Hi all,
I use the excellent packages rjags and cacheSweave, and unfortunately
seem to have found an incompatibility between them. Below are a
minimal .Rnw file and corresponding JAGS model file which illustrate
the problem:
*** JAGS model file; name=j.bug ***
model {
mu ~ dnorm(0,1.0E-
Hi All,
>
> Attempt #2. any help is much appreciated!:
> I am reading through section 2.6 (Mathematics) of the "Writing R
> Extensions" manuscript and am wondering where I can find more
> examples/documentation on the \deqn{ } function. I would like to learn how
> to display equations using th
Dear friends:
I’m very interested in the analysis of survival data of leukemic patients,
which involves competing risks and time-dependent covariates (infections or
relapse of leukemia after BMT). In my present work, I’d like to estimate the
risk of chemotherapy, BMT or gvhd on the onset and
Hi,
I notice several typos when reading documentation and I verify that they are
still typos in the current build. Where do I send corrections? Note that
most of them are minor typos.
Thanks
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Dear friends:
I’m very interested in the analysis of survival data of leukemic patients,
which involves competing risks and time-dependent covariates (infections or
relapse of leukemia after BMT). In my present work, I’d like to estimate the
risk of chemotherapy, BMT or gvhd on the onset and
On 03/26/2010 02:58 PM, Steve Powell wrote:
For psychologists like me (possibly for others) by far the most
time-consuming detail is variable labels. I need them for just about
every analysis I do. We can use special packages like Hmisc and its
function spss.get to import the labels, but then nea
Dear expeRts
Is it not the case that very many multi-level datasets have associated
sample weights? But then I don't see a way to include them in the
analyses? Or, how can I make nlme talk to the "survey" package?
Best Wishes
Steve
__
R-help@r-project.o
For psychologists like me (possibly for others) by far the most
time-consuming detail is variable labels. I need them for just about
every analysis I do. We can use special packages like Hmisc and its
function spss.get to import the labels, but then nearly all the other
packages don't respect the l
Gabor Grothendieck wrote:
On Thu, Mar 25, 2010 at 10:43 PM, Frank E Harrell Jr
wrote:
(Ted Harding) wrote:
http://www.nomogram.org
They call this sort of interface "nomogram" too, and you can have
a go at the one they offer for bladder cancer. This is not a
graphical nomogram in the sense th
On Mar 25, 2010, at 9:21 PM, tj wrote:
Anyone who can help me with this?
I have 48 observations (I dont want to alter their order). I want to
group
these observations into 16 blocks. So I should have 3 observations
for each
block. This is what I did in R, but it has warnings.
y
Keith McMillan wrote:
>
> Dear R users,
>
>
>
> Is documentation for the netlabR package available in English?
>
>
>
> If not does anyone know if or when it will be?
>
>
>
> Regards,
>
>
>
> Keith
>
I don't see a package named netlab or netlabR on CRAN, which package are you
ta
On Thu, Mar 25, 2010 at 10:28 PM, Joaquin Rapela wrote:
> Your reply was very useful Rich. Thanks!
>
> I would like to set the height of the viewport in draw.colorkey to match
> the
> height of the levelplots, as it is done when I add colorkeys to all the
> levelplots. The height of the levelplot
tj wrote:
>
> Anyone who can help me with this?
> I have 48 observations (I dont want to alter their order). I want to group
> these observations into 16 blocks. So I should have 3 observations for
> each block. This is what I did in R, but it has warnings.
>
>> y #contains my
On Thu, Mar 25, 2010 at 10:43 PM, Frank E Harrell Jr
wrote:
> (Ted Harding) wrote:
>>
>> http://www.nomogram.org
>>
>> They call this sort of interface "nomogram" too, and you can have
>> a go at the one they offer for bladder cancer. This is not a
>> graphical nomogram in the sense that you desc
> 1) I get that the kernel is a "normal function". But my understanding
> was that the kernel created a higher dimensional space than the original
> data, thus allowing the SVM to be a "pseudo-linear" classifier in that
> higher dimension.
