do I get these? Is there any list where I will get subject-wise of thread-wise archive of old emails? I understand that will be an ocean of quality information and one can learn a lot from these old mails and I don't need to keep track of my emails all the time. Kindly guide. Regards Katherine The official archives for R-Help are here: https://stat.ethz.ch/pipermail/r-help/ and these are mirrored in various locations, such as: http://www.mail-archive.com/r-help@stat.math.ethz.ch/ http://dir.gmane.org/gmane.comp.lang.r.general You can also search the archives for all R lists at: http://rseek.org/ http://finzi.psych.upenn.edu/search.html http://tolstoy.newcastle.edu.au/R/ My recommendation would be to set up a mail filter or rule (using r-help@r-project.org in the the sender and cc: address fields) so that the list e-mails are automatically moved from your main inbox to a folder just for these e-mails and you can then browse them as your schedule permits, rather than having them interspersed with other e-mails in the same location. I do this with a number of the R related lists and have a folder for each one to keep them separated. Most e-mail clients and/or online services have some type of filtering or rule configuration available to do this. Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Archieve of mails from R forum
http://r-help.markmail.org/ has a nice interface for searching the archives, too. On Wed, Mar 27, 2013 at 12:18 PM, Marc Schwartz marc_schwa...@me.comwrote: On Mar 27, 2013, at 1:58 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear R helpers, Everyday I do receive many many mails from R forum and after some period of times, INBOX is filled with numerous mails. At times if for some period of time, I haven't accessed mails, it becomes difficult to keep track of mails and many times simply due to the volume (and owing to the lack of time due to office constraints), I have to simply delete the mails without opening them and I understand this is a huge loss. If in case I wish to refer to all the old emails that have been appeared in the R forum, where do I get these? Is there any list where I will get subject-wise of thread-wise archive of old emails? I understand that will be an ocean of quality information and one can learn a lot from these old mails and I don't need to keep track of my emails all the time. Kindly guide. Regards Katherine The official archives for R-Help are here: https://stat.ethz.ch/pipermail/r-help/ and these are mirrored in various locations, such as: http://www.mail-archive.com/r-help@stat.math.ethz.ch/ http://dir.gmane.org/gmane.comp.lang.r.general You can also search the archives for all R lists at: http://rseek.org/ http://finzi.psych.upenn.edu/search.html http://tolstoy.newcastle.edu.au/R/ My recommendation would be to set up a mail filter or rule (using r-help@r-project.org in the the sender and cc: address fields) so that the list e-mails are automatically moved from your main inbox to a folder just for these e-mails and you can then browse them as your schedule permits, rather than having them interspersed with other e-mails in the same location. I do this with a number of the R related lists and have a folder for each one to keep them separated. Most e-mail clients and/or online services have some type of filtering or rule configuration available to do this. Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] conditional Dataframe filling
Dear Arun, Thank you very much for your help with this.I did not know where to start looking to solve that problem, so I truly appreciate your input. The line of code you sent seems to work but it duplicates the results. Do you know why that may happen? Below is a larger database, to which I apply your line of code. Thank you very much again, Camilo dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat2-dat1 dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ Quoting arun smartpink...@yahoo.com: HI, Just a correction: : dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum #should also work A.K. - Original Message - From: arun smartpink...@yahoo.com To: Camilo Mora cm...@dal.ca Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1- read.table(text= a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE ,sep=,header=TRUE) dat2-dat1 dat2[]-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: r-help@r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c d 0 0 0 0 1 2 3 0 1 0 1 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
This completes in about a second on my system and uses the actual matrix
dimensions you quote:
nr - 153899
nc - 3415
keep - rnorm(nr * nc, 80, 20)
dim(keep) - c(nr, nc)
shrink.to - 1000
system.time(
{
idx - rep(1:shrink.to, length.out = nr)
plot.me - sweep(rowsum(keep, idx), 1, tabulate(idx), , FUN = '/')
})
user system elapsed
0.920.000.9
Jason Law
Statistician
City of Portland
Bureau of Environmental Services
Water Pollution Control Laboratory
6543 N Burlington Avenue
Portland, OR 97203-5452
503-823-1038
jason@portlandoregon.gov
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:05 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Well my true matrix is
of
num [1:153899, 1:3415]
that I want to convert to something like
num [1:1000, 1:3415] (keeping column number the same).
I only give subsets here to allow others to run the code at their computer
Thanks a lot
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 4:45 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
system.time({
+ keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
+ ShrinkTo-500
+ idx-cut(1:nrow(keep), ShrinkTo)
+ keep.ag-aggregate(keep, list(idx), mean)
+ PlotMe-as.matrix(keep.ag[,-1])
+ })
user system elapsed
29.23 0.14 29.37
It takes 30 seconds when using 30 rows. It is not enough time to get a cup
of tee, which I do not consider ages. Maybe split lapply approach or
data.matrix or **ply could be quicker but I do not consider worth spending
hours to elaborate some solution which will spare 20 sec computing time.
Regards
Petr
Sent: Wednesday, March 27, 2013 4:12 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
I have fixed it like that:
keep-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo-500
idx-cut(1:nrow(keep), ShrinkTo)
keep.ag-aggregate(keep, list(idx), mean)
PlotMe-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
str(as.matrix(keep.ag[,-1]) ) # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:30] V1 V2 V3 V4 ...
Please read and follow what was recommended.
quote
use as.matrix(data.frame) on numeric part
^^^
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
And please can you explain how mean(rnorm(whatever)) shall be integer?
