Deepayan Sarkar deepayan.sar...@gmail.com writes:
You can specify a fixed-width fontfamily if that helps:
levelplot(matrix(seq(4,120,l=9),3,3),
colorkey = list(at = seq(0, 120, 20),
labels = list(labels = c(' 0',' 20',' 40','
60',' 80','100','120'),
Given a set of latitude and longitude coordinates pairs (stored in variables
latitudevals and longitudevals), I would like to plot them onto the image
of a equirectangular world map. I would like to plot each coordinate pair with
a red circle, if possible. Does anyone have any suggestions as to
Dear All,
I have been searching online for help increasing my R code more efficiently
for almost a whole day, however, there is no solution to my case. So if
anyone could give any clue to solve my problem, I would be very appreciate
for you help. Thanks in advance.
Here is my issue:
My desktop
I was trying to ByteCompile a package that I made. The package compiles
successfully with byte compile set to FALSE.
When I set ByteCompile to TRUE, I receive the following error message while
doing R CMD INSTALL
/usr/lib/R/bin/INSTALL: line 34: 9964 Done echo 'tools:::.install_packages()'
In the rgl package *rgl.postscript *can save 3d scatter plots you have
generated using the plot3d command .
For example
open3d()
x - sort(rnorm(1000))
y - rnorm(1000)
z - rnorm(1000) + atan2(x,y)
plot3d(x, y, z, col=rainbow(1000))
rgl.postscript(persp3dd.pdf,pdf)
--
View this message in
This solution works really nicely I learned much by working through it.
However but I am having trouble with subplot formatting; setting
main=d$Subject results in the correct title over each plot but repeated
multiple times. Also I can't seem to format the axis labels and numbers to
reduce the
Hello,
What do you mean by image? A file (jpeg, bmp,...)?
Best Regards
Le 26/06/2012 10:47, Steven Winter a écrit :
Given a set of latitude and longitude coordinates pairs (stored in variables
latitudevals and longitudevals), I would like to plot them onto the image
of a equirectangular
See the vignette for package 'parallel' to make use of your 4 cores.
On 26/06/2012 01:07, Xi wrote:
Dear All,
I have been searching online for help increasing my R code more efficiently
for almost a whole day, however, there is no solution to my case. So if
anyone could give any clue to solve
On 2012-06-25 15:30, Duncan Mackay wrote:
Hi Elliot
This works on Win 7 ver 2.15
useOuterStrips(combineLimits(
xyplot(x + y ~ d | g, groups = h, data = dat, type = 'l',
scales = list(y = list(relation = free),
x = list( at =seq(from =
Hello Xi,
If a program has input or output to disk or network, this may cause
it to wait and not use the available CPU.
Output is usually buffered, but may cause delay if the buffer gets
full (I'm not sure though whether this is an issue with plenty of
memory available)
Take care
Good morning,
Thanks for help.
I can explain better what I am trying to do.
I'm trying to read data from a file, separated by a tab, with the following
code.
Dataset-read.table(C:/Users/Administrator/Desktop/R/graph.txt,sep=\t,
quote=\,header = TRUE)
View(Dataset)
dput(Dataset)
View(Dataset)
Hi all,
I am imputing missingness of 90 columns in a data frame using mice.
But mice gives back :
Error in nnet.default(X, Y, w, mask = mask, size = 0, skip = TRUE, softmax =
TRUE, : too many (1100) weights
Any idea to solve this error is welcome,
Anera
[[alternative HTML
I have a data frame consisting of three columns(name of compund,ppm and
frequency).Name contains string values .ppm and frequency contains numeric
values with decimal points upto four digits.
I have an excel sheet which is like a library.The first column contains the
name of compounds and
Hi,
I performed an experiment where I raised different families coming from two
different source populations, where each family was split up into a
different treatments. After the experiment I measured several traits on each
individual.
To test for an effect of either treatment or source as well
Dear R users,
Recent changes to the MuMIn package now means that the model averaging command
(model.avg) no longer returns confidence intervals, but instead returns zvalues
and corresponding pvalues for fixed effects included in models.
Previously I have used this package for model
Try this alternative solution using only base functions:
# split the data into 4 data.frames
l - split(data, data$Subject)
names(l)
# set up the graph parameters
par(mfrow=n2mfrow(length(l)), mar=c(4,4,1,1), mgp = c(2, 1, 0))
# good old for loop over the subject names
for( n in names(l)){
d -
Dear all
I am now going to do some text analysis using R.
However, the data is very noisy that I need to clean it first.
I don't have much experience in the text cleaning process. Is anyone would
provide help on this?
If you are able to provide some similar code which was done before would be
On 26/06/2012 08:59, Anera Salucci wrote:
Hi all,
I am imputing missingness of 90 columns in a data frame using mice.