I've been fond of this video that shows "the point":
http
(Ted Harding) wrote:
On 25-Mar-10 17:44:55, R Heberto Ghezzo, Dr wrote:
On the same topic but from a different perspective. A Nomogram or
better a Line Aligned Nomogram is a graph with 2 or more scales,
maybe linear where by alignig values in each scale you can read
values in the other scale. Th
What is the definition of the whiskers in the ggplot2 qplot with
geom="boxplot"?
Why is it so hard to find?
Thanks,
Jim Rome
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-
Your reply was very useful Rich. Thanks!
I would like to set the height of the viewport in draw.colorkey to match the
height of the levelplots, as it is done when I add colorkeys to all the
levelplots. The height of the levelplots varies depending on the number of
levelplots that I put on the figu
Hi
Greg Snow wrote:
The subplot function in the TeachingDemos package will create a new
plot inside of the existing one, so if you can plot your one image,
the subplot function will help (using base graphics). The EBImage
package may help with the reading and plotting of the figure. (there
is
On 03/26/2010 11:36 AM, Geddes, Scott wrote:
Hi,
I wish to NEATLY store, and later display one table per file (similar to
capabilities of write.table()).
BUT BY NEATLY I MEAN:
1. Column Headings right aligned with data values.
2. Decimal points line-up per column.
3. Data values padded trailin
Anyone who can help me with this?
I have 48 observations (I dont want to alter their order). I want to group
these observations into 16 blocks. So I should have 3 observations for each
block. This is what I did in R, but it has warnings.
> y #contains my 48 observations
[1] 2.4
On Mar 25, 2010, at 8:01 PM, Rolf Turner wrote:
>
> On 26/03/2010, at 1:36 PM, Geddes, Scott wrote:
>
>>
>> Hi,
>>
>> I wish to NEATLY store, and later display one table per file (similar to
>> capabilities of write.table()).
>>
>> BUT BY NEATLY I MEAN:
>> 1. Column Headings right aligned wit
Márcio,
Think matrix!
Do you really want b to be 100 copies of the same numbers?
You are asking for a strange crossproduct with the main diagonal zeroed out.
a2 <- crossprod(a)
a2[cbind(1:1000, 1:1000)] <- 0
all.equal(a, a2)
Rich
[[alternative HTML version deleted]]
Tena koe Marcio
Seems like you are simply multiplying transpose b by b and replacing the
diagonal with 0. If this is correct, then use
a <- t(b) %*% b
diag(a) <- 0
If this is not a correct interpretation of what you are trying to do, could you
show us with a small reproducible example.
HTH .
Dear lattice::levelplot users,
I would like to use levelplot to plot uniformly sized blocks.
Here is an example:
require (lattice)
u = runif(10) * 10
v = runif(10) * 10
z = runif(10)
levelplot ( z ~ u + v, aspect="iso" )
The resultant figures have points (blocks) that are of variable size that
if you do:
fit<-survfit (Surv(DTDMRS3, DMRS3) ~ RS2540477)
fit$surv
will have the survival function, and
fit$time
will have the failure times, these should give you what you want
Hope this helps
Corey
-
Corey Sparks, PhD
Assistant Professor
Department of Demography and Organization Studies
Hi guys, I am still learning R, and not well familiar with all the apply
functions.
I am trying to find faster alternatives to replace the for cycle.
Is there a faster way to do the example below?
nm <- 1000
b <- matrix (rnorm (5000, 0, 1), nrow = 500, ncol = nm)
a <- matrix (0, nm, nm)
for (i in
Dear R users,
Is documentation for the netlabR package available in English?
If not does anyone know if or when it will be?
Regards,
Keith
This
e-mail and any files transmitted with it are confidential and
On 26/03/2010, at 1:36 PM, Geddes, Scott wrote:
>
> Hi,
>
> I wish to NEATLY store, and later display one table per file (similar to
> capabilities of write.table()).
>
> BUT BY NEATLY I MEAN:
> 1. Column Headings right aligned with data values.
> 2. Decimal points line-up per column.
> 3. Dat
Hi,
I wish to NEATLY store, and later display one table per file (similar to
capabilities of write.table()).
