Re: [R] Problem with new(er) R version's matrix package
On 25 April 2014 20:15, David Winsemius dwinsem...@comcast.net wrote: On Apr 25, 2014, at 9:17 AM, Werner W. wrote: Dear Rs, I am re-executing some older code. It does work in the ancient R 2.12.0 which I still have on my PC but with the new version R 3.1.0 it does not work any more (but some other new stuff, which won't work with 2.12). The problem arises in context with the systemfit package using the matrix package. In R 3.1.0 the following error is thrown: Error in as.matrix(solve(W, tol = solvetol)[1:ncol(xMat), 1:ncol(xMat)]) : error in evaluating the argument 'x' in selecting a method for function 'as.matrix': Error in .solve.sparse.dgC(as(a, dgCMatrix), b = b, tol = tol) : LU computationally singular: ratio of extreme entries in |diag(U)| = 7.012e-39 However, I have no clue what I can do about this. Was there some change in the defaults of the matrix package? I couldn't find anything apparent in the changelog. As the same code works in R 2.12.0, I suppose that the problem is not my data. You have not told us what version of the Matrix package you were using. As such I would suggest that you review the Changelog which is a link for the CRAN page for pkg:Matrix and go back 4 years or so since R major versions change about once a year. http://cran.r-project.org/web/packages/Matrix/ChangeLog In addition, please provide a minimal, self-contained, reproducible example. Best, Arne -- Arne Henningsen http://www.arne-henningsen.name __ 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] Within ID variable delete all rows after reaching a specific value
Jennifer:
a) Don't post in HTML... read the Posting Guide.
b) Don't make data frames by first making matrices... you rarely create
what you think you are creating. In your case, your code creates a bunch
of factor columns... use the str() function to verify that your data are
sensible before analyzing it.
c) The ave and cumsum functions are useful here:
tmp - data.frame( X1 = rbinom( 1000, 1, .03 )
, X2 = array( 1:127, c(1000,1) )
, X3 = array( format( seq( ISOdate(1990,1,1)
, by='month'
, length=56 )
, format='%d.%m.%Y')
, c( 1000, 1 ) ) )
tmp - tmp[ with( tmp, order( X2, X3 ) ), ]
tmp2 - subset( tmp
, 1 = ave( X1
, X2
, FUN=function( x ) {
cumsum( cumsum( x ) )
} ) ) )
which generates a vector of increasing values once the first nonzero value
is found in each group, and then only keeps the rows for which those
increasing values are zero or one.
On Sat, 26 Apr 2014, Jim Lemon wrote:
On 04/26/2014 12:42 PM, Jennifer Sabatier wrote:
So, I know that's a confusing Subject header.
Here's similar data:
tmp- data.frame(matrix(
c(rbinom(1000, 1, .03),
array(1:127, c(1000,1)),
array(format(seq(ISOdate(1990,1,1), by='month',
length=56), format='%d.%m.%Y'), c(1000,1))),
ncol=3))
tmp- tmp[with(tmp, order(X2, X3)), ]
table(tmp$X1)
X1 is the variable of interest - disease status. It's a survival-type of
variable, where you are 0 until you become 1.
X2 is the person ID variable.
X3 is the clinic date (here it's monthly, just for example...but in my real
data it's a bit more complicated - definitely not equally spaced nor the
same number of visits to the clinic per ID.).
Some people stay X1 = 0 for all clinic visits. Only a small proportion
become X1=1.
However, the data has errors I need to clean off. Once someone becomes
X1=1 they should have no more rows in the dataset. These are data entry
errors.
In my data I have people who continue to have rows in the data. Sometimes
the rows show X1=0 and sometimes X1=1. Sometimes there's just one more row
and sometimes there are many more rows.
How can I go through, find the first X1 = 1, and then delete any rows after
that, for each value of X2?
Thanks!
Jen
Hi Jen,
This might do what you want:
tmp$X3-as.Date(tmp$X3,%d.%m.%Y)
tmp-tmp[order(tmp$X2,tmp$X3),]
first-TRUE
for(patno in unique(tmp$X2)) {
cat(patno,\n)
tmpbit-tmp[tmp$X2 == patno,]
firstone-which(tmpbit$X1 == 1)[1]
cat(firstone,\n)
if(is.na(firstone)) firstone-dim(tmpbit)[1]
newtmpbit-tmpbit[1:firstone,]
if(first) {
newtmp-newtmpbit
first-FALSE
}
else newtmp-rbind(newtmp,newtmpbit)
}
Jim
__
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.
---
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
__
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] Within ID variable delete all rows after reaching a specific value
Hi,
You may also try:
set.seed(425)
##your code
tmp - data.frame(
#
tmp1 - tmp
str(tmp1)
#'data.frame': 1000 obs. of 3 variables:
# $ X1: Factor w/ 2 levels 0,1: 1 1 1 1 1 1 1 1 1 1 ...
# $ X2: Factor w/ 127 levels 1,10,100,..: 1 1 1 1 1 1 1 1 2 2 ...
# $ X3: Factor w/ 56 levels 01.01.1990,01.01.1991,..: 1 21 17 37 33 51 48
10 11 45 #...
tmp1 - tmp1[with(tmp1,order(X2, as.Date(X3, %d.%m.%Y))),]
tmp2 - tmp1[with(tmp1,!ave(as.numeric(as.character(X1)),X2, FUN=function(x)
cumsum(cumsum(x)) 1 )),]
###checking results with Jim's method
tmp2New - tmp2
tmp2New$X3 - as.Date(tmp2New$X3, %d.%m.%Y)
identical(tmp2New,newtmp) ##Jim's result
#[1] TRUE
A.K.
On Saturday, April 26, 2014 12:07 AM, Jim Lemon j...@bitwrit.com.au wrote:
On 04/26/2014 12:42 PM, Jennifer Sabatier wrote:
So, I know that's a confusing Subject header.
Here's similar data:
tmp- data.frame(matrix(
c(rbinom(1000, 1, .03),
array(1:127, c(1000,1)),
array(format(seq(ISOdate(1990,1,1), by='month',
length=56), format='%d.%m.%Y'), c(1000,1))),
ncol=3))
tmp- tmp[with(tmp, order(X2, X3)), ]
table(tmp$X1)
X1 is the variable of interest - disease status. It's a survival-type of
variable, where you are 0 until you become 1.
X2 is the person ID variable.
X3 is the clinic date (here it's monthly, just for example...but in my real
data it's a bit more complicated - definitely not equally spaced nor the
same number of visits to the clinic per ID.).
Some people stay X1 = 0 for all clinic visits. Only a small proportion
become X1=1.
However, the data has errors I need to clean off. Once someone becomes
X1=1 they should have no more rows in the dataset. These are data entry
errors.
In my data I have people who continue to have rows in the data. Sometimes
the rows show X1=0 and sometimes X1=1. Sometimes there's just one more row
and sometimes there are many more rows.
How can I go through, find the first X1 = 1, and then delete any rows after
that, for each value of X2?
Thanks!
Jen
Hi Jen,
This might do what you want:
tmp$X3-as.Date(tmp$X3,%d.%m.%Y)
tmp-tmp[order(tmp$X2,tmp$X3),]
first-TRUE
for(patno in unique(tmp$X2)) {
cat(patno,\n)
tmpbit-tmp[tmp$X2 == patno,]
firstone-which(tmpbit$X1 == 1)[1]
cat(firstone,\n)
if(is.na(firstone)) firstone-dim(tmpbit)[1]
newtmpbit-tmpbit[1:firstone,]
if(first) {
newtmp-newtmpbit
first-FALSE
}
else newtmp-rbind(newtmp,newtmpbit)
}
Jim
__
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] about rect.hclust
On 26 Apr 2014, at 06:41, Gregory Jefferis jeffe...@gmail.com wrote: # plot dendrogram coloured by cluster: ## library(dendroextras) cluc=colour_clusters(clu, k=3, groupLabels =TRUE) plot(cluc) There was a typo in an argument name for colour_clusters in my earlier response (thanks to Dennis Murphy for pointing this out). Correction in above. Fancier versions are also possible: plot(colour_clusters(clu, k=3, groupLabels = as.roman)) Best, Greg. [[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] Simple Loop Counter
HI,
May be this helps:
for(i in letters) {
n - n+1
x[n,] - c(i, n)
cat(loop, n, \n)
}
x
#or
for(i in seq_along(letters)) {
n - n+1
x[n,] - c(letters[i], n)
cat(loop, i, \n)
}
x
A.K.
#How can I add the loop counter in #2 to the loop in #1.?