But mice gives back :
Error in nnet.default(X, Y, w, mask = mask, size = 0, skip = TRUE, softmax =
TRUE, : too many (1100) weights
Any idea to solve this error is
1. This is not an R question; it is a statistical issue.
2. R-sig-mixed-models is the appropriate list, not r-help.
-- Bert
On Tue, Jun 26, 2012 at 3:28 AM, KL sticklena...@gmail.com wrote:
Hi,
I performed an experiment where I raised different families coming from two
different source
On 06/26/2012 06:24 PM, MSousa wrote:
Good morning,
Thanks for help.
I can explain better what I am trying to do.
I'm trying to read data from a file, separated by a tab, with the following
code.
Dataset-read.table(C:/Users/Administrator/Desktop/R/graph.txt,sep=\t,
quote=\,header = TRUE)
Hi Steve,
On Mon, Jun 25, 2012 at 9:47 PM, Steven Winter stevenwinte...@yahoo.com wrote:
Given a set of latitude and longitude coordinates pairs (stored in variables
latitudevals and longitudevals), I would like to plot them onto the image
of a equirectangular world map. I would like to plot
It would help if you provided an example for your data frame, and example
for your spreadsheet, and more information on how to judge if the ppm
values are similar. Maybe this code will help you get started ...
# Here's an example data frame
mydf - data.frame(
compound=letters[1:10],
My point was just that the situation in a cumulative link model is not
much different from a binomial glm - the binomial glm is even a
special case of the clm with only two response categories. And just
like summary(glm(, family=binomial)) reports z-values and computes
p-values by using the
Sorry I misunderstood what you wanted. Using ggplot2 and reshape2 which I
imagine you will have to install, this should give you what you want
library(ggplot2)
library(reshape2)
xx1 - melt(Dataset, id = c(Source))
p - ggplot( xx1 , aes(variable, value, fill= Source )) +
Please contact the package maintainer.
Best,
Uwe Ligges
On 26.06.2012 00:41, Ricardo Pietrobon wrote:
rrdf is incredibly helpful, but I've notice that the rrdf package for mac
hasn't been working for some time: http://goo.gl/5Ukpn . wondering if there
is still a plan to maintain that in the
On 26.06.2012 00:41, Ricardo Pietrobon wrote:
rrdf is incredibly helpful, but I've notice that the rrdf package for mac
hasn't been working for some time: http://goo.gl/5Ukpn . wondering if there
is still a plan to maintain that in the long run, or if there is some other
alternative to read
Please contact the package maintainer.
Best,
Uwe Ligges
On 26.06.2012 12:46, Robertson, Andrew wrote:
Dear R users,
Recent changes to the MuMIn package now means that the model averaging command
(model.avg) no longer returns confidence intervals, but instead returns zvalues
and
On 26.06.2012 08:54, Mayank Bansal wrote:
I was trying to ByteCompile a package that I made. The package compiles
successfully with byte compile set to FALSE.
When I set ByteCompile to TRUE, I receive the following error message while
doing R CMD INSTALL
/usr/lib/R/bin/INSTALL: line 34:
I would like to plot two data sets (frequency (y-axis) of mean values for
0-1(x=axis)) on a single histogram for comparison. The hist() only allow the
overlay of two histograms, and although barplot() allows beside=TRUE, it does
not show frequency values (like hist) but rather all of the
Hello,
I have a question about the “plot.predict” function in Frank Harrell's rms
package.
Do you know how to superpose in the same graph the prediction curve of ols
and raw data points?
Put most simply, I would like to combine these two graphs:
fit_linear - ols (y4 ~
Hey,
today I wanted to use the shapiro.test() on data containing 3
numerical values per group.
It is the first time that an NA was given back for some of the groups.
In the follwing an example of code and output is shown:
shapiro.test(c(0.000637806, 0.00175561, 0.001196708))
Hi,
I have exactly the same question (how to remove empty levels in my subset),
but in my case the factor command does not work, because my dataframe is not
atomic
Try this:
test2$a - factor(test2$a)
R gives me the error message:
Error in sort.list(y) : 'x' must be atomic for 'sort.list'
Hello.
I have a problem with 2 dataframes. There are 2 columns - value and
dates. These dataframes have different dimension. Some dates coincide.
And I need to intersect them by dates and have on output two dataframes
with identical columns dates and new dimension . value have to
recieve in
You could use points() instead of plot() for the second command.
Sarah
On Tue, Jun 26, 2012 at 8:37 AM, achaumont agnes.chaum...@live.be wrote:
Hello,
I have a question about the “plot.predict” function in Frank Harrell's rms
package.
Do you know how to superpose in the same graph the
That sounds like a job for merge(), but it's hard to be sure because
you didn't provide the information requested in the posting guide.
Sarah
On Tue, Jun 26, 2012 at 11:03 AM, Васильченко Александр
vasilchenko@gmail.com wrote:
Hello.
