It looks like a bug in R, not Rscript: what Rscript does is effectively
R --file=size.R
and you can easily test that
echo 'print("Hello world")'> size.R
R --file=size.R
give you the error while
cp size.R asize.R
R --file=asize.R
works as expected.
None of which helps you directly, but p
Hi Jason,
I did not have time to actually test this code so there may be typos
and some of it may not work as I thought. I would create a copy of
your data in a test directory and experiment with that until you are
confident you have everything working how you want . As a side note,
since you ar
Hi
On Wed, Jul 7, 2010 at 7:49 PM, jd6688 wrote:
>
> Here are what i am going to accomplish:
>
> I have 400 files named as xxx.txt. the content of the file looks like the
> following:
>
>namecount
>
> 1. aaa 100
> 2. bbb2000
> 3. ccc300
> 4. ddd 3000
>
>
> more th
On 2010-07-07 18:24, fsch wrote:
I managed to solve the problem myself using:
q=sort(unique(foram$stage[which(foram$length>0)]))
boxplot(foram$logl~foram$stage, data=foram, outline=F, at=q, axes=TRUE,
add=TRUE, col="gray82", medlwd=1)
points(foram$stage,foram$logl, cex=.1)
However, for fut
Hello R users,
I'm trying to extract random samples from a big array I have.
I have a data frame of over 40k lines and would like to produce around 50
random sample of around 200 lines each from this array.
this is the matrix
ID xxx_1c xxx__2c xxx__3c xxx__4c xxx__5T xxx__6T xxx__7T xx
On Wed, Jul 7, 2010 at 8:59 AM, Michael Friendly wrote:
> No one replied to my second question: how to get standard errors or
> confidence intervals for the
> estimated fixed effects from lme(). AFAICS, intervals() only gives CIs
> for coefficients.
Assuming you mean 'predictions based on the
Hello All,
It seems weird to me that Rscript does the following thing and enters
the R prompt mode. If I change the file name to something that doesn't
start with 'size', then Rscript runs normally. Does anybody know what
is going on?
$ Rscript size.R
WARNING: '--file=size.R' value is invalid: ig
The R Clinic is a fantastic website with many examples. There are some examples
under the graphics section.
http://biostat.mc.vanderbilt.edu/wiki/Main/RClinic
> Date: Tue, 6 Jul 2010 23:29:33 -0400
> From: bjlwilkin...@gmail.com
> To: deepayan.sar...@gmail.com
> CC: r-help@r-project.o
On Tue, Jul 6, 2010 at 8:29 PM, Ben Wilkinson wrote:
> Thanks - that helps put the data on the same scale but it doesn't actually
> add the scale as it still shows min and max on the axis.
The scale maps from (min, max) -> (0, 1). You can compute and specify
tick positions and labels accordingly.
Try scatterplot() in the car package. It draws boxplots for X & Y in
the margins, auto scaled to the axes
fsch wrote:
Hello-
I'm new to R, coding and stats. (Oh no.)
Anyway, I have about 12000 data points in a data.frame (dealing with
dimensions and geological stage information for fossil p
wow chuck. you really know how to dig up the archives. I don't know if it's
exactly relevant for what the OP is asking but i did use the
( or atleast a )Â paper by hotelling and it was titled "the selection of
variates for use in prediction with some comments on the general problem of
On Wed, Jul 7, 2010 at 10:55 PM, Charles C. Berry wrote:
>
>
> contr.mysum <- function (n, contrasts = TRUE, sparse = FALSE,
> base=length(levels))
> {
The memisc package has a contr.sum function with a base= argument.
__
R-help@r-project.org mailing l
On Jul 7, 2010, at 8:39 PM, Adam Berrones wrote:
I have looked absolutely everywhere but I cannot figure out why my
Mac will
not let me create simple data vectors.
blah
Error: object 'blah' not found
blah <- c(1, 2, 3)
Error in c(1, 2, 3) : 'file' must be a character string or connecti
Assuming you are on Windows, you can probably do this within R, without
involving the IT department.