BUT BY NEATLY I MEAN:
1. Column Headings right aligned with data values.
2. Decimal points line-up per column.
3. Data values padded trailing zeros per column (if not integers).
4. "Trick
On 26/03/2010, at 12:38 PM, jim holtman wrote:
> WHen using '==' or '%in%' it is a equality test -- it has to equal zero. If
> you want a tolerance in the test, use 'all.equal'
>
> On Thu, Mar 25, 2010 at 5:29 PM, Dimitri Liakhovitski wrote:
>
>> Hello!
>>
>> I am wondering at what point does
Euphoria wrote:
Hi all!
I have created survival vs. time plots. Now I would like to plot (1 -
Survival) vs. time.
Is there a way for me to retrieve the survival estimate information, to
which I can manually make an adjustment (ie; failure = 1 - survival) before
I re-plot this information?
Here
Hi all!
I have created survival vs. time plots. Now I would like to plot (1 -
Survival) vs. time.
Is there a way for me to retrieve the survival estimate information, to
which I can manually make an adjustment (ie; failure = 1 - survival) before
I re-plot this information?
Here is the code I use
Thanks to those who responded to my post about reading SQL server tables using
RODBC. Someone in our IT department sent me a link that contained the following
information:
SQL ODBC connection strings
Standard Security:< br> "Driver={SQLServer};Server=Your_Server_Name;Database=
Your_Database_Na
Greetings!
I have one quick question: How do you do a Tukey test on an ANCOVA?
Thanks for any tips!
-Paul
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/po
WHen using '==' or '%in%' it is a equality test -- it has to equal zero. If
you want a tolerance in the test, use 'all.equal'
On Thu, Mar 25, 2010 at 5:29 PM, Dimitri Liakhovitski wrote:
> Hello!
>
> I am wondering at what point does R consider a numeric value to be
> equal to zero - for stateme
On 26/03/2010, at 12:23 PM, Achim Zeileis wrote:
> Sorry for coming so late to this thread. One possible explanation for the
> R side is the following...
>
>
> On Thu, 25 Mar 2010, Marc Schwartz wrote:
>
>> On Mar 25, 2010, at 5:41 PM, Joshua Wiley wrote:
>>
Kind of off the thread a bit
Sorry for coming so late to this thread. One possible explanation for the
R side is the following...
On Thu, 25 Mar 2010, Marc Schwartz wrote:
On Mar 25, 2010, at 5:41 PM, Joshua Wiley wrote:
Kind of off the thread a bit, but when I do:
as.Date(40182)
I ***do not*** get "2080-01-06". I
On 03/26/2010 08:32 AM, Babaorumi wrote:
Hi,
I have a data set of points which are represented by 3 variables x,y,z where
x is the position of the point on the x-absciss and y on the y-absciss. Each
of my points has a value z, which I want to be displayed as follows: the
more z is high, the mor
On Mar 25, 2010, at 5:41 PM, Joshua Wiley wrote:
>> Kind of off the thread a bit, but when I do:
>>
>>> as.Date(40182)
>>
>> I ***do not*** get "2080-01-06". Instead I get an error:
>>
>> Error in as.Date.numeric(40182) : 'origin' must be supplied
>>
>> Am I the only user who gets picked on i
Thanks Steve,
1) I get that the kernel is a "normal function". But my understanding
was that the kernel created a higher dimensional space than the original
data, thus allowing the SVM to be a "pseudo-linear" classifier in that
higher dimension. So, if the the kernel is the dot_product do I iter
Tena korua
Also, bear in mind that colours can be specified as grey0 - grey100;
i.e., paste('grey', round(100*z/max(z), 0), sep=''). Also, the
colorspace package is worth considering if you are not restricted to
black and white.