Regards
Petr
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From:PIKAL Petr petr.pi...@precheza.cz
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alaios
Sent: Wednesday, March 27, 2013 9:13 AM
To: R help
Subject: [R] Averaging Out many rows from a column AND funtion to
string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
similar in size as this is a matrix that needs plotting (so is ok if
it is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and
averaging per column (notice that the column number should stay the
same)
Make an index variable and aggregate values according it
Something like
idx-cut(1:153899, 153)
keep.ag-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
a bit of my code that you can re run. I convert a
Re: [R] conditional Dataframe filling
Dear Camilo, How do you want to deal with the NAs? If I remove the NAs: dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat1- as.data.frame(na.omit(dat1)) dat2-dat1 dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d e f g h i j k l m #w 0 0 0 0 0 1 2 3 4 0 0 0 0 #y 1 2 3 4 5 0 0 1 2 0 0 0 1 #z 0 0 0 0 1 0 0 0 1 0 0 0 1 dat1 # a b c d e f g h i j k l m #w TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE #y FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE TRUE TRUE TRUE FALSE #z TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 3:27 PM Subject: Re: [R] conditional Dataframe filling Dear Arun, Thank you very much for your help with this.I did not know where to start looking to solve that problem, so I truly appreciate your input. The line of code you sent seems to work but it duplicates the results. Do you know why that may happen? Below is a larger database, to which I apply your line of code. Thank you very much again, Camilo dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat2-dat1 dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ Quoting arun smartpink...@yahoo.com: HI, Just a correction: : dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum #should also work A.K. - Original Message - From: arun smartpink...@yahoo.com To: Camilo Mora cm...@dal.ca Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1- read.table(text= a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE ,sep=,header=TRUE) dat2-dat1 dat2[]-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: r-help@r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c d 0 0 0 0 1 2 3 0 1 0 1 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Weighted Kaplan-Meier estimates with R
On Thu, Mar 28, 2013 at 5:07 AM, rm r...@wippies.se wrote: While testing that I get the same results with the package survey as with the package survival, I encountered the issue of how to draw survival curves. Apparently the implementations in the two packages differ, as I show below. I would very much welcome your views, since the tail of the survival curve has a major impact on the interpretation of my results. In my data, the last death occurs at 2094 days, while the last censoring time is 3297 days. If possible, I would like to say something about the probability between 2100 days and 3300 days. So, my question is that after the last observed death, in the very simple example below at 883 days, how should one draw the survival curve? The graph produced by svykm (package survey) ends at 883 days, whereas survfit (package survival) continues the graph all the way to the last censoring time, which is at 1022 days. If you don't ask for standard errors, svykm() also draws the curve all the way out to 1022 days. The problem is with the standard error estimation under complex sampling -- to save memory it computes the standard errors only at event times. It shouldn't be too hard to get it to extend that to the last censoring time, but the reason it isn't too hard is that the curve and standard error estimates don't change after the last failure time, so it won't be particularly useful. -thomas -- Thomas Lumley Professor of Biostatistics University of Auckland [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] conditional Dataframe filling
Thanks Arun, Well that is interesting. My intention was to have a dataframe with the same number of rows in the original data, and for the rows with NAs, then return NA (If there are NAs, often the entire row has NAs). What is interesting is that in your code with NAs, the row that has NAs gets NAs in the output, which is what I am looking for. I guess a solution is to subset complete rows and then run your line of code. Unless there is an alternative, to tell cumsum to leave NAs as NAs? Thanks again, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ Quoting arun smartpink...@yahoo.com: Dear Camilo, How do you want to deal with the NAs? If I remove the NAs: dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat1- as.data.frame(na.omit(dat1)) dat2-dat1 dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d e f g h i j k l m #w 0 0 0 0 0 1 2 3 4 0 0 0 0 #y 1 2 3 4 5 0 0 1 2 0 0 0 1 #z 0 0 0 0 1 0 0 0 1 0 0 0 1 dat1 # a b c d e f g h i j k l m #w TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE #y FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE TRUE TRUE TRUE FALSE #z TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 3:27 PM Subject: Re: [R] conditional Dataframe filling Dear Arun, Thank you very much for your help with this.I did not know where to start looking to solve that problem, so I truly appreciate your input. The line of code you sent seems to work but it duplicates the results. Do you know why that may happen? Below is a larger database, to which I apply your line of code. Thank you very much again, Camilo dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat2-dat1 dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ Quoting arun smartpink...@yahoo.com: HI, Just a correction: : dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum #should also work A.K. - Original Message - From: arun smartpink...@yahoo.com To: Camilo Mora cm...@dal.ca Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1- read.table(text= a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE ,sep=,header=TRUE) dat2-dat1 dat2[]-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: r-help@r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c d 0 0 0 0 1 2 3 0 1 0 1 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282
Re: [R] ggplot2: Controlling line width of panel borders
This is an old post, but I had the exact problem described here and after two days of monkeying around in theme, I wanted to provide the non-ggplot related solution. In summary, the problem is that in the faceted plots, lines on the right an bottom in the subplots appear to be thicker than faceted. This is a problem with the graphics device you are using (a windows x11 device I suspect). I have provided a slightly more concise illustration of the problem. x11() p - ggplot(mtcars, aes(mpg, wt)) + geom_point() p + facet_grid(. ~ cyl)+ theme_bw()+ #drop a mostly white theme on for contrast theme(panel.border = element_rect(fill=NA, colour = black, size=1)) #bump up the border size to 1, the minimum for many publications. Now do the same with: pdf(file=mygraphic.pdf) p - ggplot(mtcars, aes(mpg, wt)) + geom_point() p + facet_grid(. ~ cyl)+ theme_bw()+ #drop a mostly white theme on for contrast theme(panel.border = element_rect(fill=NA, colour = black, size=1)) #bump up the border size to 1, the minimum for many publications. dev.off() This will output directly to a file so you will need to hunt it down. Inspect it and you will see that there is no problem, everything looks exactly the way you coded it. In short, the problem is not with ggplot. X11(), I assume, or some unseen process calling a windows graphic device is just not visually representing the code properly. If you use a pdf graphics device to output the figure the problem goes away. However, this means you can't view it instantly... but you do have a vectorized figure. Trust your code not your graphics device. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] conditional Dataframe filling