?setInternet2
for the details and discussion. Enter the line
setInternet2(use = TRUE)
before doing the install.packages()
On Wed, Jul 7, 2010 at 7:55 PM, Katie Ashton
wrote:
> To whom this m
On Wed, 7 Jul 2010, chen jia wrote:
Hi there,
I run two regressions:
y = a1 + b1 * x + e1
y = a2 + b2 * z + e2
I want to test against the null hypothesis: b1 = b2. How do I design the test?
You are testing a non-nested hypothesis, which requires special handling.
The classical test is du
G'day Stephen,
On Wed, 7 Jul 2010 16:41:18 -0400
"Bond, Stephen" wrote:
> Please, do not post if you do not know the answer. People will see
> this has answers and skip.
>
> I tried with
> mat1=contrasts(fixw$snconv)
> mat1=mat1[,-2]
> summary(frm2sum <- glm(resp.frm ~
> C(snconv,contr=mat1)+m
>
>
> > blah
> Error: object 'blah' not found
> > blah <- c(1, 2, 3)
> Error in c(1, 2, 3) : 'file' must be a character string or connection
>
> It used to work, and 6 months later I'm back on R, and now it will not
> work.
>
> Most likely, you inadvertently masked the base function c with your own
that makes sense. thank you, guys!
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https://stat.ethz.c
I managed to solve the problem myself using:
q=sort(unique(foram$stage[which(foram$length>0)]))
boxplot(foram$logl~foram$stage, data=foram, outline=F, at=q, axes=TRUE,
add=TRUE, col="gray82", medlwd=1)
points(foram$stage,foram$logl, cex=.1)
However, for future reference since this was my fir
To whom this may concern,
I am having issues installing a program called "SAM" (Statistical Analysis of
Microarrays) which is an excel add in and requires the use of R. Due to the
firewall at my hospital, R cannot connect to a mirror and therefore cannot
install necessary files for the complete
I have looked absolutely everywhere but I cannot figure out why my Mac will
not let me create simple data vectors.
> blah
Error: object 'blah' not found
> blah <- c(1, 2, 3)
Error in c(1, 2, 3) : 'file' must be a character string or connection
It used to work, and 6 months later I'm back on R, a
Hi,
I am trying to use the glmnet package to do some simple feature selection.
However, I would ideally like to be able to specify the number of features
to return (the glmnet package, as far as I can tell, only allows
specification of a regularization parameter, lambda, that in turn returns a
m
If I surmise correctly, you want something like contr.sum but with the
option to select the level with the row of '-1' and no column. Here '3' is
that level:
contr.sum(3)
[,1] [,2]
110
201
3 -1 -1
relevel() and contr.sum() generally will not do what you want witho
I have not seen that particular subtype documented in the archives,
but others (subtpes 16 and 17 if I remember correctly) were due to
difficulties in extracting value labels from the incompletely
documented spss file structure. Have you examined the resulting data
object with str()?
-
Sorry to bother everyone. Soon after posting my question I found out what
summary.glm() will do to a glm.fit object. Gotta love having your learning
curve exposed in public :) Seth
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On Jul 7, 2010, at 9:48 PM, Seth wrote:
Hi,
I am working with a function that makes use of glm.fit. Without
modifying
the somewhat long code too much, I would like to have t-values
returned for
the predictor variables used in the fitting process. Is there a
relatively
straightforward
Hi,
I am working with a function that makes use of glm.fit. Without modifying
the somewhat long code too much, I would like to have t-values returned for
the predictor variables used in the fitting process. Is there a relatively
straightforward way to do this? Thanks, Seth Myers
--
View this
On 2010-07-03, at 17:43 , Peter Ehlers wrote:
>
> On 2010-07-03 0:05, Godfrey van der Linden wrote:
>> G'day, All.
>>
>> I have been trying to trackdown a problem in my R analysis script. I perform
>> a scale() operation on a matrix then do further work.
>>
>> Is there any way of inverting th
Hello,
I am trying to use nlm to estimate the parameters that minimize the
following function:
Predict<-function(M,c,z){
+ v = c*M^z
+ return(v)
+ }
M is a variable and c and z are parameters to be estimated.