I have a problem with 2 dataframes. There are 2 columns
Hi,
On Tue, Jun 26, 2012 at 8:06 AM, svo s.vanom...@uu.nl wrote:
Hi,
I have exactly the same question (how to remove empty levels in my subset),
but in my case the factor command does not work, because my dataframe is not
atomic
Try this:
test2$a - factor(test2$a)
R gives me the error
Hello. The data.table package is very helpful in terms of speed. But I am
having trouble actually using the output from linear regression. Is there
any way to get the data.table output to be as pretty/useful as that from
the plyr package? Below is an example.
library('data.table');
Hi. Try with following functions:
?intersection
?%in%
?[
Perhaps someone will provide you more help if you read and follow posting
guide
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
Andrija
On Tue, Jun 26, 2012 at 5:03 PM, ÷ÁÓÉÌØÞÅÎËÏ áÌÅËÓÁÎÄÒ
See
?shapiro.test
...the number of non-missing values must be between 3 and 5000.
By the way, how reasonable testing normality of 3 values?
Best
ozgur
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Why not just plot the two histograms on the same scale in a 2 panel plot?
John Kane
Kingston ON Canada
-Original Message-
From: mb...@sun.ac.za
Sent: Tue, 26 Jun 2012 15:24:55 +0200
To: r-help@r-project.org
Subject: [R] plotting two histograms on one plot with hist function
I
On Jun 26, 2012, at 11:29 AM, Sarah Goslee wrote:
You could use points() instead of plot() for the second command.
Ummm. Maybe not. I think think that plot.Predict uses lattice
graphics. You may need to use trellis.focus() followed by lpoints().
Or use the + operation with suitable
On Tue, Jun 26, 2012 at 10:02 AM, John Kane jrkrid...@inbox.com wrote:
Why not just plot the two histograms on the same scale in a 2 panel plot?
I think OP request was for comparison. Two panels may do, but why not a
barplot of the histograms in the same panel ?
barplot( rbind(
Actually, your sample size is 3. Sorry for that.
Ozgur
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__
R-help@r-project.org mailing list
On 26-06-2012 16:33, Oliver Ruebenacker wrote:
Hello Xi,
If a program has input or output to disk or network, this may cause
it to wait and not use the available CPU.
Output is usually buffered, but may cause delay if the buffer gets
full (I'm not sure though whether this is an
Hi,
Try this:
dat1-data.frame(value=c(15,20,25,30,45,50),dates=c(2005-05-25,2005-06-25,2005-07-25,2005-08-25,2005-09-25,2005-10-25))
dat2-data.frame(value=c(15,20,25,50),dates=c(2005-05-25,2005-06-25,2005-07-25,2005-10-25))
merge(dat1,dat2, by=dates)
dates value.x value.y
1 2005-05-25
Do you make any progress in solving this? I'm having the same struggle.
Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/How-To-Setup-hunspell-in-R-tp4541801p4634523.html
Sent from the R help mailing list archive at Nabble.com.
I fit a simple linear model y = bX to a data set today, and that produced 24
residuals (I have 24 data points, one for each year from 1984-2007). I would
like to test the time-independence of the residuals of my model, and I was
recommended by my supervisor to use the Ljung-Box test. The
On Jun 26, 2012, at 16:43 , r...@uni-potsdam.de wrote:
Hey,
today I wanted to use the shapiro.test() on data containing 3 numerical
values per group.
It is the first time that an NA was given back for some of the groups.
In the follwing an example of code and output is shown:
Hello Christopher,
If a process has data to write to hard disk, the data is usually
written to a buffer in memory, and from there it is written to the
hard disk independently of the CPU. Since writing to memory is much
faster than writing to hard disk, this allows the process to run
Hello:
Sorry, this might look like a beginner question, but I'm just starting to
work on the C and R interface.
I'm trying to compile a C file (with a function) to load it to an R function
but, in the command line I keep getting a lot of errors, like:
I'm finally back from vacation and looking at your email.
1. The primary mistake is in your call, where you say
fit - survfit(mod.allison.5, newdata.1, id=Id)
This will use the character string Id as the value of the identifier,
not the data. The effect is exactly the same as the
Hello,
That's a statistics question, but it's also about using an R function.
The Ljung-Box test isn't supposed to be used in such a context, to test
the residuals of an ols y = bX + e. It is used to test time independence
of the original series or of the residuals of an ARMA(p, q) fit.
In
Hello seasons R users,
Is it possible to store a complete regression result into an array? I've
already been able to save individual regression coefficients, but would like
to store the whole regression results into different arrays through a loop.
So that in under different quantiles
You can store entire regression results in a list, then use lapply()
to retrieve individual coefficients as desired.
Lists are very powerful for managing odd data formats, and no loops needed.