#***
#1.
x - matrix( , ncol = 2, nrow = 26) # empty matrix
n - 0 #set n to 0
for(i in letters) {
n - n+1
x[n,] - c(i, n)
}
x
#***
#2.
for (i in 1:10) {
cat(loop, i, \n)
}
__
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] about rect.hclust
Thank you for mentioning the dendextendhttp://cran.r-project.org/web/packages/dendextend/package Gregory. As I mention in the package, the dendextend package implements a similar function as in the dendroextra. The big difference between the two is that the dendextend package has an implementation of the cutree function for dendrograms (making it possible to not have to go through the hclust object in order to perform it). This is helpful when using non-standard dendrograms that can not be turned into an hclust object (via as.hclust.dendrogram). Also, the dendextend package includes many other functions for dendrogram manipulation which might be helpful. Here is a quick example that can give you a sense of the possibilities: install.packages(dendextend) require(dendextend) # pdf(dendextend_example.pdf) dend - as.dendrogram(hclust(dist(USArrests), ave)) d1=color_branches(dend, k=5, col = c(3,1,1,4,1)) plot(d1) # selective coloring of branches :) d2=color_branches(d1,k=5) plot(d2) # getting rainbow_hcl colors for each cluster (the default) d3=color_branches(d1,5,groupLabels=TRUE) # adding numbers to the branches plot(d3) d4=color_labels(d3, k=5, col = c(3,1,1,4,1)) # color labels by cluster plot(d4) d5=color_labels(d4, col = c(red, orange)) # color labels from left to right plot(d5) # dev.off() Here is the output: https://www.dropbox.com/s/vii83jd9xnbmuv7/dendextend_example.pdf Also, there is a (rough draft) of a vignettes for the package here: https://github.com/talgalili/dendextend/blob/master/vignettes/dendextend-tutorial.pdf?raw=true Feedback is welcome. Best, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Sat, Apr 26, 2014 at 10:36 AM, Gregory Jefferis jeffe...@gmail.comwrote: On 26 Apr 2014, at 06:41, Gregory Jefferis jeffe...@gmail.com wrote: # plot dendrogram coloured by cluster: ## library(dendroextras) cluc=colour_clusters(clu, k=3, groupLabels =TRUE) plot(cluc) There was a typo in an argument name for colour_clusters in my earlier response (thanks to Dennis Murphy for pointing this out). Correction in above. Fancier versions are also possible: plot(colour_clusters(clu, k=3, groupLabels = as.roman)) Best, Greg. [[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] purtest and missing values (plm-package)
Hey arun, I am thankful for your help and assistance! With the example it works! Have a nice day, Katie -Ursprüngliche Nachricht- Von: arun [mailto:smartpink...@yahoo.com] Gesendet: Freitag, 25. April 2014 18:01 An: r-help@r-project.org Cc: Katharina Mersmann Betreff: Re: [R] purtest and missing values (plm-package) Hi, May be this helps: library(plm) data(Grunfeld) Grunfeld$inv[c(2,8)] - NA GrunfeldNew - subset(Grunfeld, !year %in% year[is.na(inv)]) x - data.frame(split(GrunfeldNew$inv, GrunfeldNew$firm)) purtest(x, pmax = 4, exo = none, test = levinlin,lags=SIC) #Levin-Lin-Chu Unit-Root Test (ex. var. : None ) # #data: x #z.x1 = 5.0507, p-value = 4.402e-07 #alternative hypothesis: stationarity # #Warning message: #In selectT(l, theTs) : the time serie is short A.K. On Friday, April 25, 2014 10:48 AM, Katharina Mersmann kmers...@smail.uni-koeln.de wrote: Because, my inbox shows a cutted-email-version, once again: Hello R-Community, I have quarterly panel-data and not surprisingly missing values, in a few variables, which are differently distributed around the panel. Now I want to run different unit-root tests. For ADF on the pooled data set, I chose CADF-package, which can determine the number of lags automa. by SIC and handles missing values. (I hope this is the right one ) Secondly I want to run an LLC and IPS test specificially for the panel data (Levin, Lin Chu –test and Im, Pesaran Shin-test ) for which I use the purtest function, from plm-package. But I don´t know how to apply it, if my series contains missing values ( so only a error-message is created) purtest(data.plm,data=data.plm, test = levinlin,exo = none,lags =SIC) Fehler in lm.fit(X, y) : NA/NaN/Inf in 'x' Zusätzlich: Warnmeldung: In Ops.factor(object[2:length(object)], object[1:(length(object) - : - not meaningful for factors So my Question is: 1.Is there a way to handle the missing values? 2.Do I just omit them? And if, is there a way to integrate this by adding an ” na.omit” into the function ? To make it easier explaining the way of proceeding, a reproducible example could be: data(Grunfeld, package = plm) y - data.frame(split(Grunfeld$inv, Grunfeld$firm)) purtest(y, pmax = 4, exo = none, test = levinlin,lags=”SIC”) # works no missing data # add an NA data(Grunfeld, package = plm) x −data.frame(split(Grunfeldinv, Grunfeld$firm)) Grunfeldinv[2]−NA purtest(x, pmax = 4, exo = none, test = levinlin,lags=”SIC”) # Error in lm.fit(X, x) : NA/NaN/Inf in 'x' Thanks for your hints and suggestions! Katie -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von Katharina Mersmann Gesendet: Freitag, 25. April 2014 15:53 An: r-help@r-project.org Betreff: [R] purtest and missing values (plm-package) Hello R-Community, I have quarterly panel-data and not surprisingly missing values, in a few variables, which are differently distributed around the panel. Now I want to run different unit-root tests. For ADF on the pooled data set, I chose CADF-package, which can determine the number of lags automa. by SIC and handles missing values. (I hope this is the right one ) Secondly I want to run an LLC and IPS test specificially for the panel data (Levin, Lin Chu b __ 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] Metafor: How to integrate effectsizes?
At 20:34 25/04/2014, Viechtbauer Wolfgang (STAT) wrote:
If you know the d-value and the corresponding
group sizes for a study, then it's possible to
add that study to the rest of the dataset. Also,
if you only know the test statistic from an
independent samples t-test (or only the p-value
corresponding to that test), it's possible to
back-compute what the standardized mean difference is.
I added an illustration of this to the metafor package website:
http://www.metafor-project.org/doku.php/tips:assembling_data_smd
Verena might also like to look at the compute.es
package available from CRAN to see whether any of
the conversions programmed there do the job.
Best,
Wolfgang
--
Wolfgang Viechtbauer, Ph.D., Statistician
Department of Psychiatry and Psychology
School for Mental Health and Neuroscience
Faculty of Health, Medicine, and Life Sciences
Maastricht University, P.O. Box 616 (VIJV1)
6200 MD Maastricht, The Netherlands
+31 (43) 388-4170 | http://www.wvbauer.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Michael Dewey
Sent: Friday, April 25, 2014 16:23
To: Verena Weinbir
Cc: r-help@r-project.org
Subject: Re: [R] Metafor: How to integrate effectsizes?
At 12:33 25/04/2014, you wrote:
Thank you very much for your reply and the book recommendation, Michael.
Yes, I mean Cohen's d - sorry for the typo :-)
Just to make this sure for me: There is no
possibility to integrate stated Cohens' ds in an
R-Metaanalysis (or a MA at all), if there is no
further information traceable regarding SE or the like?
If there is really no other information like
sample sizes, significance level, value of some
significance test then you would have to impute a
value from somewhere. That would seem a last resort.
I have cc'ed this back to the list, please keep
it on the list so others may benefit and contribute.
best regards,
Verena
On Fri, Apr 25, 2014 at 1:21 PM, Michael Dewey
mailto:i...@aghmed.fsnet.co.uki...@aghmed.fsnet.co.uk wrote:
At 13:15 24/04/2014, Verena Weinbir wrote:
Hello!
I am using the metafor package for my master's thesis as an R-newbie.
While
calculating effectsizes from my dataset (mean values and
standarddeviations) using escalc shouldn't be a problem (I hope ;-)),
I
wonder how I could at this point integrate additional studies, which
only
state conhens d (no information about mean value and sds available), to
calculate an overall analysis. Â I would be very grateful for your
support!
You mean Cohen's d I think.
You will need some more information to enable
you to calculate its standard error. Have a look at Rosenthal's chapter
in
@book{cooper94,
  author = {Cooper, H and Hedges, L V},
  title = {A handbook of research synthesis},
  year = {1994},
  publisher = {Russell Sage},
  address = {New York},
  keywords = {meta-analysis}
}
(There is an updated edition)
This gives you more information about converting
effect sizes and extracting them from unpromising beginnings.
It often requires some ingenuity to get the
information you need so have a go and then get
back here with more details if you run into problems
Best regards,
Verena
Michael Dewey
i...@aghmed.fsnet.co.uk
http://www.aghmed.fsnet.co.uk/home.html
__
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] Cramer Rao upper/lower bounds- No Comments ?