I then write the negative loglikelihood function assuming normal errors:
nll<-function
On 2010-07-08, at 10:33 , Duncan Murdoch wrote:
> On 07/07/2010 7:32 PM, Gabor Grothendieck wrote:
>> On Wed, Jul 7, 2010 at 7:22 PM, Duncan Murdoch
>> wrote:
>>
>>> On 07/07/2010 5:58 PM, karena wrote:
>>>
Hi, I am a newbie of R, and playing with the "ifelse" statement.
I
On 07/07/2010 7:32 PM, Gabor Grothendieck wrote:
On Wed, Jul 7, 2010 at 7:22 PM, Duncan Murdoch wrote:
On 07/07/2010 5:58 PM, karena wrote:
Hi, I am a newbie of R, and playing with the "ifelse" statement.
I have the following codes:
## first,
for(i in 1:3) {
for(j in 2:4) {
cor.temp
Hello r-help,
I am having trouble installing the add-on package gstat. I suspect (but
I'm not sure) that the trouble may be that it runs all the checks for
gcc, then uses Intel's compiler to actually compile. I have not been
able to successfully force it to use gcc to test that theory. R CMD
IN
Would it really be so hard to
provide commented, minimal, self-contained, _reproducible_ code
?
-Peter Ehlers
On 2010-07-07 15:16, fsch wrote:
Hello-
I'm new to R, coding and stats. (Oh no.)
Anyway, I have about 12000 data points in a data.frame (dealing with
dimensions and geological s
On 2010-07-07 16:52, Jim Bouldin wrote:
I'm trying to obtain the mean of the middle 95% of the values from each row
of a matrix (that is, the highest and lowest 2.5% of values in each row
are removed before calculating the mean). I am having all sorts of
problems with this; for example the com
Yi wrote:
Hi, folks,
Suppose the data is as follows:
###
x=1:10
y=2:11
z=3:12
data1=data.frame(x,y)
data2=data.frame(x,z)
##
In real case, how to check data1$x and data2$x are the same or not? I am
using for-loop to do this.
Is there a quicker way?
all(data1$x == data2$x)
or
i
On Wed, Jul 7, 2010 at 7:22 PM, Duncan Murdoch wrote:
> On 07/07/2010 5:58 PM, karena wrote:
>>
>> Hi, I am a newbie of R, and playing with the "ifelse" statement.
>>
>> I have the following codes:
>> ## first,
>>
>> for(i in 1:3) {
>> for(j in 2:4) {
>> cor.temp <- cor(iris.allnum[,i], iris.allnu
On Jul 7, 2010, at 6:52 PM, Jim Bouldin wrote:
I'm trying to obtain the mean of the middle 95% of the values from
each row
of a matrix (that is, the highest and lowest 2.5% of values in each
row
are removed before calculating the mean).
A winsorized.mean?
I am having all sorts of
p
Hi, folks,
Suppose the data is as follows:
###
x=1:10
y=2:11
z=3:12
data1=data.frame(x,y)
data2=data.frame(x,z)
##
In real case, how to check data1$x and data2$x are the same or not? I am
using for-loop to do this.
Is there a quicker way?
Thanks!
Yi
[[alternative HTML ver
On 07/07/2010 5:58 PM, karena wrote:
Hi, I am a newbie of R, and playing with the "ifelse" statement.
I have the following codes:
## first,
for(i in 1:3) {
for(j in 2:4) {
cor.temp <- cor(iris.allnum[,i], iris.allnum[,j])
if(i==1 & j==2) corr.iris <- cor.temp
else corr.iris <- c(corr.iris, cor.
On Jul 7, 2010, at 6:32 PM, thmsfuller...@gmail.com wrote:
Hi All,
I meant to take the min row by row. But the result is apparently not
what I want. Changing min to pmin solve the problem.
df=data.frame(X=1:10, Y=1:10)
transform(df, Z=min(X,10-Y))
X Y Z
1 1 1 0
2 2 2 0
3 3 3 0
Hi, all
I am doing a optimization problem using nlminb. It seems to me that the
result is kind of sensitive to the starting value. (It happened that the
resulting parameter doesn't change from the starting value.) Any suggestions
on giving the starting value ? or any other function that can do
opt
apply(matrix1, 1, quantile, probs = c(0.05, 0.9), na.rm = TRUE)
may be what you are looking for.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jim Bouldin
Sent: Thursday, 8 July 2010 8:52 AM
To: R help
Subject: [R] quantiles on
I'm trying to obtain the mean of the middle 95% of the values from each row
of a matrix (that is, the highest and lowest 2.5% of values in each row
are removed before calculating the mean). I am having all sorts of
problems with this; for example the command:
apply(matrix1,1,function(x) quantil
Hi All,
I meant to take the min row by row. But the result is apparently not
what I want. Changing min to pmin solve the problem.