Dear Camilo, You can do this: dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat1- as.data.frame(dat1) dat2-dat1 dat2[rowSums(is.na(dat2))==0,]- t(apply(!dat1[rowSums(is.na(dat1))==0,],1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d e f g h i j k l m #w 0 0 0 0 0 1 2 3 4 0 0 0 0 #x NA NA NA NA NA NA NA NA NA NA NA NA NA #y 1 2 3 4 5 0 0 1 2 0 0 0 1 #z 0 0 0 0 1 0 0 0 1 0 0 0 1 Suppose if NAs are there but not for the entire row (if I understand correctly), you wanted to have the whole row NA, right. datNew- structure(list(a = c(TRUE, NA, FALSE, TRUE, TRUE), b = c(TRUE, NA, FALSE, TRUE, TRUE), c = c(TRUE, NA, FALSE, TRUE, FALSE), d = c(TRUE, NA, FALSE, TRUE, FALSE), e = c(TRUE, NA, FALSE, FALSE, NA), f = c(FALSE, NA, TRUE, TRUE, NA), g = c(FALSE, NA, TRUE, TRUE, TRUE), h = c(FALSE, NA, FALSE, TRUE, FALSE ), i = c(FALSE, NA, FALSE, FALSE, NA), j = c(TRUE, NA, TRUE, TRUE, TRUE), k = c(TRUE, NA, TRUE, TRUE, FALSE), l = c(TRUE, NA, TRUE, TRUE, FALSE), m = c(TRUE, NA, FALSE, FALSE, TRUE )), .Names = c(a, b, c, d, e, f, g, h, i, j, k, l, m), row.names = c(w, x, y, z, u), class = data.frame) datNew # a b c d e f g h i j k l m #w TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE #x NA NA NA NA NA NA NA NA NA NA NA NA NA #y FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE TRUE TRUE TRUE FALSE #z TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE #u TRUE TRUE FALSE FALSE NA NA TRUE FALSE NA TRUE FALSE FALSE TRUE dat2New- datNew dat2New[rowSums(is.na(dat2New))==0,]-t(apply(!datNew[rowSums(is.na(datNew))==0,],1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2New[rowSums(is.na(dat2New))!=0 rowSums(is.na(dat2New))!=ncol(dat2New),]-NA dat2New # a b c d e f g h i j k l m #w 0 0 0 0 0 1 2 3 4 0 0 0 0 #x NA NA NA NA NA NA NA NA NA NA NA NA NA #y 1 2 3 4 5 0 0 1 2 0 0 0 1 #z 0 0 0 0 1 0 0 0 1 0 0 0 1 #u NA NA NA NA NA NA NA NA NA NA NA NA NA A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 4:10 PM Subject: Re: [R] conditional Dataframe filling Thanks Arun, Well that is interesting. My intention was to have a dataframe with the same number of rows in the original data, and for the rows with NAs, then return NA (If there are NAs, often the entire row has NAs). What is interesting is that in your code with NAs, the row that has NAs gets NAs in the output, which is what I am looking for. I guess a solution is to subset complete rows and then run your line of code. Unless there is an alternative, to tell cumsum to leave NAs as NAs? Thanks again, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ Quoting arun smartpink...@yahoo.com: Dear Camilo, How do you want to deal with the NAs? If I remove the NAs: dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat1- as.data.frame(na.omit(dat1)) dat2-dat1 dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d e f g h i j k l m #w 0 0 0 0 0 1 2 3 4 0 0 0 0 #y 1 2 3 4 5 0 0 1 2 0 0 0 1 #z 0 0 0 0 1 0 0 0 1 0 0 0 1 dat1 # a b c d e f g h i j k l m #w TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE #y FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE TRUE TRUE TRUE FALSE #z TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 3:27 PM Subject: Re: [R] conditional Dataframe
Re: [R] conditional Dataframe filling
Nice!. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ Quoting arun smartpink...@yahoo.com: Dear Camilo, You can do this: dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat1- as.data.frame(dat1) dat2-dat1 dat2[rowSums(is.na(dat2))==0,]- t(apply(!dat1[rowSums(is.na(dat1))==0,],1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d e f g h i j k l m #w 0 0 0 0 0 1 2 3 4 0 0 0 0 #x NA NA NA NA NA NA NA NA NA NA NA NA NA #y 1 2 3 4 5 0 0 1 2 0 0 0 1 #z 0 0 0 0 1 0 0 0 1 0 0 0 1 Suppose if NAs are there but not for the entire row (if I understand correctly), you wanted to have the whole row NA, right. datNew- structure(list(a = c(TRUE, NA, FALSE, TRUE, TRUE), b = c(TRUE, NA, FALSE, TRUE, TRUE), c = c(TRUE, NA, FALSE, TRUE, FALSE), d = c(TRUE, NA, FALSE, TRUE, FALSE), e = c(TRUE, NA, FALSE, FALSE, NA), f = c(FALSE, NA, TRUE, TRUE, NA), g = c(FALSE, NA, TRUE, TRUE, TRUE), h = c(FALSE, NA, FALSE, TRUE, FALSE ), i = c(FALSE, NA, FALSE, FALSE, NA), j = c(TRUE, NA, TRUE, TRUE, TRUE), k = c(TRUE, NA, TRUE, TRUE, FALSE), l = c(TRUE, NA, TRUE, TRUE, FALSE), m = c(TRUE, NA, FALSE, FALSE, TRUE )), .Names = c(a, b, c, d, e, f, g, h, i, j, k, l, m), row.names = c(w, x, y, z, u), class = data.frame) datNew # a b c d e f g h i j k l m #w TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE #x NA NA NA NA NA NA NA NA NA NA NA NA NA #y FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE TRUE TRUE TRUE FALSE #z TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE #u TRUE TRUE FALSE FALSE NA NA TRUE FALSE NA TRUE FALSE FALSE TRUE dat2New- datNew dat2New[rowSums(is.na(dat2New))==0,]-t(apply(!datNew[rowSums(is.na(datNew))==0,],1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2New[rowSums(is.na(dat2New))!=0 rowSums(is.na(dat2New))!=ncol(dat2New),]-NA dat2New # a b c d e f g h i j k l m #w 0 0 0 0 0 1 2 3 4 0 0 0 0 #x NA NA NA NA NA NA NA NA NA NA NA NA NA #y 1 2 3 4 5 0 0 1 2 0 0 0 1 #z 0 0 0 0 1 0 0 0 1 0 0 0 1 #u NA NA NA NA NA NA NA NA NA NA NA NA NA A.K. - Original Message - From: Camilo Mora cm...@dal.ca To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Wednesday, March 27, 2013 4:10 PM Subject: Re: [R] conditional Dataframe filling Thanks Arun, Well that is interesting. My intention was to have a dataframe with the same number of rows in the original data, and for the rows with NAs, then return NA (If there are NAs, often the entire row has NAs). What is interesting is that in your code with NAs, the row that has NAs gets NAs in the output, which is what I am looking for. I guess a solution is to subset complete rows and then run your line of code. Unless there is an alternative, to tell cumsum to leave NAs as NAs? Thanks again, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone: Country code: 57 Provider code: 313 Phone 776 2282 From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ Quoting arun smartpink...@yahoo.com: Dear Camilo, How do you want to deal with the NAs? If I remove the NAs: dat1 - structure(list( w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE), x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA), y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE), z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)), row.names = c(NA, -13L), class = data.frame) dat1-t(dat1) colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m) dat1- as.data.frame(na.omit(dat1)) dat2-dat1 dat2[]-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum dat2 # a b c d e f g h i j k l m #w 0 0 0 0 0 1 2 3 4 0 0 0 0 #y 1 2 3 4 5 0 0 1 2 0 0 0 1 #z 0 0 0 0 1 0 0 0 1 0 0 0 1 dat1 # a b c d e f g h i j k l m #w TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE #y FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE
[R] prop.test correct true and false gives same answer
All, How come both of these are the same. Both say 1-sample proportions test without continuity correction. I would suspect one would say without and one would say with. prop.test(118,236,.5,correct=FALSE,conf.level=0.95) 1-sample proportions test without continuity correction data: 118 out of 236, null probability 0.5 X-squared = 0, df = 1, p-value = 1 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.4367215 0.5632785 sample estimates: p 0.5 prop.test(118,236,.5,correct=TRUE,conf.level=0.95) 1-sample proportions test without continuity correction data: 118 out of 236, null probability 0.5 X-squared = 0, df = 1, p-value = 1 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.4367215 0.5632785 sample estimates: p 0.5 -- View this message in context: http://r.789695.n4.nabble.com/prop-test-correct-true-and-false-gives-same-answer-tp4662659.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] conditional Dataframe filling
Hi Camilo,
No problem.
In case, you wanted to process the partial NA rows, this could help:
datNew- structure(list(a = c(TRUE, NA, FALSE, TRUE, TRUE), b = c(TRUE,
NA, FALSE, TRUE, TRUE), c = c(TRUE, NA, FALSE, TRUE, FALSE),
d = c(TRUE, NA, FALSE, TRUE, FALSE), e = c(TRUE, NA, FALSE,
FALSE, NA), f = c(FALSE, NA, TRUE, TRUE, NA), g = c(FALSE,
NA, TRUE, TRUE, TRUE), h = c(FALSE, NA, FALSE, TRUE, FALSE
), i = c(FALSE, NA, FALSE, FALSE, NA), j = c(TRUE, NA, TRUE,
TRUE, TRUE), k = c(TRUE, NA, TRUE, TRUE, FALSE), l = c(TRUE,
NA, TRUE, TRUE, FALSE), m = c(TRUE, NA, FALSE, FALSE, TRUE
)), .Names = c(a, b, c, d, e, f, g, h, i,
j, k, l, m), row.names = c(w, x, y, z, u), class =
data.frame)
dat2New- datNew
dat2New[rowSums(is.na(dat2New))==0 | rowSums(is.na(dat2New))!=ncol(dat2New),]-
t(apply(!datNew[rowSums(is.na(datNew))==0 |
rowSums(is.na(datNew))!=ncol(datNew),],1,function(x) {x[!is.na(x)]-
unlist(lapply(split(x[!is.na(x)],cumsum(c(0,abs(diff(x[!is.na(x)]),cumsum));x}))
dat2New
# a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1
#u 0 0 1 2 NA NA 0 1 NA 0 1 2 0
datNew
# a b c d e f g h i j k l m
#w TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE TRUE TRUE TRUE FALSE
#z TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE
#u TRUE TRUE FALSE FALSE NA NA TRUE FALSE NA TRUE FALSE FALSE TRUE
A.K.
- Original Message -
From: Camilo Mora cm...@dal.ca
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org
Sent: Wednesday, March 27, 2013 4:49 PM
Subject: Re: [R] conditional Dataframe filling
Nice!.