Hello, Is this homework? There's a no homework policy here. Em 25-04-2014 12:40, Mohammed Ouassou escreveu: Dear R users; I have a question about Cramer Rao upper/lower bounds Cramer Rao _lower_ bound, not upper. Is it possible to compute Crammer Rao upper/lower bounds from residuals and corresponding covariance matrices ? Residuals of what? Hope this helps, Rui Barradas Any suggestions will be appreciated, thanks in advance. M.O __ 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] Read.table mucks up headers
Hello Uwe Others Thanks for all your help, I figured out what the problem was. It wasn't working with old R version. Once I updated it to the latest version, it seemed to work. Thank you -- View this message in context: http://r.789695.n4.nabble.com/Read-table-mucks-up-headers-tp4688742p4689519.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] [R-sig-DB] Update results not being written to existing data frame when using sqldf UPDATE
On Sat, Apr 26, 2014 at 10:06 AM, Christopher Lowenkamp clowenk...@gmail.com wrote: RStudio Version 0.98.501 R version 3.1.0 Mac OSX 10.9.2 Packages loaded: sqldf gsubfn proto RSQLite DBI RSQLite.extfuns tcltk Good morning: I am trying to run an sqldf update with two tables. Both tables contain a variable called ‘off_id’. I am trying to update a variable in tablea (v2) with the number of times each record in tablea appears in tableb. ## tablea - data.frame(off_id = c(12, 14, 16, 17, 18, 22, 1, 5, 7, 44, 4, 3), v2 = 0) tableb - data.frame(off_id = c(12, 12, 14, 14, 14, 14, 16, 17, 12, 12, 1, 18, 18, 5, 7, 3, 16, 1, 1, 3, 3, 3, 1)) sql1 - UPDATE tablea SET v2 = (SELECT count(*) FROM tableb WHERE tableb.off_id = tablea.off_id) sql2 - “SELECT * FROM tablea” #The following code returns NULL sqldf(sql1, sql2) #When I run the following I do get back the data but tablea$v2 still does not update sqldf(c(sql1, sql2), method = raw) #If I run the following I get the expected results in tablec$v2, but tablea$v2 does not update tablec - as.data.frame(sqldf(c(sql1, sql2))) ## I am wondering what I am doing wrong. Is there a way to get tablea$v2 to update? I did check at https://code.google.com/p/sqldf/ (and have read through FAQ 8 a number of times) but don't see an answer to the problem I am having. sqldf never modies any object in your R work space. It did update the table in the main sqlite database but its up to you if you want to write it back to R. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at 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] Faster way to transform vector [3 8 4 6 1 5] to [2 6 3 5 1 4]
Hi,
could anybody help me to find a fast way to fix the following question?
Given a verctor of length N, for example bo = [3 8 4 6 1 5],
I want to drive a vector whose elements are 1, 2, ..., N and the order of
elements is the same as that in verctor bo.
In this example, the result is supposed to be bt = [2 6 3 5 1 4].
I used the following code to solove this:
bo - c(3, 8, 4, 6, 1, 5)
N - length(bo)
bt - rep(0, N)
M - max(bo)
temp - bo
for(i in 1 : N)
{
min - M
i_min - 0
for(j in 1 : N)
{
if(min = temp[j])
{
min - temp[j]
i_min -j
}
}
bt[i_min] - i
temp[i_min] - M+ 1
}
bt
[1] 2 6 3 5 1 4
However, the time complexity is O(N2).
When N is larger than 100, it takes too much time.
Is there any faster way to fix it?
best
Xueming
xmliu1...@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.
[R] Food grade Peroxide for treatment life threatening chronic ailments.
Forget quil eggs. 35% food grade hydrogen peroxide has brought about relief to tens of thousands of patients suffering from disastrous diseases like CANCER, LEUKEMIA, HEPATITIS, KIDNEY FAIURE, HEART PROBEMS, LIVER DISORDERS. ASTHMA, ALLERGIES, CHRONIC ULCERS, HIGH BLOOD PRESSURE, CANDIDIASIS, FAILURE TO CONCEIVE IN LADIES, SINUSES, and many more. Over 6,100 articles in European scientific literature have attested to its effectiveness in not only killing diseased cells but also simultaneously revitalizing and rejuvenating healthy cells, thereby creating vibrant energy and well-being. It is safe, inexpensive and powerful healing modality and has been administered by an estimated 15,000 European doctors, naturopaths and homeopaths to more than 10 million patients in the past 70 years to successfully treat practically every known disease. 35% Food grade hydrogen peroxide is now availble at our clinic for treatment of most diseases. Kindly call us for more information. 0710 100 199 or Visit our website http://www.secretcures.co.ke/_item?item_id=5700735861784576 for more information. We are located at Nyota house. 3rd floor room 304 along accra road Nairobi cbd. Regards, Kenya Natural Alternative Medicine Initiative CLINIC. --- This email is free from viruses and malware because avast! Antivirus protection is active. __ 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] Update results not being written to existing data frame when using sqldf UPDATE
RStudio Version 0.98.501 R version 3.1.0 Mac OSX 10.9.2 Packages loaded: sqldf gsubfn proto RSQLite DBI RSQLite.extfuns tcltk Good morning: I am trying to run an sqldf update with two tables. Both tables contain a variable called âoff_idâ. I am trying to update a variable in tablea (v2) with the number of times each record in tablea appears in tableb. ## tablea - data.frame(off_id = c(12, 14, 16, 17, 18, 22, 1, 5, 7, 44, 4, 3), v2 = 0) tableb - data.frame(off_id = c(12, 12, 14, 14, 14, 14, 16, 17, 12, 12, 1, 18, 18, 5, 7, 3, 16, 1, 1, 3, 3, 3, 1)) sql1 - UPDATE tablea SET v2 = (SELECT count(*) FROM tableb WHERE tableb.off_id = tablea.off_id) sql2 - âSELECT * FROM tableaâ #The following code returns NULL sqldf(sql1, sql2) #When I run the following I do get back the data but tablea$v2 still does not update sqldf(c(sql1, sql2), method = raw) #If I run the following I get the expected results in tablec$v2, but tablea$v2 does not update tablec - as.data.frame(sqldf(c(sql1, sql2))) ## I am wondering what I am doing wrong. Is there a way to get tablea$v2 to update? I did check at https://code.google.com/p/sqldf/ (and have read through FAQ 8 a number of times) but don't see an answer to the problem I am having. Thanks for your time and consideration. Christopher Lowenkamp Administrative Office US Courts University of Missiouri-Kansas City [[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] Can't plot transparency images in R-3.1.0
It seems to have something to do with my other library dependencies that has broken the plotting. I tried to re-install R-3.0.2 again and I am getting the same error for that R version too. Been scouring the web for advice: http://stackoverflow.com/questions/10777008/how-to-set-cairo-as-default-backend-for-x11-in-r http://r.789695.n4.nabble.com/ggplot-2-semi-transparency-error-td902379.html But it seems that these solutions have to do with saving the plot. I am just trying to plot it in the R Console... On Fri, Apr 25, 2014 at 7:06 PM, Ulises M. Alvarez u...@sophie.unam.mxwrote: On 04/25/2014 07:26 PM, Fong Chun Chan wrote: Hi all, I recently upgraded to R-3.1.0 from R-3.0.2. Things seem to be fine, but when trying to plot a ggplot image with transparency, I get the following issue: Warning message: In grid.Call.graphics(L_polygon, x$x, x$y, index) : semi-transparency is not supported on this device: reported only once per page This only appears to affect my R-3.1.0 installation and not my R-3.0.2 as I can still plot normally with it. Has anyone else experienced this problem? Thanks, Hi: After the upgrade to 3.1.0, did you execute? update.packages(checkBuilt=TRUE) I did it, and I have no problem running the code that you provide. sessionInfo() # R version 3.1.0 (2014-04-10) # Platform: x86_64-pc-linux-gnu (64-bit) # # locale: # [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C # [3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8 # [5] LC_MONETARY=en_US.utf8LC_MESSAGES=en_US.utf8 # [7] LC_PAPER=en_US.utf8 LC_NAME=C # [9] LC_ADDRESS=C LC_TELEPHONE=C # [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C # # attached base packages: # [1] stats graphics grDevices utils datasets methods # base # # other attached packages: # [1] ggplot2_0.9.3.1 -- Ulises M. Alvarez http://sophie.unam.mx/ [[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] Faster way to transform vector [3 8 4 6 1 5] to [2 6 3 5 1 4]
Look into the rank function.