> df=data.frame(X=1:10, Y=1:10)
> transform(df, Z=min(X,10-Y))
X Y Z
1 1 1 0
2 2 2 0
3 3 3 0
4 4 4 0
5 5 5 0
6 6 6 0
7 7 7 0
8 8 8 0
9 9
Hello Sarah
I am unable to register for this event as the link you had supplied goes
nowhere. Any other method of registering pls?
Thx
Raghu
On Wed, Jul 7, 2010 at 8:08 PM, Sarah Lewis-2 [via R] <
ml-node+2281370-1994523266-309...@n4.nabble.com
> wrote:
> I am pleased to announce to agenda for
Hi there,
I run two regressions:
y = a1 + b1 * x + e1
y = a2 + b2 * z + e2
I want to test against the null hypothesis: b1 = b2. How do I design the test?
I think I can add two equations together and divide both sides by 2:
y = 0.5*(a1+a2) + 0.5*b1 * x + 0.5*b2 * z + e3, where e3 = 0.5*(e1 + e2
yeah, both work. Thank you, guys...
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R-help@r-project.
Thank you! That was the appropriate adjustment - it's working as I need it
to! =)
-CC
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___
Hello-
I'm new to R, coding and stats. (Oh no.)
Anyway, I have about 12000 data points in a data.frame (dealing with
dimensions and geological stage information for fossil protists) and have
plotted them in a basic scatter plot. I also added a boxplot to overlay
these points. Each worked fine i
Hi, I am a newbie of R, and playing with the "ifelse" statement.
I have the following codes:
## first,
for(i in 1:3) {
for(j in 2:4) {
cor.temp <- cor(iris.allnum[,i], iris.allnum[,j])
if(i==1 & j==2) corr.iris <- cor.temp
else corr.iris <- c(corr.iris, cor.temp)
}
}
this code is working fine.
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
I have imported a spss-data set with the read.spss command and received the
following warning
messages. Does this mean that something did not work correctly? What am I to
make out of these messages?
Thanks for your help, Katharina
Warning messages:
On Jul 7, 2010, at 6:05 PM, addi wei wrote:
Hello, I have tried multiple times to unsubscribe from the general
mailing list to no avail. I only wish to receive emails when someone
posts a response in my thread. How do I do this?
I suspect many people send a copy of their response to the O
Hello, I have tried multiple times to unsubscribe from the general
mailing list to no avail. I only wish to receive emails when someone
posts a response in my thread. How do I do this? Thanks.
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A.Z.,
You could recreate the data (I assumed that the values are discrete integers
and not a bin for a continuous variable) ...
value <- 1:6
count <- c(10,8,12,9,14,7)
d <- rep(value,count)
table(d)
... and then do what you want with the data in the d vector ...
summary(d)
hist(d)
> On 07-J
Please do not post with ambiguous questions lacking sufficient R code
and data to represent the problem.
On Jul 7, 2010, at 4:41 PM, Bond, Stephen wrote:
Please, do not post if you do not know the answer. People will see
this has answers and skip.
More likely people will see that it has in
See Teds answer for histogram (I'd go with barplot).
For most statistical procedures there is a weighted version (e.g.
weighted.mean() for the mean). Your counts are valid weights for most
procedures.
Cheers
Joris
On Wed, Jul 7, 2010 at 10:39 PM, Andrei Zorine wrote:
> Hello,
> I just need a hi
On 07-Jul-10 20:39:32, Andrei Zorine wrote:
> Hello,
> I just need a hint here:
> Suppose I have no raw data, but only a frequency table I have, and I
> want to run basic statistical procedures with it, like histogram,
> descriptive statistics, etc. How do I do this with R?