Thanks,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone: Country code: 57
Provider code: 313
Phone 776 2282
From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun smartpink...@yahoo.com:
Dear Camilo,
You can do this:
dat1 - structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = data.frame)
dat1-t(dat1)
colnames(dat1)-c(a,b,c,d,e,f,g,h,i,j,k, l,m)
dat1- as.data.frame(dat1)
dat2-dat1
dat2[rowSums(is.na(dat2))==0,]-
t(apply(!dat1[rowSums(is.na(dat1))==0,],1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum
dat2
# a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1
Suppose if NAs are there but not for the entire row (if I understand
correctly), you wanted to have the whole row NA, right.
datNew- structure(list(a = c(TRUE, NA, FALSE, TRUE, TRUE), b = c(TRUE,
NA, FALSE, TRUE, TRUE), c = c(TRUE, NA, FALSE, TRUE, FALSE),
d = c(TRUE, NA, FALSE, TRUE, FALSE), e = c(TRUE, NA, FALSE,
FALSE, NA), f = c(FALSE, NA, TRUE, TRUE, NA), g = c(FALSE,
NA, TRUE, TRUE, TRUE), h = c(FALSE, NA, FALSE, TRUE, FALSE
), i = c(FALSE, NA, FALSE, FALSE, NA), j = c(TRUE, NA, TRUE,
TRUE, TRUE), k = c(TRUE, NA, TRUE, TRUE, FALSE), l = c(TRUE,
NA, TRUE, TRUE, FALSE), m = c(TRUE, NA, FALSE, FALSE, TRUE
)), .Names = c(a, b, c, d, e, f, g, h, i,
j, k, l, m), row.names = c(w, x, y, z, u), class =
data.frame)
datNew
# a b c d e f g h i j
k l m
#w TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE TRUE
TRUE TRUE
#x NA NA NA NA NA NA NA NA NA NA
NA NA NA
#y FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE TRUE TRUE
TRUE FALSE
#z TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE TRUE
TRUE FALSE
#u TRUE TRUE FALSE FALSE NA NA TRUE FALSE NA TRUE FALSE
FALSE TRUE
dat2New- datNew
dat2New[rowSums(is.na(dat2New))==0,]-t(apply(!datNew[rowSums(is.na(datNew))==0,],1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x),cumsum
dat2New[rowSums(is.na(dat2New))!=0
rowSums(is.na(dat2New))!=ncol(dat2New),]-NA
dat2New
# a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1
#u NA NA NA NA NA NA NA NA
Re: [R] prop.test correct true and false gives same answer
?prop.test is helpful. Continuity correction is used only if it does not exceed the difference between sample and null proportions in absolute value. albyn On Wed, Mar 27, 2013 at 02:04:51PM -0700, David Arnold wrote: All, How come both of these are the same. Both say 1-sample proportions test without continuity correction. I would suspect one would say without and one would say with. prop.test(118,236,.5,correct=FALSE,conf.level=0.95) 1-sample proportions test without continuity correction data: 118 out of 236, null probability 0.5 X-squared = 0, df = 1, p-value = 1 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.4367215 0.5632785 sample estimates: p 0.5 prop.test(118,236,.5,correct=TRUE,conf.level=0.95) 1-sample proportions test without continuity correction data: 118 out of 236, null probability 0.5 X-squared = 0, df = 1, p-value = 1 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.4367215 0.5632785 sample estimates: p 0.5 -- View this message in context: http://r.789695.n4.nabble.com/prop-test-correct-true-and-false-gives-same-answer-tp4662659.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Albyn Jones Reed College jo...@reed.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Archieve of mails from R forum
Dear Sir, Thanks a lot for the input. I am sure it will go a long way for me to understand R. Thanks again. Regards Katherine --- On Wed, 27/3/13, Mason ma...@verbasoftware.com wrote: From: Mason ma...@verbasoftware.com Subject: Re: [R] Archieve of mails from R forum To: Marc Schwartz marc_schwa...@me.com Cc: Katherine Gobin katherine_go...@yahoo.com, r-help@r-project.org help r-help@r-project.org Date: Wednesday, 27 March, 2013, 7:30 PM http://r-help.markmail.org/ has a nice interface for searching the archives, too. On Wed, Mar 27, 2013 at 12:18 PM, Marc Schwartz marc_schwa...@me.com wrote: On Mar 27, 2013, at 1:58 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear R helpers, Everyday I do receive many many mails from R forum and after some period of times, INBOX is filled with numerous mails. At times if for some period of time, I haven't accessed mails, it becomes difficult to keep track of mails and many times simply due to the volume (and owing to the lack of time due to office constraints), I have to simply delete the mails without opening them and I understand this is a huge loss. If in case I wish to refer to all the old emails that have been appeared in the R forum, where do I get these? Is there any list where I will get subject-wise of thread-wise archive of old emails? I understand that will be an ocean of quality information and one can learn a lot from these old mails and I don't need to keep track of my emails all the time. Kindly guide. Regards Katherine The official archives for R-Help are here: https://stat.ethz.ch/pipermail/r-help/ and these are mirrored in various locations, such as: http://www.mail-archive.com/r-help@stat.math.ethz.ch/ http://dir.gmane.org/gmane.comp.lang.r.general You can also search the archives for all R lists at: http://rseek.org/ http://finzi.psych.upenn.edu/search.html http://tolstoy.newcastle.edu.au/R/ My recommendation would be to set up a mail filter or rule (using r-help@r-project.org in the the sender and cc: address fields) so that the list e-mails are automatically moved from your main inbox to a folder just for these e-mails and you can then browse them as your schedule permits, rather than having them interspersed with other e-mails in the same location. I do this with a number of the R related lists and have a folder for each one to keep them separated. Most e-mail clients and/or online services have some type of filtering or rule configuration available to do this. Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A simple perceptron neural network (nnet)
can u explain me, how it works your code??? please. i´m also doing a simple perceptron for homework on R and i dont know where to start. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] prop.test correct true and false gives same answer
On 27-Mar-2013 21:04:51 David Arnold wrote: All, How come both of these are the same. Both say 1-sample proportions test without continuity correction. I would suspect one would say without and one would say with. prop.test(118,236,.5,correct=FALSE,conf.level=0.95) 1-sample proportions test without continuity correction data: 118 out of 236, null probability 0.5 X-squared = 0, df = 1, p-value = 1 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.4367215 0.5632785 sample estimates: p 0.5 prop.test(118,236,.5,correct=TRUE,conf.level=0.95) 1-sample proportions test without continuity correction data: 118 out of 236, null probability 0.5 X-squared = 0, df = 1, p-value = 1 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.4367215 0.5632785 sample estimates: p 0.5 Note what is said (admittedly somewhat deeply tucked away) under Details in ?prop.test: Continuity correction is used only if it does not exceed the difference between sample and null proportions in absolute value. In your example, the sample proportion exactly matches the null-hypothesis proportion (0.5). Confirmation: [A] Your same example: prop.test(118,236,.5,correct=TRUE,conf.level=0.95) # 1-sample proportions test without continuity correction # data: 118 out of 236, null probability 0.5 # X-squared = 0, df = 1, p-value = 1 # alternative hypothesis: true p is not equal to 0.5 # 95 percent confidence interval: # 0.4367215 0.5632785 # sample estimates: # p # 0.5 [B1] Slightly change x, but keep correct=TRUE: prop.test(117,236,.5,correct=TRUE,conf.level=0.95) # 1-sample proportions test with continuity correction # data: 117 out of 236, null probability 0.5 # X-squared = 0.0042, df = 1, p-value = 0.9481 # alternative hypothesis: true p is not equal to 0.5 # 95 percent confidence interval: # 0.4304724 0.5611932 # sample estimates: # p # 0.4957627 [B2] Slightly change x, but now correct=FALSE: prop.test(117,236,.5,correct=FALSE,conf.level=0.95) # 1-sample proportions test without continuity correction # data: 117 out of 236, null probability 0.5 # X-squared = 0.0169, df = 1, p-value = 0.8964 # alternative hypothesis: true p is not equal to 0.5 # 95 percent confidence interval: # 0.4325543 0.5591068 # sample estimates: # p # 0.4957627 So it doesn't do the requested continuity correction in [A] because there is no need to. But in [B1] it makes a difference (compare with [B2]), so it does it. Hoping this helps, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 27-Mar-2013 Time: 21:21:39 This message was sent by XFMail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odd graphic device behavior
On Mar 27, 2013, at 19:26 , Thomas Adams wrote: Peter, Thank you. When I run: X11(width=7, height=7), I get the same full-screen graphics device window. Running dev.list() gives me X11cairo Running system(xdpyinfo) looks reasonable I tried running options(device=x11) at the R prompt, but this did not seem to change anything. And X11.options() said what? When I started R with R --vanilla and then did: require(stats) plot(cars) I got: Error in plot.new() : figure margins too large ... but the window was a reasonable size. However, after I resized the graphics device window modestly and re-ran plot(cars), the plot was generated in the resized window, but the lettering was large as before and the lines were thick as they were previously. Something's clearly up with the device's notion of pointsize. Or inches. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ::manzhang::
http://www.signdesignbooks.com/acsw/df.km?flnu Man Zhang 3/27/2013 11:31:04 PM bf iopo rarv rsh atzwnxfqoijhcwndgcyyyvtllyxaqkmiftoh hhtshnfaeaoeqixpkgebigw cjbvhl [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Passing arguments between apply and l(s)apply functions vs. nested for loop
Hi R community,
I have a question concerning passing arguments between apply and lapply? Or
maybe, once my problem is explained, the question is really about how to best
transform my nested for loops into list/matrix operations; I am just beginning
this transformation away from nested for loops, so I beg of you to have some
lenience regarding my ignorance.