If there are duplicated values in the input vector its 'ties' argument
says how to deal with them. If there are ties I think your algorithm
puts the last one in the first position, e.g., it maps
c(101,101,101,102,102) to c(3,2,1,5,4). rank does not include this
option, but if that is really what you want to do you can use
myRank - function (x) rev(rank(rev(x), ties = first))
On Sat, Apr 26, 2014 at 2:54 AM, xmliu1...@gmail.com
xmliu1...@gmail.com wrote:
Hi,
could anybody help me to find a fast way to fix the following question?
Given a verctor of length N, for example bo = [3 8 4 6 1 5],
I want to drive a vector whose elements are 1, 2, ..., N and the order of
elements is the same as that in verctor bo.
In this example, the result is supposed to be bt = [2 6 3 5 1 4].
I used the following code to solove this:
bo - c(3, 8, 4, 6, 1, 5)
N - length(bo)
bt - rep(0, N)
M - max(bo)
temp - bo
for(i in 1 : N)
{
min - M
i_min - 0
for(j in 1 : N)
{
if(min = temp[j])
{
min - temp[j]
i_min -j
}
}
bt[i_min] - i
temp[i_min] - M+ 1
}
bt
[1] 2 6 3 5 1 4
However, the time complexity is O(N2).
When N is larger than 100, it takes too much time.
Is there any faster way to fix it?
best
Xueming
xmliu1...@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.
--
Bill Dunlap
TIBCO Software
wdunlap tibco.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] Faster way to transform vector [3 8 4 6 1 5] to [2 6 3 5 1 4]
Hi Xueming,
Try
(1:length(bo))[rank(bo)]
In a function the above would be
f - function(x){
N - length(x)
(1:N)[rank(x)]
}
f(bo)
# [1] 2 6 3 5 1 4
HTH,
Jorge.-
On Sat, Apr 26, 2014 at 7:54 PM, xmliu1...@gmail.com xmliu1...@gmail.comwrote:
Hi,
could anybody help me to find a fast way to fix the following question?
Given a verctor of length N, for example bo = [3 8 4 6 1 5],
I want to drive a vector whose elements are 1, 2, ..., N and the order of
elements is the same as that in verctor bo.
In this example, the result is supposed to be bt = [2 6 3 5 1 4].
I used the following code to solove this:
bo - c(3, 8, 4, 6, 1, 5)
N - length(bo)
bt - rep(0, N)
M - max(bo)
temp - bo
for(i in 1 : N)
{
min - M
i_min - 0
for(j in 1 : N)
{
if(min = temp[j])
{
min - temp[j]
i_min -j
}
}
bt[i_min] - i
temp[i_min] - M+ 1
}
bt
[1] 2 6 3 5 1 4
However, the time complexity is O(N2).
When N is larger than 100, it takes too much time.
Is there any faster way to fix it?
best
Xueming
xmliu1...@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.
[[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] mapply echoes function call when browser() is called from within FUN
When mapply is applied to a function that has a call to browser() within
it, the result can be a disastrous amount of feedback.
To clarify this situation please consider the following function,
containing a call to browser within it:
plus = function(a, b) {browser(); a + b}
A plain vanilla call to plus() yields the following:
LAPTOP_32G_01 plus(1,2)
Called from: plus(1, 2)
Browse[1]
[1] 3
Now consider the following application of mapply to plus:
LAPTOP_32G_01 mapply(plus, 1:2, 1:2)
Called from: (function (a, b)
{
browser()
a + b
})(dots[[1L]][[1L]], dots[[2L]][[1L]])
Browse[1]
Called from: (function (a, b)
{
browser()
a + b
})(dots[[1L]][[2L]], dots[[2L]][[2L]])
Browse[1]
[1] 2 4
Notice that at each step, after the browser is called, mapply prints out
the function call including its arguments:
Called from: (function (a, b)
{
browser()
a + b
})(dots[[1L]][[1L]], dots[[2L]][[1L]])
etc.
In the present case this does no harm except to make things a little harder
to read. However, if one of the inputs happens to be a data frame with a
million rows, the entire million rows are printed to the screen. I have
been bitten by this, which is why I am writing this note. I have a question
and a request:
Question: Is there some way to prevent mapply (or browser) from echoing the
function call when browser is called from within FUN?
Request: If not, could the ability to turn off this echoing be provided. As
things stand, calling browser from within FUN, when FUN is a realistically
big function or has realistically big arguments, is a disaster.
Thanks.
[[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] mapply echoes function call when browser() is called from within FUN
On 26/04/2014, 11:42 AM, Rguy wrote:
When mapply is applied to a function that has a call to browser() within
it, the result can be a disastrous amount of feedback.
To clarify this situation please consider the following function,
containing a call to browser within it:
plus = function(a, b) {browser(); a + b}
A plain vanilla call to plus() yields the following:
LAPTOP_32G_01 plus(1,2)
Called from: plus(1, 2)
Browse[1]
[1] 3
Now consider the following application of mapply to plus:
LAPTOP_32G_01 mapply(plus, 1:2, 1:2)
Called from: (function (a, b)
{
browser()
a + b
})(dots[[1L]][[1L]], dots[[2L]][[1L]])
Browse[1]
Called from: (function (a, b)
{
browser()
a + b
})(dots[[1L]][[2L]], dots[[2L]][[2L]])
Browse[1]
[1] 2 4
Notice that at each step, after the browser is called, mapply prints out
the function call including its arguments:
Called from: (function (a, b)
{
browser()
a + b
})(dots[[1L]][[1L]], dots[[2L]][[1L]])
etc.
In the present case this does no harm except to make things a little harder
to read. However, if one of the inputs happens to be a data frame with a
million rows, the entire million rows are printed to the screen. I have
been bitten by this, which is why I am writing this note. I have a question
and a request:
I don't see the argument values being printed in your example, and if I
replace them with dataframes, I still don't see them. So it's not quite
as simple as you describe to get the voluminous output.
Reproducible examples are needed if you want something fixed.
Question: Is there some way to prevent mapply (or browser) from echoing the
function call when browser is called from within FUN?
Yes, use the skipCalls argument to browser.
Duncan Murdoch
Request: If not, could the ability to turn off this echoing be provided. As
things stand, calling browser from within FUN, when FUN is a realistically
big function or has realistically big arguments, is a disaster.
Thanks.
[[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.
[R] lattice plot formatting: pch, abbreviation and labels
Dear all, I am trying to use the lattice plot, but the syntax is quite difficult. Specifically I have eight variables (1 to 8) each of them further subdivided in two classes (negative=0 and positive=1). I am using the stripplot() to represent these values. I would like to represent the negative and positive values with black and white dots so I have tried to use the argument pch=c(16, 1) both in the main stripplot function and embedded in the scale argument. However the resulting plot shows that some points are drawn with the pch=16 and other with pch=1 irrespective of their class. Is there a way to draw the values for the variable positivity = 0 with pch=16 and those with positivity = 1 with pch=0? In addition I would like to change the labels under the axis from 0,1 to N,P. However when I placed labels=c(N, P) or labels=list(N, P) in the main stripplot() I did not obtained any difference and when placed within the scale argument also the -labels were modified. Is there a way to change the label 0 with N and the label 1 with P? final problem: I would like to abbreviate the Unstimulated box label with Unst.. I have tried the abbreviate=TRUE argument but again the syntax is too complex for me and it did not work. Is there a way to abbreviate the variables names? Thank you very much for your help. Best wishes, Luigi CODE: ### open plot library library(lattice) my.data-structure(list( column_1 = 1:120, column_2 = structure(c( 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = structure(c( 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 0,0,0,0,0,0,0,0)), column_4 = c( 192.0519108,183.6403531,53.46798757,83.60638077,69.60749873,159.4706861,256.8765622,499.2899303, 2170.799076,1411.349719,2759.472348,2098.973397,2164.739515,1288.676574,1611.486543,6205.229575, 870.7424981,465.9967135,191.8962375,864.0937485,2962.693675,1289.259137,2418.651212,7345.712517, 0,168.1198893,674.4342961,101.1575401,47.81596237,0,0,1420.793922, 142.6871331,5.466468742,291.9564635,80.73914133,73.02239621,64.47806871,144.3543635,3167.959757, 3164.748333,1092.634557,28733.20269,1207.87783,729.6090973,151.8706088,241.2466141,9600.963594, 1411.718287,12569.96285,1143.254476,6317.378481,16542.27718,79.68025792,1958.495138,7224.503437, 208.4382941,69.48609769,656.691151,0.499017582,7114.910926,187.6296174,41.73980805,8930.784541, 4.276752185,0.432300363,60.89228665,1.103924786,0.490686366,1.812993239,7.264531581,1518.610307, 2172.051528,595.8513744,17141.84336,589.6565971,1340.287628,117.350942,593.7034054,24043.61463, 0,81.83292179,1539.864321,36.41722958,8.385131047,161.7647376,65.21615696,7265.573875, 97.84753179,154.051827,0.613835842,10.06138851,45.04879285,176.8284258,18795.75462,30676.769, 5780.34957,944.2200834,2398.235596,1083.393165,2541.714557,1251.670895,1547.178549,1792.679176, 3067.988416,8117.210173,23676.02226,8251.937547,17360.80494,18563.61561,16941.865,31453.96708, 2767.493803,4796.33016,12292.93705,3864.657567,9380.673835,14886.44683,8457.88646,26050.47191)), .Names = c(row, stimulation, positivity, copy), row.names = c(NA, -120L), class = data.frame) attach(my.data) stripplot(my.data$copy ~ factor(my.data$positivity)|factor(my.data$stimulation, levels = c(Unstimulated, ESAT6,CFP10,Rv3615c, Rv2654, Rv3879, Rv3873,PHA)), my.data, hor=F, layout = c(8,1), scales = list(relation = same), jitter.data=TRUE, alpha=1, pch=c(16,1), col=black, ylab=expression(bold(Copy)), xlab=expression(bold(Stimulation)), main=Plot, par.settings = list(strip.background=list(col=white)), par.strip.text=list(font=2)) __ 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] Faster way to transform vector [3 8 4 6 1 5] to [2 6 3 5 1 4]
Hi,
Perhaps,
rank(bo)
#[1] 2 6 3 5 1 4
A.K.