> For example, how do I
On Wed, 7 Jul 2010, Michael Friendly wrote:
Tal Galili wrote:
Hello David,
Thanks to your posting I started looking at the function in the arm
package.
It appears this function is quite mature, and offers (for example) the
ability to easily overlap coefficients from several models.
I update
To be correct, everything is written to all folders according to my testing.
There is absolutely no need whatsoever to use l_ply. And in any case,
take as much as possible outside the loop, like the library statement
and the max.col.
Following is _a_ solution, not the most optimal, but as close a
I have imported a spss-data set with the read.spss command and received the
following warning
messages. Does this mean that something did not work correctly? What am I to
make out of these messages?
Thanks for your help, Katharina
Warning messages:
1: In read.spss("/Daten/ESS4DE.spss/ESS4DE.sav"
Thank you, this has been really helpful, I think I have managed to
produce the desired effect.
However, I am having trouble with generating and passing subscripts
from the main xyplot call to panel.superpose. My code resembles the
examples (e.g. pp 71-73) in your 2008 book --- though I cannot refe
On Wed, Jul 7, 2010 at 12:17 PM, DrHatch wrote:
> I expected this script to show nine panels, each with a plot of the
> function.
> But when I run it, only the diagonal panels have content.
> Executing > print(trel_1) shows what I expected in the upper left corner.
> I am using R ver. 2.11.0 and l
Hello,
I just need a hint here:
Suppose I have no raw data, but only a frequency table I have, and I
want to run basic statistical procedures with it, like histogram,
descriptive statistics, etc. How do I do this with R?
For example, how do I plot a histogram for this table for a sample of size 60?
Please, do not post if you do not know the answer. People will see this has
answers and skip.
I tried with
mat1=contrasts(fixw$snconv)
mat1=mat1[,-2]
summary(frm2sum <- glm(resp.frm ~
C(snconv,contr=mat1)+mprime+mshape,data=fixw,family="quasibinomial"))
the unwanted level is still there. Unbel
On 7/7/2010 1:11 PM, Ian Bentley wrote:
Hi all,
I'm trying to use ggplot to make a boxplot of some data, but I can't seem to
figure out how to make it use the data I'm giving it.
The data is in a data.frame so that it has two columns:
meltl
value L1
1234 1
1234 1
1235 1
...
1255 1
2335
On Tue, Jul 6, 2010 at 2:47 AM, Coen van Hasselt
wrote:
> Hello,
>
> I would like to know how I can filter out empty plots in xyplot, when
> stratifying on some variables.
>
> Example:
>
> I have a dataset in which I plot CONC ~ TIME, stratified for patient
> ID(1,2,..,100), FORM(1,2) and BOOST (1
Hello,
I am having some trouble using a model I created from plsr (of train) to
analyze each invididual R^2 of the 10 components against the test data. For
example:
mice1 <- plsr(response ~factors, ncomp=10 data=MiceTrain)
R2(mice1)##this provides the correct R2 for the Train data for 1
2010/7/5 László Sándor :
> Hi all,
>
> Back in 2007, Deepayan and Patrick had an exchange about how to modify
> axes for lattice plots (pasted below). I need something similar, but I
> also need to produce ticks on the axes. Deepayan quickly coded up
> substitute gridlines because they needed to ma
What about this:
> testdata <- cbind.data.frame(value = runif(100), L1 = rep(1:10, each=10))
> head(testdata)
value L1
1 0.3370902 1
2 0.6766098 1
3 0.2433171 1
4 0.8848674 1
5 0.5253600 1
6 0.4067238 1
> tail(testdata)
value L1
95 0.8149121 10
96 0.5186669 10
97 0.1080695 1
On Jul 7, 2010, at 4:04 PM, Bond, Stephen wrote:
Clarifying my question:
options(contrasts = c("contr.sum", "contr.poly"))
contrasts()
[,1] [,2] [,3] [,4] [,5]
0110000
0301000
0500100
0600010
07000
Hi all,
I'm trying to use ggplot to make a boxplot of some data, but I can't seem to
figure out how to make it use the data I'm giving it.