HTH
Peter Alspach
> -Original Message-
> From: r-he
On 26/03/2010, at 11:21 AM, Nordlund, Dan (DSHS/RDA) wrote:
> Rolf,
>
> I tried the same thing at first, and got the same error. So I suspect Anna
> didn't really use that code either. :-)
Thanks! That helps to sooth my paranoia. :-)
cheers,
Rolf
##
On 25-Mar-10 22:17:23, Rolf Turner wrote:
>
> I think that
>
> riffle <- function (a,b) {
> n <- min(length(a),length(b))
> p1 <- as.vector(rbind(a[1:n],b[1:n]))
> p2 <- c(a[-(1:n)],b[-(1:n)])
> c(p1,p2)
> }
>
> does the trick, and is pretty simple
> cheers,
> Rolf Turne
Hi,
On Thu, Mar 25, 2010 at 5:32 PM, Babaorumi wrote:
>
> Hi,
>
> I have a data set of points which are represented by 3 variables x,y,z where
> x is the position of the point on the x-absciss and y on the y-absciss. Each
> of my points has a value z, which I want to be displayed as follows: the
> Kind of off the thread a bit, but when I do:
>
>> as.Date(40182)
>
> I ***do not*** get "2080-01-06". Instead I get an error:
>
> Error in as.Date.numeric(40182) : 'origin' must be supplied
>
> Am I the only user who gets picked on in this way, or does it
> happen to others as well? The help on
On Mar 25, 2010, at 4:58 PM, Douglas Bates wrote:
> Thanks very much, Marc.
>
> So did you really read 110 pages of the User Guide to find this out or
> did you do something more clever?
You are welcome Doug.
Nah...not so much clever as persistent.
I actually started by Googling for the file
Rolf,
I tried the same thing at first, and got the same error. So I suspect Anna
didn't really use that code either. :-)
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA
> 1) how calculation of the kernel happens.
The kernel is just a "normal function" (though not every function is a
proper kernel function): it takes two values (each value being a
vector (or something) representing an example) and returns a real
valued answer.
> 2) how to calculate the predicted
I think that
riffle <- function (a,b) {
n <- min(length(a),length(b))
p1 <- as.vector(rbind(a[1:n],b[1:n]))
p2 <- c(a[-(1:n)],b[-(1:n)])
c(p1,p2)
}
does the trick, and is pretty simple
cheers,
Rolf Turner
###
Try this also:
riffle3 <- function(a, b) {
mNrow <- nrow(cbind(a, b))
m <- as.data.frame(cbind(a[1:mNrow], b[1:mNrow]))
as.numeric(na.exclude(unlist(lapply(split(m, 1:mNrow), as.numeric
}
On Thu, Mar 25, 2010 at 6:10 PM, Jeff Brown wrote:
>
> I just had to solve this
## this is much easier to read
riffle3 <- function(a, b) {
mlab <- min(length(a), length(b))
seqmlab <- seq(length=mlab)
c(rbind(a[seqmlab], b[seqmlab]), a[-seqmlab], b[-seqmlab])
}
riffle3((1:10),(50:55))
## [1] 1 50 2 51 3 52 4 53 5 54 6 55 7 8 9 10
riffle3((50:55),(1:10))
##
Kind of off the thread a bit, but when I do:
> as.Date(40182)
I ***do not*** get "2080-01-06". Instead I get an error:
Error in as.Date.numeric(40182) : 'origin' must be supplied
Am I the only user who gets picked on in this way, or does it
happen to others as well? The help on as.Date() cl
On 25-Mar-10 21:10:17, Jeff Brown wrote:
> I just had to solve this problem for myself, after not having
> luck with the code posted above. I'm posting in case others
> need a completely general function.
>
> riffle <- function (a,b) {
> # Interleave a & b, starting with a, without repeating
Thanks very much, Marc.
So did you really read 110 pages of the User Guide to find this out or
did you do something more clever?
On Thu, Mar 25, 2010 at 4:38 PM, Marc Schwartz wrote:
> On Mar 25, 2010, at 3:54 PM, Douglas Bates wrote:
>
>> The TIMSS2007 database http://timss.bc.edu/TIMSS2007/idb
On Mar 25, 2010, at 3:54 PM, Douglas Bates wrote:
> The TIMSS2007 database http://timss.bc.edu/TIMSS2007/idb_ug.html seems
> to provide "both kinds" of universal data formats - either SPSS saved
> data sets or SAS saved data sets. (Yes, I am being sarcastic.)
> These, of course, are accompanied b
Hi,
I have a data set of points which are represented by 3 variables x,y,z where
x is the position of the point on the x-absciss and y on the y-absciss. Each
of my points has a value z, which I want to be displayed as follows: the
more z is high, the more the color on the map is dark.
How can I
Hello!
I am wondering at what point does R consider a numeric value to be
equal to zero - for statements of the type x==0 and x %in% 0.
Thank you very much!