On Saturday, April 26, 2014 11:00 AM, xmliu1...@gmail.com
xmliu1...@gmail.com wrote:
Hi,
could anybody help me to find a fast way to fix the following question?
Given a verctor of length N, for example bo = [3 8 4 6 1 5],
I want to drive a vector whose elements are 1, 2, ..., N and the order of
elements is the same as that in verctor bo.
In this example, the result is supposed to be bt = [2 6 3 5 1 4].
I used the following code to solove this:
bo - c(3, 8, 4, 6, 1, 5)
N - length(bo)
bt - rep(0, N)
M - max(bo)
temp - bo
for(i in 1 : N)
{
min - M
i_min - 0
for(j in 1 : N)
{
if(min = temp[j])
{
min - temp[j]
i_min -j
}
}
bt[i_min] - i
temp[i_min] - M+ 1
}
bt
[1] 2 6 3 5 1 4
However, the time complexity is O(N2).
When N is larger than 100, it takes too much time.
Is there any faster way to fix it?
best
Xueming
xmliu1...@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.
__
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] GLM using truncated lognormal distribution
Dear honorable list-members, I know how to fit a truncated lognormal distribution (or Gaussian) (example here: http://max2.ese.u-psud.fr/epc/conservation/Girondot/Publications/Blog_r/Entrees/2012/5/24_Adjust_a_truncated_lognormal_distribution.html ) but I would like to use it in the context of glm. Rather than using family=gaussian(), ideally I would like to have a family=truncated_gaussian(). I see using fix(gaussian) how is organized the gaussian() function. It is not 100% clear now but I think I could manage to change it to do a family=truncated_gaussian(). But before to do it, perhaps it exists already. I find the package truncnorm but it does not do this function. Thanks a lot for any advice, Marc __ 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] writing a package with doParallel and compiled C
Hi all, Any tips on how to use doParallel as part of a function in a package that I am writing? The function calls compiled C code, and (as far as I know), doParallel requires you to load C functions for each core that you use (either with the .packages argument to foreach(), or by explicitly loading the object for each loop). However - the package doesn't pass the check command when I try to build it. I suspect that this is because the package isn't yet loaded into the namespace while it is being checked, and doParallel can therefore not find it. Does anyone know whether my suspicion is wrong, or whether there is a better way to handle doParallel in this case? Many thanks, -- Adam Clark PhD Candidate, Dept. Ecology, Evolution, and Behavior University of Minnesota, Twin Cities 100 Ecology Building, 1987 Upper Buford Circle, St. Paul, MN 55108 atcl...@umn.edu, (857)-544-6782, www.cbs.umn.edu/lab/tilman/adamclark [[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] writing a package with doParallel and compiled C
I am going to go out on a limb and suggest that you not do this at all. Packages are good places for algorithms, and parallel processing is an infrastructure optimization that a) is not always an efficiency win, and b) can be quite sensitive to the actual infrastructure that is available (best solutions for Windows and *nix platforms often being noticeably different). If you have considered this already and still intend to proceed, I have to defer to someone else for an answer to your actual question. --- 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. On April 26, 2014 11:52:18 AM PDT, Adam Clark atcl...@umn.edu wrote: Hi all, Any tips on how to use doParallel as part of a function in a package that I am writing? The function calls compiled C code, and (as far as I know), doParallel requires you to load C functions for each core that you use (either with the .packages argument to foreach(), or by explicitly loading the object for each loop). However - the package doesn't pass the check command when I try to build it. I suspect that this is because the package isn't yet loaded into the namespace while it is being checked, and doParallel can therefore not find it. Does anyone know whether my suspicion is wrong, or whether there is a better way to handle doParallel in this case? Many 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] Problem with new(er) R version's matrix package
Arne Henningsen arne.henning...@gmail.com on Sat, 26 Apr 2014 08:15:37 +0200 writes: On 25 April 2014 20:15, David Winsemius dwinsem...@comcast.net wrote: On Apr 25, 2014, at 9:17 AM, Werner W. wrote: Dear Rs, I am re-executing some older code. It does work in the ancient R 2.12.0 which I still have on my PC but with the new version R 3.1.0 it does not work any more (but some other new stuff, which won't work with 2.12). The problem arises in context with the systemfit package using the matrix package. In R 3.1.0 the following error is thrown: Error in as.matrix(solve(W, tol = solvetol)[1:ncol(xMat), 1:ncol(xMat)]) : error in evaluating the argument 'x' in selecting a method for function 'as.matrix': Error in .solve.sparse.dgC(as(a, dgCMatrix), b = b, tol = tol) : LU computationally singular: ratio of extreme entries in |diag(U)| = 7.012e-39 However, I have no clue what I can do about this. Was there some change in the defaults of the matrix package? I couldn't find anything apparent in the changelog. As the same code works in R 2.12.0, I suppose that the problem is not my data. You have not told us what version of the Matrix package you were using. As such I would suggest that you review the Changelog which is a link for the CRAN page for pkg:Matrix and go back 4 years or so since R major versions change about once a year. http://cran.r-project.org/web/packages/Matrix/ChangeLog In addition, please provide a minimal, self-contained, reproducible example. Yes, please do. As maintainer of the Matrix package, I'm willing to look into the situation of course. As was mentioned, many things have changed in 4 years. The error message above looks like you'd want to invert a (very close to) singular matrix, and there could be quite few reasons why parts of the older code gave slightly different answers. Without a reproducible example, we can't get started though. Best regards, Martin Maechler, ETH Zurich __ 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] For loop processing too slow - pre-format data.frame?
Hi,
I am bootstrapping, but my loops are taking way too long I need to make it
faster. Looking on the R-help archive I suspect it may be due to not
specifying the size of my data.frame, mainly because I don't know in advance
how large it has to be. Can anyone help?
My data looks like this (first 5 entries of 'SpeyBay'):
Year JulianDay Hour Day Month Quarter Season SeaState Visibility TideState
1 2005916 1 4 2 22 2 2.18
2 2005917 1 4 2 22 2 1.53
3 2005919 1 4 2 22 3 0.80
4 200591 11 1 4 2 22 4 0.96
5 200591 14 1 4 2 21 6 2.25
TideHeight CetPres Segment
1 2 0 1
2 3 0 1
3 5 0 2
4 -5 0 3
5 -2 0 4
I am bootstrapping 1000 times but re-sampling on segment (since my data is
autocorrelated), which means I am trying to reconstruct my data based on
random segments e.g. segment 3, then segment 1, each of which may include
from 1-14 data rows. So I don't know how many rows I am going to get in
advance.
When I run my for loop, I just use rbind with undefined size of the new
variable e.g. 'tempD2', and I suspect it is this that is slowing down the
whole process (probably partly due to having a for loop within a for loop).
Can anyone give me any advice on how to pre-define a data frame (if this is
what the data shown above is) that can have an undefined size - or how to
make it big enough to take all the data?). I've been trying to figure this
out for ages with no luck sure it's something simple!
Code shown below - any tips on making the code faster would be greatly
appreciated - the last run took several hours which is just not practical!