The data is in a data.frame so that it has two columns:
>meltl
value L1
1234 1
1234 1
1235 1
...
1255 1
2335 2
3444 2
...
10001 50
12311 50
...
The fi
Hi,
I recommend that you look at the following help pages and experiment a
little (maybe create a toy directory with only three or four files with
a few lines each):
?files
?dir
?grep
?strsplit
Good luck!
Stephan
jd6688 schrieb:
Here are what i am going to accomplish:
I have 400 files na
Clarifying my question:
options(contrasts = c("contr.sum", "contr.poly"))
> contrasts(fixw$snconv)
[,1] [,2] [,3] [,4] [,5]
0110000
0301000
0500100
0600010
0700001
09 -1 -1 -1 -1 -1
I
Here are what i am going to accomplish:
I have 400 files named as xxx.txt. the content of the file looks like the
following:
namecount
1. aaa 100
2. bbb2000
3. ccc300
4. ddd 3000
more that 1000 rows in each files.
these are the areas i need help:
1. how can i on
On Fri, Jul 2, 2010 at 11:40 AM, Gregory Gentlemen
wrote:
> Fellow R-users,
>
> I have a longitudinal data set with missing values in it. I would like to
> produce a residual plot for each time using panel.xyplot function but I get
> an error message. Here's a simple example,
>
> library(nlme)
>
Hi Joshua:
Here are what i am going to accomplish:
I have 400 files named as xxx.txt. the content of the file looks like the
following:
namecount
1. aaa 100
2. bbb2000
3. ccc300
4. ddd 3000
more that 1000 rows in each files.
these are the areas i need help:
1.
Try write.table instead.
On Wed, Jul 7, 2010 at 1:56 PM, thmsfuller...@gmail.com <
thmsfuller...@gmail.com> wrote:
> Hello All,
>
> I'm trying to pass the argument col.names to write.csv using '...'.
> But I got the following warnings. Maybe it is very simple. But I'm not
> sure what I am wrong.
On Jul 7, 2010, at 3:13 PM, Bond, Stephen wrote:
I need to use contr.sum and observe that some levels are not
statistically different from the overall mean of zero.
What is the proper way of forcing the zero estimate? It seems the
column corresponding to that level should become a column of
On Jul 7, 2010, at 12:56 PM, thmsfuller...@gmail.com wrote:
Hello All,
I'm trying to pass the argument col.names to write.csv using '...'.
But I got the following warnings. Maybe it is very simple. But I'm not
sure what I am wrong. Could you please help point to me what the
problem is?
#
Hi!
I want to write portions of my data (3573 columns at a time) to twenty
folders I have available titled "A_1" to "A_20" such that the first 3573
columns will go to folder A_1, next 3573 to folder A_2 and so on.
This code below ensures that the data is written into all 20 folders, but
only the
Hi,
You might want to check out ?system ?read.table and ?subset
and also ?sub and maybe ?strsplit
Reading one of the good intro to R documents out there would
also help you get started with this fairly straightforward task
Beyond that, you did a good job of explaining your problem, but
this is o
On Wed, 7 Jul 2010, thmsfuller...@gmail.com wrote:
Hello All,
I'm trying to pass the argument col.names to write.csv using '...'.
But I got the following warnings. Maybe it is very simple. But I'm not
sure what I am wrong. Could you please help point to me what the
problem is?
Its not a _prob
On Jul 7, 2010, at 1:26 PM, Gabor Grothendieck wrote:
On Wed, Jul 7, 2010 at 1:25 PM, Gabor Grothendieck
wrote:
On Wed, Jul 7, 2010 at 1:15 PM, Immanuel
wrote:
Hey,
big help, thanks!
One little question remains, if I create
more then one string and table ...
-
# gener
On 07/07/10 18:52, Paul Johnson wrote:
> [...]
> 1:
> axis(1, line=6, at=mu+dividers*sigma,
> labels=as.expression(c(b1,b2,b3,b4,b5), padj=-1))
>
>
> 2:
> axis(1, line=9, at=mu+dividers*sigma,
> labels=c(as.expression(b1),b2,b3,b4,b5), padj=-1)
>
> This second one shouldn't work, I think.
?list.files
In particular look at pattern argument.