Part I:
I used a set of nested for loops for a computation, which works as fine as it
is slow--very slow. My needs are on the order of about 2 iterations; in
nested loop format this is millions of calculations. To give you a sense of
what I'm trying to do, it may help to first see the nested for loop (this code
is run-able via copy-and-paste):
#START CODE SNIPPET
#LIST AND VECTOR
rm(list=ls())
l - list(1:3,2:3,4:10,7:9)
v - 1:3
#USED IN j loop to catch values
catch.mat - matrix(NA,nrow=length(v),ncol=length(l))
#LOOPS
for (i in 1:length(v)){
for (j in 1:length(l)) {
catch.mat[i,j] - sum(l[[j]]==i)
}
}
#SIMPLY APPLY OVER catch.mat
catch.all - apply(catch.mat,1,sum)
catch.mat
catch.all
#END CODE SNIPPET
This does exactly what I want, catch.all provides the number of elements in
l for which each element of v is a member, given the constraint that
within each element in l the sub-elements are unique.
Part II:
However, for my data set it takes about 3 days to run. So, I stumbled onto
list and matrix operations (apply family of functions) and have been working to
coerce my code above into an apply-like format. Here is my best example (of
several failings of different sorts), after trying for several hours and
reading much on the web and in some books:
#THIS CODE SNIPPET IS PASTE-ABLE AFTER RUNNING FIRST SNIPPET ABOVE
#SIMPLE TEST TO SHOW THAT apply is passing elements from mat.1 to f.1
f.1 - function(x){
i - x
print(x)
print(i)
}
mat.1 - matrix(1:2318,nrow=1,ncol=2318)
apply(mat.1,1,f.1)
#THAT GAVE EXPECTED RESULTS
#THEN I ADD A NEW FUNCTION
f.2 - function(x,l=l){
i - x
rm(x)
return(sum(lapply(l,function(x) sum(x==i
}
mat.1 - matrix(1:2318,nrow=1,ncol=2318)
apply(mat.1,1,f.2)
#BUT GET ERROR
# apply(mat.1,1,f.2)
#Error in lapply(l, function(x) sum(x == i)) :
# promise already under evaluation: recursive default argument reference or
earlier problems?
#BUT I KNOW THAT the lapply in f.2 works
sapply(l,function(x) sum(x==2))
sum(sapply(l,function(x) sum(x==2)))
#END CODE SNIPPET
I Am totally stuck, as I really don't understand the internals of R or S well
enough to get a sense of what is the root of this problem in my code.
I would appreciate some guidance by example, not by reference. I don't think
further reading of existing texts, unless extremely basic, is going to help me.
Thanks,
Mark Orr
Mark G. Orr, Ph.D.
Epidemiology Merit Fellow
Assoc. Research Scientist
Columbia Univ.-Mailman Sch. Public Health
Department of Epidemiology
722 W. 168th St., RM 528
New York, NY
T: 212-305-3815
F: 212-342-5168
mo2...@columbia.edu
http://chbdlab.org/
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] lmer, p-values and all that
Dear Help: I am trying to follow Professor Bates' recommendation, quoted by Professor Crawley in The R Book, p629, to determine whether I should model data using the 'plain old' lm function or the mixed model function lmer by using the syntax anova(lmModel,lmerModel). Apparently I've not understood the recommendation or the proper likelihood ratio test in question (or both) for I get this error message: Error: $ operator not defined for this S4 class. Would someone be kind enough to point out my blunder? Thank you, Michael Grant [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plotting interactive networks in R
Hi all, I was wondering if there was package/ tutorial somewhere so that I can plot INTERACTIVE networks in R. What I mean by interactive is that you can zoom in, twist and rotate, and if necessary move nodes around. Any thoughts? Thanks, Sachin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error after R and Twitter Successfull Handshake
After installing updated twitteR and ROAuth, I am still getting the same error as stated in these links. As there anyone who has been able to solve the prolem? Suggested solution do not work. http://stackoverflow.com/questions/9916283/twitter-roauth-and-windows-register-ok-but-certificate-verify-failed http://stackoverflow.com/questions/7411215/oauth-authentication-with-twitter-api-failed Peter Maclean Department of Economics UDSM [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Archieve of mails from R forum
On Mar 27, 2013, at 12:18 PM, Marc Schwartz wrote: On Mar 27, 2013, at 1:58 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear R helpers, Everyday I do receive many many mails from R forum and after some period of times, INBOX is filled with numerous mails. At times if for some period of time, I haven't accessed mails, it becomes difficult to keep track of mails and many times simply due to the volume (and owing to the lack of time due to office constraints), I have to simply delete the mails without opening them and I understand this is a huge loss. If in case I wish to refer to all the old emails that have been appeared in the R forum, where do I get these? Is there any list where I will get subject-wise of thread-wise archive of old emails? I understand that will be an ocean of quality information and one can learn a lot from these old mails and I don't need to keep track of my emails all the time. Kindly guide. Regards Katherine The official archives for R-Help are here: https://stat.ethz.ch/pipermail/r-help/ and these are mirrored in various locations, such as: http://www.mail-archive.com/r-help@stat.math.ethz.ch/ http://dir.gmane.org/gmane.comp.lang.r.general You can also search the archives for all R lists at: http://rseek.org/ http://finzi.psych.upenn.edu/search.html Sadly, the finzi.psych repository is no longer archiving rhelp and hasn't been doing so for several years. http://tolstoy.newcastle.edu.au/R/ I was glad you didn't link to Nabble, which is neither an archive and has been a portal for spam. My recommendation would be to set up a mail filter or rule (using r-help@r-project.org in the the sender and cc: address fields) so that the list e-mails are automatically moved from your main inbox to a folder just for these e-mails and you can then browse them as your schedule permits, rather than having them interspersed with other e-mails in the same location. I do this with a number of the R related lists and have a folder for each one to keep them separated. Most e-mail clients and/or online services have some type of filtering or rule configuration available to do this. Regards, Marc Schwartz -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Archieve of mails from R forum
On Mar 27, 2013, at 7:53 PM, David Winsemius dwinsem...@comcast.net wrote: On Mar 27, 2013, at 12:18 PM, Marc Schwartz wrote: On Mar 27, 2013, at 1:58 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear R helpers, Everyday I do receive many many mails from R forum and after some period of times, INBOX is filled with numerous mails. At times if for some period of time, I haven't accessed mails, it becomes difficult to keep track of mails and many times simply due to the volume (and owing to the lack of time due to office constraints), I have to simply delete the mails without opening them and I understand this is a huge loss. If in case I wish to refer to all the old emails that have been appeared in the R forum, where do I get these? Is there any list where I will get subject-wise of thread-wise archive of old emails? I understand that will be an ocean of quality information and one can learn a lot from these old mails and I don't need to keep track of my emails all the time. Kindly guide. Regards Katherine The official archives for R-Help are here: https://stat.ethz.ch/pipermail/r-help/ and these are mirrored in various locations, such as: http://www.mail-archive.com/r-help@stat.math.ethz.ch/ http://dir.gmane.org/gmane.comp.lang.r.general You can also search the archives for all R lists at: http://rseek.org/ http://finzi.psych.upenn.edu/search.html