Many thanks in advance,
Clare Embling
CODE:
SpringWatch - 504
SummerWatch - 704
AutumnWatch - 392
MaxSample - 704
signif - 0
for(j in 1:1000){
# resampling 2 different years (D E) in 3 different seasons (2, 3 4)
separately
D2S - sample(D2Start:D2Stop,MaxSample,replace=T)
D3S - sample(D3Start:D3Stop,MaxSample,replace=T)
D4S - sample(D4Start:D4Stop,MaxSample,replace=T)
E2S - sample(E2Start:E2Stop,MaxSample,replace=T)
E3S - sample(E3Start:E3Stop,MaxSample,replace=T)
E4S - sample(E4Start:E4Stop,MaxSample,replace=T)
# Creating new data frames with the first sampled segment
TempD2 - SpeyBay[(Segment==D2S[1]),]
TempD3 - SpeyBay[(Segment==D3S[1]),]
TempD4 - SpeyBay[(Segment==D4S[1]),]
TempE2 - SpeyBay[(Segment==E2S[1]),]
TempE3 - SpeyBay[(Segment==E3S[1]),]
TempE4 - SpeyBay[(Segment==E4S[1]),]
# loop to add together all the rows of data for each segment sampled
for(i in 2:MaxSample) {
TempD2 - rbind(TempD2,SpeyBay[(Segment==D2S[i]),])
TempD3 - rbind(TempD3,SpeyBay[(Segment==D3S[i]),])
TempD4 - rbind(TempD4,SpeyBay[(Segment==D4S[i]),])
TempE2 - rbind(TempE2,SpeyBay[(Segment==E2S[i]),])
TempE3 - rbind(TempE3,SpeyBay[(Segment==E3S[i]),])
TempE4 - rbind(TempE4,SpeyBay[(Segment==E4S[i]),])
}
# But actually I only want a certain number of rows of data...
NewD2 - TempD2[1:SpringWatch,]
NewD3 - TempD3[1:SummerWatch,]
NewD4 - TempD4[1:AutumnWatch,]
NewE2 - TempE2[1:SpringWatch,]
NewE3 - TempE3[1:SummerWatch,]
NewE4 - TempE4[1:AutumnWatch,]
# then combine together (could do this in one step!
NewD - rbind(NewD2,NewD3,NewD4)
NewE - rbind(NewE2,NewE3,NewE4)
CompDE - rbind(NewD,NewE)
#Run a GLM-GEE on the resampled distributions to see if there is a
statistical difference between years
NewGLMGEE1 -
geeglm(CetPres~Year++SeaState,data=CompDE,family=binomial,id=Segment,corstr=ar1)
pv - summary(NewGLMGEE1)$coefficients[, Pr(|W|)] ## will extract
them
signif[j] - pv[2] # only interested in the significance of Year in the
model
}
--
View this message in context:
http://r.789695.n4.nabble.com/For-loop-processing-too-slow-pre-format-data-frame-tp4689543.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.
[R] help using extrafont package | R graphics
Greetings -- Submitted this a little while ago -- for some reason, still being held up by the moderator. Trying again... For a host of reasons, I need to use/embed Garamond font with various R graphics for a particular publication. I've figured out how to more or less get there from here, using the following sequence: library(extrafont) Then, I need to import the system fonts (Windoze box, R 3.1.0). So, I use font_import() But, this takes a *huge* amount of time, and often throws errors as it chokes on various fonts (I have probably 250+ fonts installed). I only want *one* font (Garamond). But, for the life of me, I can't figure out how to get font_import to select only the single font I want. In theory font_import(paths = NULL, recursive = TRUE, prompt = TRUE, pattern = NULL) as defaults, where pattern is a regex that font filenames must match. The file name for Garamong is gara.ttf, so I tried font_import(pattern=gara) R responds with 'Importing fonts make take a few minutes, depending on the...etc, etc'. Continue? [y/n] Hit 'y', and am presented with Scanning ttf files in C:\Windows\Fonts ... Extracting .afm files from .ttf files... Error in data.frame(fontfile = ttfiles, FontName = , stringsAsFactors = FALSE) : arguments imply differing number of rows: 0, 1 I have no idea what to do with this. Suggestions/pointers to the obvious welcome. And I thought futzing with fonts in LaTeX was fun! ;-) [[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] average and median values for each of the class
Dear all, I have a matrix (dimension, 16 x 12) where 2nd column represents class (1,1,1,1,1,2,2,2, etc) information. I want to estimate average and median values for each of the class and add this information as a row at end of the each classes. for example: dput(dat) structure(list(class = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L), name1 = c(2.554923977, 2.371586762, 2.497293431, 2.464827875, 2.981934845, 2.228995664, 2.099640729, 1.900314302, 2.630005966, 2.632590262, 2.581887814, 2.408797563, 2.098761103, 3.070460716, 1.436980716, 1.645121806), name2 = c(1.297412278, 1.104804244, 1.30621114, 1.126009533, 1.466740841, 1.012041118, 0.923466541, 0.840575023, 1.285530176, 1.041909333, 1.194917856, 1.085015826, 1.047492703, 1.587558217, 0.593340012, 0.723630088 ), name3 = c(0.587160798, 0.596127884, 0.623760721, 0.549016135, 0.686642084, 0.487523394, 0.458620467, 0.397974913, 0.615928976, 0.546005649, 0.657383069, 0.546613129, 0.476503461, 0.749062102, 0.304160587, 0.29037358), name4 = c(2.833441759, 2.713374426, 2.532626548, 2.409093102, 3.014912721, 2.113507947, 2.017291324, 1.667744912, 2.602560666, 2.31649643, 2.761204809, 2.433963493, 2.229911767, 3.191646399, 1.269919241, 1.387479858), name5 = c(2.172365295, 1.955695471, 2.141072829, 1.975743278, 2.377018372, 1.791300389, 1.669079382, 1.500209628, 2.164401874, 1.830038378, 2.106750025, 1.92888294, 1.707217549, 2.585082653, 1.114841754, 1.315712452 ), name6 = c(0.715129844, 0.688186262, 0.70133748, 0.709362008, 0.712145174, 0.563593885, 0.532109761, 0.472197304, 0.690165016, 0.65635473, 0.615835066, 0.64310098, 0.562974891, 0.900622255, 0.408546784, 0.416284408), name7 = c(1.995505133, 1.860095899, 1.843151597, 1.709861774, 2.155993511, 1.506409746, 1.315405587, 1.234544153, 1.96629927, 1.74879757, 1.93994009, 1.660173854, 1.556735295, 2.355723318, 0.866634243, 1.013367677), name8 = c(0.275484997, 0.233856392, 0.294021245, 0.315504347, 0.251906585, 0.250263636, 0.348599173, 0.273806933, 0.32067937, 0.278581115, 0.293726291, 0.308350808, 0.201297444, 0.351927886, 0.204230625, 0.185681471 ), name9 = c(2.461066627, 2.210756164, 2.289047888, 2.253988252, 2.668184733, 1.911697836, 1.793443775, 1.560027186, 2.36941155, 1.96191, 2.391501376, 2.002215107, 1.932144233, 2.73705052, 1.15580754, 1.807697999), name10 = c(0.723025351, 0.613147422, 0.805399925, 0.65651577, 0.779389048, 0.54260459, 0.492283542, 0.507969501, 0.749700016, 0.644231327, 0.810319215, 0.620331891, 0.600240557, 0.884775748, 0.40006142, 0.391661912), name11 = c(0.308565619, 0.453808281, 0.363716904, 0.376332596, 0.324998876, 0.361013073, 0.430744786, 0.468818055, 0.166072668, 0.369262627, 0.297666411, 0.256091173, 0.123021464, 0.308188684, 0.646436241, 0.722972632 )), .Names = c(class, name1, name2, name3, name4, name5, name6, name7, name8, name9, name10, name11), class = data.frame, row.names = c(ara1, ara2, ara3, ara4, ara5, ara6, ara7, ara8, ara9, ara10, ara11, ara12, ara13, ara14, ara15, ara16 )) I wrote this: avg-as.data.frame(aggregate(dat[,2:dim(dat)[2]], dat[class], function(x) mean(x,na.rm=T)) ) med-as.data.frame(aggregate(dat[,2:dim(dat)[2]], dat[class], function(x) median(x,na.rm=T)) ) # avg # classname1 name2 name3name4name5 name6name7 name#8name9name10name11 #1 1 2.574113 1.2602356 0.6085415 2.700690 2.124379 0.7052322 1.912922 #0.2741547 2.376609 0.7154955 0.3654845 #2 2 2.214739 1.0154032 0.4900119 2.100276 1.781248 0.5645165 1.505665 #0.2983373 1.908645 0.5731394 0.3566621 #3 3 2.541092 1.1072810 0.589 2.503888 1.955224 0.6384303 1.782971 #0.2935527 2.118543 0.6916275 0.3076734 #4 4 2.202068 1.0761303 0.5099087 2.230492 1.802381 0.6240480 1.593031 #0.2524853 1.941667 0.6283592 0.3592155 #5 5 1.645122 0.7236301 0.2903736 1.387480 1.315712 0.4162844 1.013368 #0.1856815 1.807698 0.3916619 0.7229726 # med # classname1 name2 name3name4name5 name6name7 name#8name9name10name11 #1 1 2.497293 1.2974123 0.5961279 2.713374 2.141073 0.7093620 1.860096 #0.2754850 2.289048 0.7230254 0.3637169 #2 2 2.164318 0.9677538 0.4730719 2.065400 1.730190 0.5478518 1.410908 #0.2972432 1.852571 0.5252870 0.3958789 #3 3 2.581888 1.0850158 0.5466131 2.433963 1.928883 0.6431010 1.748798 #0.2937263 2.002215 0.6442313 0.2976664 #4 4 2.098761 1.0474927 0.4765035 2.229912 1.707218 0.5629749 1.556735 #0.2042306 1.932144 0.6002406 0.3081887 #5 5 1.645122 0.7236301 0.2903736 1.387480 1.315712 0.4162844 1.013368 #0.1856815 1.807698 0.3916619 0.7229726 But I do not know how can I add this information in the original data? For example, for class 1, the output will look like this: dput(res1) structure(list(class = c(1L, 1L, 1L, 1L, 1L, 1L, 1L), name1 = c(2.554923977, 2.371586762, 2.497293431, 2.464827875, 2.981934845, 2.574113378, 2.497293431), name2 = c(1.297412278,