?file.rename
?lapply
?read.table
?"["
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Jul 7, 2010, at 1:11 PM, jd6688 wrote:
Here are what i am going to accomplish:
I have 400
I expected this script to show nine panels, each with a plot of the
function.
But when I run it, only the diagonal panels have content.
Executing > print(trel_1) shows what I expected in the upper left corner.
I am using R ver. 2.11.0 and lattice ver. 0.18-8 on Windows XP under
Eclipse and StatE
I need to use contr.sum and observe that some levels are not statistically
different from the overall mean of zero.
What is the proper way of forcing the zero estimate? It seems the column
corresponding to that level should become a column of zeros.
Is there a way to achieve that without me const
Here are what i am going to accomplish:
I have 400 files named as xxx.txt. the content of the file looks like the
following:
namecount
1. aaa 100
2. bbb2000
3. ccc300
4. ddd 3000
more that 1000 rows in each files.
these are the areas i need help:
1. how can i on
Here are what i am going to accomplish:
I have 400 files named as xxx.txt. the content of the file looks like the
following:
namecount
1. aaa 100
2. bbb2000
3. ccc300
4. ddd 3000
more that 1000 rows in each files.
these are the areas i need help:
1. ho
Hello All,
I'm trying to pass the argument col.names to write.csv using '...'.
But I got the following warnings. Maybe it is very simple. But I'm not
sure what I am wrong. Could you please help point to me what the
problem is?
#
fun=function(x, ...) {
fr=parent.frame()
tm
I am pleased to announce to agenda for next weeks LondonR meeting:
LondonR meeting - 13th July 2010
Date: Tuesday 13th July 2010
Time: 6pm - 9pm
Venue: The Shooting Star
125 - 129 Middlesex Street
E1 7JF
(N
On Wed, Jul 7, 2010 at 5:41 AM, Duncan Murdoch wrote:
>>>
>>> You want "as.expression(b1)", not "expression(b1)". The latter means
>>> "the
>>> expression consisting of the symbol b1". The former means "take the
>>> object
>>> stored in b1, and convert it to an expression.".
>>>
Thanks to
It may not help the original poster, but here's a solution based on what Greg
said above:
# Load plotrix
library(plotrix)
# Create a new layout to divide the graphics in 2, the first one (displaying
the persp() graph) being 4 times larger than the second one (displying the
legend)
layout(matrix
ggplot2
ggplot2 is a plotting system for R, based on the grammar of graphics,
which tries to take the good parts of base and lattice graphics and
avoid bad parts. It takes care of many of the fiddly details
that make plotting a hassle (l
So, the problem was that R exports only double sized floats (double), and
Paraview requires single sized floats. The solution was just to write :
writeBin(data,bfile_celldata,endian="swap",size=4)
Special Thanks to Sebastian Gibb :
http://www.mail-archive.com/r-help@r-project.org/msg100994.htm
Hi,
I am trying to superimpose (overlay) regression lines to scatter plots
by groups with xyplot (dysfunctional code below). However, my call of
panel.superpose breaks down because of the subscripts requirement. I
tried to research the documentation and examples, but I cannot figure
out how to mak
Hey,
saved my day.
Now can watch the football semi-final
thanks
> Turn them into factors with the appropriate levels before counting
> them with table:
>
> # generate an input string n long
> set.seed(123)
> n <- 300
> lets_1 <- paste(sample(letters[1:5], n, replace = TRUE), collapse = "")
> lets_2
> fortune(197)
If anything, there should be a Law: Thou Shalt Not Even Think Of Producing A
Graph That Looks Like Anything From A Spreadsheet.
-- Ted Harding (in a discussion about producing graphics)
R-help (August 2007)
Also read the discussion started with:
http://tolstoy.newcastle.ed
On Wed, 2010-07-07 at 17:31 +0200, Sergey Goriatchev wrote:
> Hello, everyone
>
> F# is now public. Compiled code should run faster than R.
>
> Anyone has opinion on F# vs. R? Just curious
>
> Best,
> S
Sergey,
F# is public, but is not open source.
F# run in windows but run in AIX, linux, M
Check out the high performance computing task view on CRAN.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of M
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