Sadly, the finzi.psych repository is no longer archiving rhelp and hasn't been doing so for several years. Ah, ok. Interesting. I had not used it in quite some time, in deference to using rseek.org. Thanks for the heads up David. http://tolstoy.newcastle.edu.au/R/ I was glad you didn't link to Nabble, which is neither an archive and has been a portal for spam. Yeah. As R-Devel co-moderator, I am well aware of that. No reasons to drive more traffic there... :-) Regards, Marc My recommendation would be to set up a mail filter or rule (using r-help@r-project.org in the the sender and cc: address fields) so that the list e-mails are automatically moved from your main inbox to a folder just for these e-mails and you can then browse them as your schedule permits, rather than having them interspersed with other e-mails in the same location. I do this with a number of the R related lists and have a folder for each one to keep them separated. Most e-mail clients and/or online services have some type of filtering or rule configuration available to do this. Regards, Marc Schwartz -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmer, p-values and all that
Michael Grant michael.grant at colorado.edu writes: Dear Help: I am trying to follow Professor Bates' recommendation, quoted by Professor Crawley in The R Book, p629, to determine whether I should model data using the 'plain old' lm function or the mixed model function lmer by using the syntax anova(lmModel,lmerModel). Apparently I've not understood the recommendation or the proper likelihood ratio test in question (or both) for I get this error message: Error: $ operator not defined for this S4 class. I don't have the R Book handy (some more context would be extremely useful! I would think it would count as fair use to quote the passage you're referring to ...) Would someone be kind enough to point out my blunder? You should probably repost this to the r-sig-mixed-mod...@r-project.org mailing list. My short answer would be: (1) I don't think you can actually use anova() to compare likelihoods between lm() and lme()/lmer() fits in the way that you want: *maybe* for lme() [don't recall], but almost certainly not for lmer(). See http://glmm.wikidot.com/faq for methods for testing significance/inclusion of random factors (short answer: you should *generally* try to make the decision whether to include random factors or not on _a priori_ grounds, not on the basis of statistical tests ...) Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmer, p-values and all that
On Mar 27, 2013, at 7:00 PM, Ben Bolker wrote: Michael Grant michael.grant at colorado.edu writes: Dear Help: I am trying to follow Professor Bates' recommendation, quoted by Professor Crawley in The R Book, p629, to determine whether I should model data using the 'plain old' lm function or the mixed model function lmer by using the syntax anova(lmModel,lmerModel). Apparently I've not understood the recommendation or the proper likelihood ratio test in question (or both) for I get this error message: Error: $ operator not defined for this S4 class. I don't have the R Book handy (some more context would be extremely useful! I would think it would count as fair use to quote the passage you're referring to ...) This is the quoted Rhelp entry: http://tolstoy.newcastle.edu.au/R/help/05/01/10006.html (I'm unable to determine whether it applies to the question at hand.) Would someone be kind enough to point out my blunder? You should probably repost this to the r-sig-mixed-mod...@r-project.org mailing list. My short answer would be: (1) I don't think you can actually use anova() to compare likelihoods between lm() and lme()/lmer() fits in the way that you want: *maybe* for lme() [don't recall], but almost certainly not for lmer(). See http://glmm.wikidot.com/faq for methods for testing significance/inclusion of random factors (short answer: you should *generally* try to make the decision whether to include random factors or not on _a priori_ grounds, not on the basis of statistical tests ...) Ben Bolker -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmer, p-values and all that
i literally just ran one. when i ran one of mine and the did summary(mod) i get the following: mod - lmer(dem ~ xbar + cpi + (1 | country), data=wvsAB) summary(mod) Linear mixed model fit by REML Formula: dem ~ xbar + cpi + (1 | country) Data: wvsAB AIC BIC logLik deviance REMLdev 34383 34418 -1718734355 34373 Random effects: Groups NameVariance Std.Dev. with a bunch more stuff below. On Mar 27, 2013, at 10:00 PM, Ben Bolker wrote: Michael Grant michael.grant at colorado.edu writes: Dear Help: I am trying to follow Professor Bates' recommendation, quoted by Professor Crawley in The R Book, p629, to determine whether I should model data using the 'plain old' lm function or the mixed model function lmer by using the syntax anova(lmModel,lmerModel). Apparently I've not understood the recommendation or the proper likelihood ratio test in question (or both) for I get this error message: Error: $ operator not defined for this S4 class. I don't have the R Book handy (some more context would be extremely useful! I would think it would count as fair use to quote the passage you're referring to ...) Would someone be kind enough to point out my blunder? You should probably repost this to the r-sig-mixed-mod...@r-project.org mailing list. My short answer would be: (1) I don't think you can actually use anova() to compare likelihoods between lm() and lme()/lmer() fits in the way that you want: *maybe* for lme() [don't recall], but almost certainly not for lmer(). See http://glmm.wikidot.com/faq for methods for testing significance/inclusion of random factors (short answer: you should *generally* try to make the decision whether to include random factors or not on _a priori_ grounds, not on the basis of statistical tests ...) Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmer, p-values and all that
i meant to add, i am not sure if an example lmer model woulf be helpful, since i can't see the OP. On Mar 27, 2013, at 10:13 PM, Nicole Ford wrote: i literally just ran one. when i ran one of mine and the did summary(mod) i get the following: mod - lmer(dem ~ xbar + cpi + (1 | country), data=wvsAB) summary(mod) Linear mixed model fit by REML Formula: dem ~ xbar + cpi + (1 | country) Data: wvsAB AIC BIC logLik deviance REMLdev 34383 34418 -1718734355 34373 Random effects: Groups NameVariance Std.Dev. with a bunch more stuff below. On Mar 27, 2013, at 10:00 PM, Ben Bolker wrote: Michael Grant michael.grant at colorado.edu writes: Dear Help: I am trying to follow Professor Bates' recommendation, quoted by Professor Crawley in The R Book, p629, to determine whether I should model data using the 'plain old' lm function or the mixed model function lmer by using the syntax anova(lmModel,lmerModel). Apparently I've not understood the recommendation or the proper likelihood ratio test in question (or both) for I get this error message: Error: $ operator not defined for this S4 class. I don't have the R Book handy (some more context would be extremely useful! I would think it would count as fair use to quote the passage you're referring to ...) Would someone be kind enough to point out my blunder? You should probably repost this to the r-sig-mixed-mod...@r-project.org mailing list. My short answer would be: (1) I don't think you can actually use anova() to compare likelihoods between lm() and lme()/lmer() fits in the way that you want: *maybe* for lme() [don't recall], but almost certainly not for lmer(). See http://glmm.wikidot.com/faq for methods for testing significance/inclusion of random factors (short answer: you should *generally* try to make the decision whether to include random factors or not on _a priori_ grounds, not on the basis of statistical tests ...) Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmer, p-values and all that