Re: [R] lattice plot formatting: pch, abbreviation and labels
Hi Luigi You are typing things unnecessarily: do not use the attach command unless absolutely necessary - it has unfortunate consequences. It is better to use with or within. alpha is not available with some devices with bad consequences if used; its default is 1 anyway. Once you have stated a data.frame as the data object it is usually not necessary to use the data.frame$ sign in to signify column names: use the column names. data = a data.frame is the equivalent to with You are reordering the levels of stimulation and changing the name of 1 so I thought it was easiest to make a column stim and do things there otherwise it the relevelling could be done in the data argument using the subset argument. Then the change to Unst could be done by using strip= strip.custom(factor.levels = ...), To change for the -/+ I made a group - the easiest way. Have a look a changing the negative symbol to 20 or something else as it is hard to visualise. You may vary things with changing cex (remember to have 2 values 1 for each group). If you do str(xyplot object) you will get a big print of the object and within that the x limits are shown as 0, 1 which means that the x values are 1 and 2 There is a command to get these things but I have forgotten it my.data$stimulation - factor(levels = c(Unstimulated, ESAT6,CFP10,Rv3615c,Rv2654, Rv3879, Rv3873,PHA)) my.data$stim - factor(my.data$stimulation, labels = c(Unst., ESAT6,CFP10,Rv3615c,Rv2654, Rv3879, Rv3873,PHA)) my.data$pos - ifelse(sign(my.data$copy) 0, 2,1) stripplot(copy ~ factor(positivity)|stim,my.data, groups = pos, hor = F, layout = c(8,1), scales = list(x = list(at = c(1,2), labels = c(N,P))), jitter.x = TRUE, amount = 2, pch = c(16,1), col = black, ylab = expression(bold(Copy)), xlab = expression(bold(Stimulation)), main=Plot, par.settings = list(strip.background=list(col=white)), par.strip.text=list(font=2) ) Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Luigi Marongiu Sent: Sunday, 27 April 2014 02:07 To: r-help@r-project.org Subject: [R] lattice plot formatting: pch, abbreviation and labels Dear all, I am trying to use the lattice plot, but the syntax is quite difficult. Specifically I have eight variables (1 to 8) each of them further subdivided in two classes (negative=0 and positive=1). I am using the stripplot() to represent these values. I would like to represent the negative and positive values with black and white dots so I have tried to use the argument pch=c(16, 1) both in the main stripplot function and embedded in the scale argument. However the resulting plot shows that some points are drawn with the pch=16 and other with pch=1 irrespective of their class. Is there a way to draw the values for the variable positivity = 0 with pch=16 and those with positivity = 1 with pch=0? In addition I would like to change the labels under the axis from 0,1 to N,P. However when I placed labels=c(N, P) or labels=list(N, P) in the main stripplot() I did not obtained any difference and when placed within the scale argument also the -labels were modified. Is there a way to change the label 0 with N and the label 1 with P? final problem: I would like to abbreviate the Unstimulated box label with Unst.. I have tried the abbreviate=TRUE argument but again the syntax is too complex for me and it did not work. Is there a way to abbreviate the variables names? Thank you very much for your help. Best wishes, Luigi CODE: ### open plot library library(lattice) my.data-structure(list( column_1 = 1:120, column_2 = structure(c( 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8, 1,2,3,4,5,6,7,8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = structure(c( 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 0,0,0,0,0,0,0,0)), column_4 = c( 192.0519108,183.6403531,53.46798757,83.60638077,69.60749873,159.4706861,256. 8765622,499.2899303, 2170.799076,1411.349719,2759.472348,2098.973397,2164.739515,1288.676574,1611 .486543,6205.229575, 870.7424981,465.9967135,191.8962375,864.0937485,2962.693675,1289.259137,2418 .651212,7345.712517,
Re: [R] average and median values for each of the class
Hi,
Your dput() suggests dat as data.frame.
##Using the results you got,
res2 - do.call(rbind,lapply(unique(dat$class),function(i) {x1
-rbind(dat[dat$class==i,], avg[avg$class==i,], med[med$class==i,]);
rownames(x1)[!grepl(ara,rownames(x1))] - paste0(c(Avg, Med), i); x1}))
A.K.
On Saturday, April 26, 2014 8:39 PM, Nico Met nicome...@gmail.com wrote:
Dear all,
I have a matrix (dimension, 16 x 12) where 2nd column represents class
(1,1,1,1,1,2,2,2, etc) information. I want to estimate average and median
values for each of the class and add this information as a row at end of
the each classes.
for example:
dput(dat)
structure(list(class = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
3L, 3L, 3L, 4L, 4L, 4L, 5L), name1 = c(2.554923977, 2.371586762,
2.497293431, 2.464827875, 2.981934845, 2.228995664, 2.099640729,
1.900314302, 2.630005966, 2.632590262, 2.581887814, 2.408797563,
2.098761103, 3.070460716, 1.436980716, 1.645121806), name2 = c(1.297412278,
1.104804244, 1.30621114, 1.126009533, 1.466740841, 1.012041118,
0.923466541, 0.840575023, 1.285530176, 1.041909333, 1.194917856,
1.085015826, 1.047492703, 1.587558217, 0.593340012, 0.723630088
), name3 = c(0.587160798, 0.596127884, 0.623760721, 0.549016135,
0.686642084, 0.487523394, 0.458620467, 0.397974913, 0.615928976,
0.546005649, 0.657383069, 0.546613129, 0.476503461, 0.749062102,
0.304160587, 0.29037358), name4 = c(2.833441759, 2.713374426,
2.532626548, 2.409093102, 3.014912721, 2.113507947, 2.017291324,
1.667744912, 2.602560666, 2.31649643, 2.761204809, 2.433963493,
2.229911767, 3.191646399, 1.269919241, 1.387479858), name5 = c(2.172365295,
1.955695471, 2.141072829, 1.975743278, 2.377018372, 1.791300389,
1.669079382, 1.500209628, 2.164401874, 1.830038378, 2.106750025,
1.92888294, 1.707217549, 2.585082653, 1.114841754, 1.315712452
), name6 = c(0.715129844, 0.688186262, 0.70133748, 0.709362008,
0.712145174, 0.563593885, 0.532109761, 0.472197304, 0.690165016,
0.65635473, 0.615835066, 0.64310098, 0.562974891, 0.900622255,
0.408546784, 0.416284408), name7 = c(1.995505133, 1.860095899,