--
Dimitri Liakhovitski
Ninah.com
dimitri.liakhovit...@ninah.com
__
R-help@r-project.org mail
Hello,
Hosack, Michael wrote:
Hello,
I need to create a dataframe containing all possible combinations of
three variables: SITE (101,102,103,104), WDAY (MON,TUE,WED,THR,FRI),
and TOD (MORN, AFTN). There should be a total of 40 unique
combinations in my dataframe. I used expand.grid() successful
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Joshua Wiley
> Sent: Thursday, March 25, 2010 1:48 PM
> To: anna
> Cc: r-help@r-project.org
> Subject: Re: [R] Convert number to Date
>
> Dear Anna,
>
> Rolf's explanation not wi
Hi
Could somebody help me how to get degree index in network, where I have
weighted edges?
Thanks
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___
I just had to solve this problem for myself, after not having luck with the
code posted above. I'm posting in case others need a completely general
function.
riffle <- function (a,b) {
# Interleave a & b, starting with a, without repeating.
x <- NULL; count = 1;
You are quite welcome. After poking around a bit more, I can offer a
more detailed explanation on Excel. It does treat 1 January 1900 as
the origin. However, while R treats origin as 0, Excel treats it as
1. This explains 1 of the two day change needed for R to get the same
results as Excel. T
Hi all,
I have simple x/y data from screen recording in a sequence:
number,x,y
1,10,30
1,20,
1,43,110
1,74,18
1,88,112
and would like to create a 3d histogram data structure that i can use
to create a 3d histogram or, more likely a heatmap. The unterlying
data structure therefor
Hi,
I'm learning more about SVMs and kernels in general. I've gotten used
to using the svm function in the e1071 package. It works great.
Now, I want to do/learn some more interesting stuff. (Perhaps my own
kernel and/or scoring system). So I want to better understand
1) how calculation of the
The subplot function in the TeachingDemos package will create a new plot inside
of the existing one, so if you can plot your one image, the subplot function
will help (using base graphics). The EBImage package may help with the reading
and plotting of the figure. (there is also grImport, but i
Hi Josh! yes definitely it makes sense as I got to retrieve a date, but a
different one! thanks a lot for the explanation )
-
Anna Lippel
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The TIMSS2007 database http://timss.bc.edu/TIMSS2007/idb_ug.html seems
to provide "both kinds" of universal data formats - either SPSS saved
data sets or SAS saved data sets. (Yes, I am being sarcastic.)
These, of course, are accompanied by massive codebooks explaining the
nature of each of the fi
Nicola,
I found its easier if you convert your SAS data to the XPORT format (a SAS
transport format). Here's instructions for creating the transport file in SAS
- use xport not cport.
http://support.sas.com/documentation/cdl/en/movefile/59598/HTML/default/a002575816.htm
I also found the impo
Dear Anna,
Rolf's explanation not withstanding, it has to do with differences in
how R and Excel treat dates. If you use
as.Date(40182, origin="1899-12-30")
you will get the same date as Excel. You can look at:
http://office.microsoft.com/training/training.aspx?AssetID=RC102786151033&CTT=6&Or
Joaquin,
p <-levelplot(frame, colorkey=TRUE)
print(update(p, legend=NULL), position=c(0/4,0,1/4,1), more=TRUE)
print(update(p, legend=NULL), position=c(1/4,0,2/4,1), more=TRUE)
print(update(p, legend=NULL), position=c(2/4,0,3/4,1), more=TRUE)
draw.colorkey(p$legend$right$args$key, draw=TRUE,
Hi Mike,
the following works for me:
SITE <- ordered(c(101,102,103,104))
WDAY <-
ordered(c("MON","TUE","WED","THR","FRI"),levels=c("MON","TUE","WED","THR","FRI"))
TOD <- ordered(c("MORN","AFTN"),levels=c("MORN","AFTN"))
foo <- expand.grid(SITE=SITE,WDAY=WDAY,TOD=TOD)
foo[order(foo$SITE),]
If
On 26/03/2010, at 9:33 AM, anna wrote:
>
> Hello, I have a date value in excel: 1/4/2010 which in number format gives me
> 40182. When I read this with read.xls from R I get same 40182 so what I do
> is that I use the as.Date() function but here is what the as.Date() function
> returns me:
>> as
On 26/03/2010, at 9:22 AM, Hosack, Michael wrote:
> Hello,
>
> I need to create a dataframe containing all possible combinations of
> three variables: SITE (101,102,103,104), WDAY (MON,TUE,WED,THR,FRI),
> and TOD (MORN, AFTN). There should be a total of 40 unique combinations
> in my dataframe.