i did find this for you, down towards the end, they discuss the anova method. i am on my way to a bayesian analysis/lmer is a step towards that- so i won't be doing anova. i can't be of much specific help with that question, but here you go. https://stat.ethz.ch/pipermail/r-sig-mixed-models/2011q1/015591.html On Mar 27, 2013, at 10:13 PM, Nicole Ford wrote: i literally just ran one. when i ran one of mine and the did summary(mod) i get the following: mod - lmer(dem ~ xbar + cpi + (1 | country), data=wvsAB) summary(mod) Linear mixed model fit by REML Formula: dem ~ xbar + cpi + (1 | country) Data: wvsAB AIC BIC logLik deviance REMLdev 34383 34418 -1718734355 34373 Random effects: Groups NameVariance Std.Dev. with a bunch more stuff below. On Mar 27, 2013, at 10:00 PM, Ben Bolker wrote: Michael Grant michael.grant at colorado.edu writes: Dear Help: I am trying to follow Professor Bates' recommendation, quoted by Professor Crawley in The R Book, p629, to determine whether I should model data using the 'plain old' lm function or the mixed model function lmer by using the syntax anova(lmModel,lmerModel). Apparently I've not understood the recommendation or the proper likelihood ratio test in question (or both) for I get this error message: Error: $ operator not defined for this S4 class. I don't have the R Book handy (some more context would be extremely useful! I would think it would count as fair use to quote the passage you're referring to ...) Would someone be kind enough to point out my blunder? You should probably repost this to the r-sig-mixed-mod...@r-project.org mailing list. My short answer would be: (1) I don't think you can actually use anova() to compare likelihoods between lm() and lme()/lmer() fits in the way that you want: *maybe* for lme() [don't recall], but almost certainly not for lmer(). See http://glmm.wikidot.com/faq for methods for testing significance/inclusion of random factors (short answer: you should *generally* try to make the decision whether to include random factors or not on _a priori_ grounds, not on the basis of statistical tests ...) Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmer, p-values and all that
On 13-03-27 10:10 PM, David Winsemius wrote: On Mar 27, 2013, at 7:00 PM, Ben Bolker wrote: Michael Grant michael.grant at colorado.edu writes: Dear Help: I am trying to follow Professor Bates' recommendation, quoted by Professor Crawley in The R Book, p629, to determine whether I should model data using the 'plain old' lm function or the mixed model function lmer by using the syntax anova(lmModel,lmerModel). Apparently I've not understood the recommendation or the proper likelihood ratio test in question (or both) for I get this error message: Error: $ operator not defined for this S4 class. I don't have the R Book handy (some more context would be extremely useful! I would think it would count as fair use to quote the passage you're referring to ...) This is the quoted Rhelp entry: http://tolstoy.newcastle.edu.au/R/help/05/01/10006.html (I'm unable to determine whether it applies to the question at hand.) OK, I misspoke -- sorry. I think the lmer()/lme() likelihoods are actually comparable; it's GLMMs (glmer(), with no analogue in lme()-land except for MASS::glmmPQL, which doesn't give you log-likelihoods at all) where the problem arises. You can (1) use lme(), (2) look at http://glmm.wikidot.com/faq for suggestions about testing random-effects terms (including perhaps don't do it at all), or (3) construct the likelihood ratio test yourself as follows: library(nlme) data(Orthodont) fm1 - lme(distance~age,random=~1|Subject,data=Orthodont) fm0 - lm(distance~age,data=Orthodont) anova(fm1,fm0)[[p-value]] detach(package:nlme,unload=TRUE) library(lme4.0) ## stable version of lme4 fm2 - lmer(distance~age+(1|Subject),data=Orthodont,REML=FALSE) anova(fm2,fm0) ## fails ddiff - -2*c(logLik(fm0)-logLik(fm2)) pchisq(ddiff,1,lower.tail=FALSE) ## not identical to above but close enough Would someone be kind enough to point out my blunder? You should probably repost this to the r-sig-mixed-mod...@r-project.org mailing list. My short answer would be: (1) I don't think you can actually use anova() to compare likelihoods between lm() and lme()/lmer() fits in the way that you want: *maybe* for lme() [don't recall], but almost certainly not for lmer(). See http://glmm.wikidot.com/faq for methods for testing significance/inclusion of random factors (short answer: you should *generally* try to make the decision whether to include random factors or not on _a priori_ grounds, not on the basis of statistical tests ...) Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Passing arguments between apply and l(s)apply functions vs. nested for loop
Mark Orr mo2259 at columbia.edu writes:
Hi R community,
I have a question concerning passing arguments between apply and lapply?
[snip]
#START CODE SNIPPET
#LIST AND VECTOR
rm(list=ls())
l - list(1:3,2:3,4:10,7:9)
v - 1:3
#USED IN j loop to catch values
catch.mat - matrix(NA,nrow=length(v),ncol=length(l))
#LOOPS
for (i in 1:length(v)){
for (j in 1:length(l)) {
catch.mat[i,j] - sum(l[[j]]==i)
}
}
#SIMPLY APPLY OVER catch.mat
catch.all - apply(catch.mat,1,sum)
catch.mat
catch.all
Given the constraint you state about uniqueness,
table(factor(unlist(l),v))
1 2 3
1 2 2
gives the same answer as catch.all --- up to names which you can
remove with unname(table(factor(unlist(l),v)))
[deleted]
#THEN I ADD A NEW FUNCTION
f.2 - function(x,l=l){
i - x
rm(x)
return(sum(lapply(l,function(x) sum(x==i
}
mat.1 - matrix(1:2318,nrow=1,ncol=2318)
apply(mat.1,1,f.2)
#BUT GET ERROR
# apply(mat.1,1,f.2)
#Error in lapply(l, function(x) sum(x == i)) :
# promise already under evaluation: recursive default argument reference or
earlier problems?
Always helps to ponder the error message --- 'recursive default argument'.
Here is an example:
foo - function(x=x) x
foo() ## here it comes again!
Error in foo() :
promise already under evaluation: recursive default argument reference
or earlier problems?
foo(1) # this works
[1] 1
The default is what you use when the argument is not given in the call.
And x=x confuses the evaluator.
You can avoid confusion by choosing a different name like this:
foo - function(xx=x) xx
x - 3
foo()
[1] 3
[rest deleted]
HTH,
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] how to call R in PHP
Hi, I met tough problems when calling R in PHP. I have tried several ways, but none of them succeed. first of all, I tried Rserve, but I failed to connect to it. Then, I tried cmd.exe. I get the result when I use R CMD BATCH --vanilla... in cmd, but I have no idea how to call cmd in php. I know system() is good to call other systems, I triedbut I get no result in my html page. if you could provide some suggestion or examples, i would appreciate your help. the code in my php isï¼ $cmd=R CMD BATCH--vanilla --slave delete.r; $res=system($cmd); the code in the delete.r file is: setwd(c://wamp//www//analysis); x-4; y-3; z-x*y; sink(125.txt); png(file=125.png); barplot(z,border=dark blue) title(main=list(earning of comparison,font=4)); plot(x,z); dev.off() [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] CRAN R Help
Dear Katherine,
For your first question:
If you are creating these files in a specific folder, then
list.files()
#[1] Individual_Present_Value_BONDS.csv Individual_Present_Value_Equity.csv
#[3] Individual_Present_Value_FOREX.csv
gives you which files are present. But, suppose you have other files too in
the folder and you want to check only the above mentioned files.