1.843151597, 1.709861774, 2.155993511, 1.506409746, 1.315405587,
1.234544153, 1.96629927, 1.74879757, 1.93994009, 1.660173854,
1.556735295, 2.355723318, 0.866634243, 1.013367677), name8 = c(0.275484997,
0.233856392, 0.294021245, 0.315504347, 0.251906585, 0.250263636,
0.348599173, 0.273806933, 0.32067937, 0.278581115, 0.293726291,
0.308350808, 0.201297444, 0.351927886, 0.204230625, 0.185681471
), name9 = c(2.461066627, 2.210756164, 2.289047888, 2.253988252,
2.668184733, 1.911697836, 1.793443775, 1.560027186, 2.36941155,
1.96191, 2.391501376, 2.002215107, 1.932144233, 2.73705052,
1.15580754, 1.807697999), name10 = c(0.723025351, 0.613147422,
0.805399925, 0.65651577, 0.779389048, 0.54260459, 0.492283542,
0.507969501, 0.749700016, 0.644231327, 0.810319215, 0.620331891,
0.600240557, 0.884775748, 0.40006142, 0.391661912), name11 = c(0.308565619,
0.453808281, 0.363716904, 0.376332596, 0.324998876, 0.361013073,
0.430744786, 0.468818055, 0.166072668, 0.369262627, 0.297666411,
0.256091173, 0.123021464, 0.308188684, 0.646436241, 0.722972632
)), .Names = c(class, name1, name2, name3, name4, name5,
name6, name7, name8, name9, name10, name11), class = data.frame,
row.names = c(ara1,
ara2, ara3, ara4, ara5, ara6, ara7, ara8, ara9,
ara10, ara11, ara12, ara13, ara14, ara15, ara16
))
I wrote this:
avg-as.data.frame(aggregate(dat[,2:dim(dat)[2]], dat[class],
function(x) mean(x,na.rm=T)) )
med-as.data.frame(aggregate(dat[,2:dim(dat)[2]], dat[class], function(x)
median(x,na.rm=T)) )
# avg
# class name1 name2 name3 name4 name5 name6 name7
name#8 name9 name10 name11
#1 1 2.574113 1.2602356 0.6085415 2.700690 2.124379 0.7052322 1.912922
#0.2741547 2.376609 0.7154955 0.3654845
#2 2 2.214739 1.0154032 0.4900119 2.100276 1.781248 0.5645165 1.505665
#0.2983373 1.908645 0.5731394 0.3566621
#3 3 2.541092 1.1072810 0.589 2.503888 1.955224 0.6384303 1.782971
#0.2935527 2.118543 0.6916275 0.3076734
#4 4 2.202068 1.0761303 0.5099087 2.230492 1.802381 0.6240480 1.593031
#0.2524853 1.941667 0.6283592 0.3592155
#5 5 1.645122 0.7236301 0.2903736 1.387480 1.315712 0.4162844 1.013368
#0.1856815 1.807698 0.3916619 0.7229726
# med
# class name1 name2 name3 name4 name5 name6 name7
name#8 name9 name10 name11
#1 1 2.497293 1.2974123 0.5961279 2.713374 2.141073 0.7093620 1.860096
#0.2754850 2.289048 0.7230254 0.3637169
#2 2 2.164318 0.9677538 0.4730719 2.065400 1.730190 0.5478518 1.410908
#0.2972432 1.852571 0.5252870 0.3958789
#3 3 2.581888 1.0850158 0.5466131 2.433963 1.928883 0.6431010 1.748798
#0.2937263 2.002215 0.6442313 0.2976664
#4 4 2.098761 1.0474927 0.4765035 2.229912 1.707218 0.5629749 1.556735
#0.2042306 1.932144 0.6002406 0.3081887
#5 5 1.645122 0.7236301 0.2903736 1.387480 1.315712 0.4162844 1.013368
#0.1856815
Re: [R] average and median values for each of the class
On Apr 26, 2014, at 5:37 PM, Nico Met wrote: Dear all, I have a matrix (dimension, 16 x 12) where 2nd column represents class (1,1,1,1,1,2,2,2, etc) information. I want to estimate average and median values for each of the class and add this information as a row at end of the each classes. Well it does have a dimension attribute but it is a data.frame, NOT a matrix. The term class is a reserved word in R. What is it that you mean by that word? if it is for each column then: sapply( dat, function(x) c( mean(x), median(x)) ) sapply( dat, function(x) c( mean_x = mean(x), median_x = median(x)) ) classname1name2 name3name4name5 name6 mean_x 2.4375 2.350258 1.102291 0.5358036 2.343448 1.895963 0.6242466 median_x 2. 2.436813 1.094910 0.5478146 2.421528 1.942289 0.6497279 name7 name8name9name10name11 mean_x 1.67054 0.2742449 2.094122 0.6388536 0.3736069 median_x 1.72933 0.2770331 2.106486 0.6322816 0.3623650 -- David. for example: dput(dat) structure(list(class = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L), name1 = c(2.554923977, 2.371586762, 2.497293431, 2.464827875, 2.981934845, 2.228995664, 2.099640729, 1.900314302, 2.630005966, 2.632590262, 2.581887814, 2.408797563, 2.098761103, 3.070460716, 1.436980716, 1.645121806), name2 = c(1.297412278, 1.104804244, 1.30621114, 1.126009533, 1.466740841, 1.012041118, 0.923466541, 0.840575023, 1.285530176, 1.041909333, 1.194917856, 1.085015826, 1.047492703, 1.587558217, 0.593340012, 0.723630088 ), name3 = c(0.587160798, 0.596127884, 0.623760721, 0.549016135, 0.686642084, 0.487523394, 0.458620467, 0.397974913, 0.615928976, 0.546005649, 0.657383069, 0.546613129, 0.476503461, 0.749062102, 0.304160587, 0.29037358), name4 = c(2.833441759, 2.713374426, 2.532626548, 2.409093102, 3.014912721, 2.113507947, 2.017291324, 1.667744912, 2.602560666, 2.31649643, 2.761204809, 2.433963493, 2.229911767, 3.191646399, 1.269919241, 1.387479858), name5 = c(2.172365295, 1.955695471, 2.141072829, 1.975743278, 2.377018372, 1.791300389, 1.669079382, 1.500209628, 2.164401874, 1.830038378, 2.106750025, 1.92888294, 1.707217549, 2.585082653, 1.114841754, 1.315712452 ), name6 = c(0.715129844, 0.688186262, 0.70133748, 0.709362008, 0.712145174, 0.563593885, 0.532109761, 0.472197304, 0.690165016, 0.65635473, 0.615835066, 0.64310098, 0.562974891, 0.900622255, 0.408546784, 0.416284408), name7 = c(1.995505133, 1.860095899, 1.843151597, 1.709861774, 2.155993511, 1.506409746, 1.315405587, 1.234544153, 1.96629927, 1.74879757, 1.93994009, 1.660173854, 1.556735295, 2.355723318, 0.866634243, 1.013367677), name8 = c(0.275484997, 0.233856392, 0.294021245, 0.315504347, 0.251906585, 0.250263636, 0.348599173, 0.273806933, 0.32067937, 0.278581115, 0.293726291, 0.308350808, 0.201297444, 0.351927886, 0.204230625, 0.185681471 ), name9 = c(2.461066627, 2.210756164, 2.289047888, 2.253988252, 2.668184733, 1.911697836, 1.793443775, 1.560027186, 2.36941155, 1.96191, 2.391501376, 2.002215107, 1.932144233, 2.73705052, 1.15580754, 1.807697999), name10 = c(0.723025351, 0.613147422, 0.805399925, 0.65651577, 0.779389048, 0.54260459, 0.492283542, 0.507969501, 0.749700016, 0.644231327, 0.810319215, 0.620331891, 0.600240557, 0.884775748, 0.40006142, 0.391661912), name11 = c(0.308565619, 0.453808281, 0.363716904, 0.376332596, 0.324998876, 0.361013073, 0.430744786, 0.468818055, 0.166072668, 0.369262627, 0.297666411, 0.256091173, 0.123021464, 0.308188684, 0.646436241, 0.722972632 )), .Names = c(class, name1, name2, name3, name4, name5, name6, name7, name8, name9, name10, name11), class = data.frame, row.names = c(ara1, ara2, ara3, ara4, ara5, ara6, ara7, ara8, ara9, ara10, ara11, ara12, ara13, ara14, ara15, ara16 )) I wrote this: avg-as.data.frame(aggregate(dat[,2:dim(dat)[2]], dat[class], function(x) mean(x,na.rm=T)) ) med-as.data.frame(aggregate(dat[,2:dim(dat)[2]], dat[class], function(x) median(x,na.rm=T)) ) # avg # classname1 name2 name3name4name5 name6name7 name#8name9name10name11 #1 1 2.574113 1.2602356 0.6085415 2.700690 2.124379 0.7052322 1.912922 #0.2741547 2.376609 0.7154955 0.3654845 #2 2 2.214739 1.0154032 0.4900119 2.100276 1.781248 0.5645165 1.505665 #0.2983373 1.908645 0.5731394 0.3566621 #3 3 2.541092 1.1072810 0.589 2.503888 1.955224 0.6384303 1.782971 #0.2935527 2.118543 0.6916275 0.3076734 #4 4 2.202068 1.0761303 0.5099087 2.230492 1.802381 0.6240480 1.593031 #0.2524853 1.941667 0.6283592 0.3592155 #5 5 1.645122 0.7236301 0.2903736 1.387480 1.315712 0.4162844 1.013368 #0.1856815 1.807698 0.3916619 0.7229726 # med # classname1 name2 name3name4name5 name6name7 name#8name9name10name11 #1 1 2.497293