Hello, I have a date value in excel: 1/4/2010 which in number format gives me
40182. When I read this with read.xls from R I get same 40182 so what I do
is that I use the as.Date() function but here is what the as.Date() function
returns me:
> as.Date(40182)
[1] "2080-01-06"
Why don't I get the sa
Hi Mike,
Try this:
SITE <- c(101,102,103,104)
WDAY <- c('MON','TUE','WED','THR','FRI')
TOD <- c('MORN', 'AFTN')
out <- expand.grid(SITE, WDAY, TOD)
out
HTH,
Jorge
On Thu, Mar 25, 2010 at 4:21 PM, Hosack, Michael <> wrote:
> Hello,
>
> I need to create a dataframe containing all possible combi
Hello,
I need to create a dataframe containing all possible combinations of
three variables: SITE (101,102,103,104), WDAY (MON,TUE,WED,THR,FRI),
and TOD (MORN, AFTN). There should be a total of 40 unique combinations
in my dataframe. I used expand.grid() successfully(?) to create my
dataframe, b
Hello,
I need to create a dataframe containing all possible combinations of three
variables: SITE (101,102,103,104), WDAY (MON,TUE,WED,THR,FRI), and TOD (MORN,
AFTN). There should be a total of 40 unique combinations in my dataframe. I
used expand.grid() successfully(?) to create my dataframe,
I want to create a simple plot containing three levelplots with one colorbar.
I used the "Three levelplots" code below, but the third levelplot is drawn
smaller than the first two. However, if I try the "Two levelplots" code below
it works well. Can anybody tell me how could I draw three levelplot
Hello Charlie,
Thanks for your valuable help. You are right. So I changed
tau_i=c(100,200,300,400) to tau_i=c(0,100,200,300) because "400" can be defined
by another term.
I appreciate for your time and knowledge.
Best,
JinNational Renewable Energy Lab.Golden CO, 80401(303)
275-4642jinsuk@
Dear R-List,
I am working with binary data that I want to store in a PostgreSQL
DataBase. I decided to use a TXT field. I read my binary file with
readBin function, I succeed in my data storage in the database but I
have some trouble to extract the data : the correct amount of bytes is
sto
Wow, you guys are awesome. Thanks!
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Hi,
On 24 March 2010 23:22, Paul Murrell wrote:
> Hi
>
> baptiste auguie wrote:
>>
>> Thanks Felix and Paul. I had overlooked grid.grabExpr, assuming that
>> one had to draw on a device before grabbing the output.
>>
>> Now I'm not sure if there's any difference between either solution,
>> I'll g
On 03/26/2010 12:41 AM, Hrishi Mittal wrote:
Hi Jim, what's the tab.title function you are using?
It's in the plotrix package. The panes function will be in the next version.
Jim
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On Thu, Mar 25, 2010 at 3:19 PM, Ted Harding
wrote:
> One needs to be very circumspect with this sort of thing! For instance,
> experimenting with simplifications of Jeff's expression:
>
> paste(
> rep( ".", 2 ),
> "a string",
> rep( ".", 3 )
> )
>
> # [1] ". a strin
Dear R-List,
I am working with binary data that I want to store in a PostgreSQL
DataBase. I decided to use a TXT field. I read my binary file with
readBin function, I succeed in my data storage in the database but I
have some trouble to extract the data : the correct amount of bytes is
sto
On Mar 25, 2010, at 7:01 AM, Steve Murray wrote:
Sorry, that was a poorly worded question. You're right in that the
gaps are in fact NAs and I would be proposing to remove these
entirely. So it's really a case of filling up a 720 x 360 grid (by
row) based on the data in the 18556 rows x
cat() isn't doing what you think, and as.character()
is unnecessary here because paste() takes care of
that.