I added one more file in the working folder.
list.files()
#[1] Individual_Present_Value_BONDS.csv Individual_Present_Value_Equity.csv
#[3] Individual_Present_Value_FOREX.csv test.txt
list.files()[grep(BONDS|Equity|FOREX, list.files())] #only the concerned
files could be seen
#[1] Individual_Present_Value_BONDS.csv Individual_Present_Value_Equity.csv
#[3] Individual_Present_Value_FOREX.csv
#Removed ..._BONDS.csv from the folder.
list.files()[grep(BONDS|Equity|FOREX, list.files())]
#[1] Individual_Present_Value_Equity.csv Individual_Present_Value_FOREX.csv
Second question: (added the ..._BONDS.csv into the folder)
#Read the 3 files in a list
lst1- lapply(list.files()[grep(BONDS|Equity|FOREX,
list.files())],function(x) {x1-read.csv(x,sep=,,stringsAsFactors=FALSE);
colnames(x1)[1]- simulations;x1}) # one of the dataset had Simulations as
first column
#Merge with a subset of data
library(plyr)
lstSub-
lapply(list.files()[grep(BONDS|Equity|FOREX,list.files())],function(x)
{x1-head(read.csv(x,sep=,,stringsAsFactors=FALSE)); colnames(x1)[1]-
simulations;x1}) #subset
join_all(lstSub,type=inner,by=simulations) # depending on your needs, you
can change to type=full
head(join_all(lstSub,type=inner,by=simulations),2) #looks like columns are
in the same order as in the individual dataset
# simulations T_Bond_14.50_2011A T_Bond_14.50_2012B T_Bond_15.50_2010D
#1 Current_PV 184300.7 178906.3 104177.5
#2 Simualtion_no_1 184545.6 178712.6 104431.7
# T_Bond_15.50_2010E T_Bond_15.50_2011B Equity_A Equity_B Equity_C Equity_D
#1 225164.0 1006044 113718.3 392252.5 344494.2 6074407
#2 225727.8 1006456 114838.4 397470.2 346675.5 5943718
# Trans_no.1.Sell.USD Trans_no.2.Sell.USD Trans_no.3.Buy.AUD
#1 -5819794 -15122103 78567139
#2 -6937125 -17368874 78217006
# Trans_no.4.Sell.USD Trans_no.5.Buy.EURO Trans_no.6.Sell.USD
#1 -18621066 10340611 -24023002
#2 -21423091 16831727 -27356539
# Trans_no.7.Buy.EURO Trans_no.8.Sell.USD Trans_no.9.Buy.GBP
#1 10340611 -24023002 -20486447
#2 16831727 -27356539 -18408896
# Trans_no.10.Sell.USD Trans_no.11.Buy.CHF Trans_no.12.Sell.USD
#1 -48273317 5435002 -4647643
#2 -54892776 5652627 -5279506
# Trans_no.13.Sell.USD
#1 -8657418
#2 -9780783
I hope this helps.
A.K.
From: Katherine Gobin katherine_go...@yahoo.com
To: smartpink...@yahoo.com
Sent: Wednesday, March 27, 2013 6:17 PM
Subject: CRAN R Help
Dear Mr Arun,
Sorry to bother you again. Last week you had given me the
question about what will happen to my R – code if say bonds or equity or forex
something
is missing?.
Your this question
has acted like CLUE to me and accordingly I had developed the R –code. And I
think now it’s working (at least as of
now as R has always thrown some surprises ahead of me and hence I am using the
phrase
“at least as of now”).
However, although I had managed say about 95% of the job, I
need one small favor from you if possible.
I have three output csv files and the names are as given
below.
individual_Present_Value_BONDS.csv
individual_Present_Value_Equity.csv
individual_Present_Value_FOREX.csv
I have attached these files in zip format for your
reference. Each of these files have first column as common column. So I can
combine these files (after reading these as data.frames) using merge command.
However, there are two problems.
(1) I am not sure if the file exists or not.
Depending on the presence in portfolio, file is created. E.g. if there is no
Equity, ‘individual_Present_Value_Equity.csv’ will not be created. So how do I
merge these files only after testing if the file exists or not.
(2) Also, assuming the files are merged, then the
columns in the merged data.frame are not in the same order as they are in the
individual files. i.e. the columns get mixed up in the
For me the first task is more
important. Will it be possible for you to help me out. Sorry again for writing
directly to you. I
wrote directly to you because you are already aware of my problem and secondly,
I don’t know if we can attach the files to the R forum mails.
Sorry to bother you once again.
Regards
Katherine
__
R-help@r-project.org mailing
Re: [R] how to call R in PHP
While someone around here might know about PHP (I don't h, the fact that you don't know how to call system suggests that you should be asking in a PHP help forum. Rserve can be used for interactive sessions, but web services usually aren't designed that way, so you probably want to call R with batch arguments the same way you would call it from the command interpreter, but not actually via the command interpreter. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Lauren Zhang zxj19880...@yahoo.cn wrote: Hi, I met tough problems when calling R in PHP. I have tried several ways, but none of them succeed.�� first of all, I tried Rserve, but I failed to connect to it. Then, I tried cmd.exe. I get the result when I use��R CMD BATCH --vanilla... in cmd, but I have no idea how to call cmd in php. I know system() is good to call other systems, I triedbut I get no result in my html page. if you could provide some suggestion or examples, i would appreciate your help. the code in my php is��� $cmd=R CMD BATCH--vanilla --slave delete.r; $res=system($cmd); the code in the delete.r file is: setwd(c://wamp//www//analysis); x-4; y-3; z-x*y; sink(125.txt); png(file=125.png); barplot(z,border=dark blue) title(main=list(earning of comparison,font=4)); plot(x,z); dev.off() [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] new question
Hi, Regarding the first question: ListFacGroup- lapply(ListFacGroup,unique) Spec(ListFacGroup,0.05) # # Seq Mod z a2 c2 c3 t2 #1 aAATATAGPR 1-n_acPro/ 2 1 0 0 1 #2 aAAASSPVGVGQR 1-n_acPro/ 2 1 0 0 1 #3 aAGAAGGR 1-n_acPro/ 2 1 1 0 1 #4 AAALQAK 2 1 0 1 1 #5 aAGAGPEMVR 1-n_acPro/ 2 2 2 1 2 #6 aEQQQFYLLLGNLLSPDNVVR 1-_Carbamoylation/ 2 1 0 0 1 #7 aEQQQFYLLLGNLLSPDNVVR 1-_Carbamoylation/ 3 1 0 0 1 #8 aEQQQFYLLLGNLLSPDNVVR 1-n_acPro/ 2 0 1 0 0 #9 aEQQQFYLLLGNLLSPDNVVR 1-n_acPro/ 3 1 2 2 1 #10 APGTAEK 2 0 1 0 0 #11 aSAPQQLSDEELFSQLR 1-n_acPro/ 2 1 0 0 1 #12 aVGNAVPCGAR 1-n_acPro/ 2 1 1 1 1 #13 AWEEPSSGNGTAR 2 1 1 1 1 #14 aAAAELSLLEK 1-n_acPro/ 1 1 0 0 1 #15 aAAAELSLLEK 1-n_acPro/ 2 1 1 1 1 #16 EVLGLILR 2 1 1 1 1 #17 aAAAGAAAEGEAPAEMGALLLEK 1-n_acPro/ 3 1 1 1 1 #18 aAAAPGTAVGATGSGIAGLAVYR 1-_Carbamoylation/ 3 0 0 1 0 #19 aAAAPGTAVGATGSGIAGLAVYR 1-n_acPro/ 3 1 0 0 1 #20 aAAANSGSSLPLFDCPTWAGKPPPGLHLDVVK 1-n_acPro/ 3 1 0 0 1 #21 AAAkAAK 8-K_ac/ 2 0 1 0 0 #22 aAAAVGAGHGAGGPGAASSSGGAR 1-n_acPro/ 2 0 1 1 0 #23 aAAAVGAGHGAGGPGAASSSGGAR 1-n_acPro/ 3 0 0 1 0 #24 aAADGDDSLYPIAVLIDELR 1-n_acPro/ 2 0 0 1 0 2nd question: I am not sure I understand it correctly. A.K. From: Vera Costa veracosta...@gmail.com To: arun smartpink...@yahoo.com Sent: Wednesday, March 27, 2013 7:07 PM Subject: Re: new question Hi. With the data in attach,the results has duplicated. I think it's better only one rows with the information. The 2nd question,about t test,it's what you did,but for frequencies directly. You understand? Thank you very much __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