A couple of more elegant solutions have already
appeared, but here's one that parallels your version:
paste(c(
rep( ".", 2 ),
"a string",
rep( ".", 3 )),
collapse=" ")
On 25-Mar-10 19:07:23, Erik Iverson wrote:
> Hello,
>
> Jeff Brown wrote:
>> I would expect the following:
>>
>> paste(
>> as.character( cat( rep( ".", 2 ) ) ),
>> "a string",
>> as.character( cat( rep( ".", 3 ) ) )
>> );
>>
>> to yield this string: ". . a string . . .", but in
Erik Iverson wrote:
Hello,
Jeff Brown wrote:
I would expect the following:
paste(
as.character( cat( rep( ".", 2 ) ) ),
"a string",
as.character( cat( rep( ".", 3 ) ) )
);
to yield this string: ". . a string . . .", but instead it yields this:
. .. . .[1] " a string "
Maybe:
gsub("^(.*)$", "..\\1...", "a string")
On Thu, Mar 25, 2010 at 3:53 PM, Jeff Brown wrote:
>
> Hi,
>
> I would expect the following:
>
> paste(
> as.character( cat( rep( ".", 2 ) ) ),
> "a string",
> as.character( cat( rep( ".", 3 ) ) )
> );
>
> to yield this string: "
Hello,
Jeff Brown wrote:
I would expect the following:
paste(
as.character( cat( rep( ".", 2 ) ) ),
"a string",
as.character( cat( rep( ".", 3 ) ) )
);
to yield this string: ". . a string . . .", but instead it yields this:
. .. . .[1] " a string "
cat is writin
On Mar 25, 2010, at 10:14 AM, Florian Burkart wrote:
> Hi,
>
> following issue:
>
> Sweave is running the R code to generate
> figures three times under standard behaviour,
> twice if run with eps=FALSE, once if run
> with eps=FALSE,pdf=FALSE.
>
> However, ideally, I'd want it to run once to
On 25-Mar-10 17:44:55, R Heberto Ghezzo, Dr wrote:
> On the same topic but from a different perspective. A Nomogram or
> better a Line Aligned Nomogram is a graph with 2 or more scales,
> maybe linear where by alignig values in each scale you can read
> values in the other scale. The relationship c
Hi,
I would expect the following:
paste(
as.character( cat( rep( ".", 2 ) ) ),
"a string",
as.character( cat( rep( ".", 3 ) ) )
);
to yield this string: ". . a string . . .", but instead it yields this:
> . .. . .[1] " a string "
The third argument has been stuck im
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Nicola Sturaro Sommacal
> Sent: Thursday, March 25, 2010 9:16 AM
> To: r-help@r-project.org
> Subject: [R] Read SAS data
>
> Hi!
>
> I need to import in R some SAS dataset (sas7b
Thanks, that solved my problem!
-Lauri
2010/3/25 Gabor Grothendieck :
> Try new.row.names = 1:150 as an arg to reshape.
>
> On Thu, Mar 25, 2010 at 2:04 PM, Lauri Nikkinen wrote:
>> Hi,
>>
>> I have a data.frame in wide format which I would like to reshape into
>> a long format:
>>
>> example (n
Try new.row.names = 1:150 as an arg to reshape.
On Thu, Mar 25, 2010 at 2:04 PM, Lauri Nikkinen wrote:
> Hi,
>
> I have a data.frame in wide format which I would like to reshape into
> a long format:
>
> example (nonsense) data:
>
>> dput(perus2)
> structure(list(id = c(30L, 38L, 21L, 12L, 22L, 2
Hi,
I have a data.frame in wide format which I would like to reshape into
a long format:
example (nonsense) data:
> dput(perus2)
structure(list(id = c(30L, 38L, 21L, 12L, 22L, 28L, 31L, 44L,
8L, 47L, 23L, 20L, 41L, 42L, 29L, 50L, 5L, 33L, 4L, 17L, 11L,
1L, 18L, 6L, 9L, 32L, 16L, 14L, 39L, 48L, 3
On the same topic but from a different perspective. A Nomogram or better a Line
Aligned Nomogram is a graph with 2 or more scales, maybe linear where by
alignig values in each scale you can read values in the other scale. The
relationship can be linear, then the scales a straght lines or functio
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