Sarah
On Tue, Jun 26, 2012 at 4:19 PM, Kevin Chang kchan...@uoguelph.ca wrote:
Hello seasons R users,
Dear list!
I would like to calculate chisq.test on simple data set with 70 observations,
but the output is ''Warning message:''
Warning message:
In chisq.test(tabele) : Chi-squared approximation may be incorrect
Here is an example:
tabele - matrix(c(11, 3, 3, 18, 3, 6, 5, 21), ncol
Hello,
I have count data that illustrate the presence or absence of individuals in
my study population. I created a grid cell across the study area and
calcuated a count value for each individual per season per year for each
grid cell. The count value is the number of time an individual was
Dear list,
I am currently scraping some text data from several PDFs using the
readPDF() function in the tm package. This all works very well and in most
cases the encoding seems to be latin1 - in some, however, it is not. Is
there a good way in R to check character encodings? I found the
On 2012-06-26 11:27, Omphalodes Verna wrote:
Dear list!
I would like to calculate chisq.test on simple data set with 70 observations,
but the output is ''Warning message:''
Warning message:
In chisq.test(tabele) : Chi-squared approximation may be incorrect
Here is an example:
The warning means that you have many cells with expected values less than 5
(4 of 8 cells in this case) so that the chi square estimate may be inflated.
The good news is that the probability of the inflated chi square is .0978
which you probably would not consider to be significant anyway. If you
I am looking for a function to flatten a list to a list of only 1
level deep. Very similar to unlist, however I don't want to turn it
into a vector because then everything will be casted to character
vectors:
x - list(name=Jeroen, age=27, married=FALSE,
home=list(country=Netherlands,
Oh, I had not thought of it in those terms. It does make sense now.
John Kane
Kingston ON Canada
-Original Message-
From: ke...@math.montana.edu
Sent: Tue, 26 Jun 2012 10:57:31 -0600
To: jrkrid...@inbox.com
Subject: Re: [R] plotting two histograms on one plot
On Jun 26, 2012, at 2:10 PM, SSimek wrote:
Hello,
I have count data that illustrate the presence or absence of individuals in
my study population. I created a grid cell across the study area and
calcuated a count value for each individual per season per year for each
grid cell. The count
On Tue, 26 Jun 2012, Marc Schwartz wrote:
On Jun 26, 2012, at 2:10 PM, SSimek wrote:
Hello,
I have count data that illustrate the presence or absence of individuals in
my study population. I created a grid cell across the study area and
calcuated a count value for each individual per season
do.call(c, x)
maybe?
On Tue, Jun 26, 2012 at 02:25:40PM -0700, Jeroen Ooms wrote:
I am looking for a function to flatten a list to a list of only 1
level deep. Very similar to unlist, however I don't want to turn it
into a vector because then everything will be casted to character
vectors:
Hmm that doesn't seem to work if the original list is nested more than
2 levels deep. I should have probably given a better example:
x - list(name=Jeroen, age=27, married=FALSE,
home=list(country=list(name=Netherlands, short=NL), city=Utrecht))
On Tue, Jun 26, 2012 at 3:04 PM, Neal Fultz
Alright, but I need something recursive for lists with arbitrary deepness.
On Tue, Jun 26, 2012 at 3:37 PM, arun smartpink...@yahoo.com wrote:
Hi,
Try:
do.call(c,do.call(c,x))
x1-do.call(c,do.call(c,x))
x2-flatlist(x)
identical(x1,x2)
[1] TRUE
A.K.
- Original Message
On Jun 26, 2012, at 2:27 PM, Omphalodes Verna wrote:
Dear list!
I would like to calculate chisq.test on simple data set with 70
observations, but the output is ''Warning message:''
Warning message:
In chisq.test(tabele) : Chi-squared approximation may be incorrect
Here is an example:
Dear Duncan
Thanks for your suggestion, but I really need sparse matrices: I have
implemented various graph algorithms based on adjacency matrices. For large
graphs, storing all the 0's in an adjacency matrices become uneconomical, and
therefore I thought I would use sparse matrices but the
Hi!
Any ideas on which package (e.g. mixdist, flexmix, etc) how I could fit a
mixture of say 3 Gaussian functions where 2 have their proportions, means,
and sigmas, and the third has a mean, sigma but a negative proportion.
Basically I'm trying to fit a mixture model to a distribution that
I know
Frankly, I'm not sure what you mean, but presumably
unlist(yourlist, recurs=FALSE)
is not it, right?
-- Bert
On Tue, Jun 26, 2012 at 2:25 PM, Jeroen Ooms jeroen.o...@stat.ucla.eduwrote:
I am looking for a function to flatten a list to a list of only 1
level deep. Very similar to unlist,
Duncan,
I should probably add that I am aware that my code is not the solution and also
that the relative gain of my code probably decreases with the problem size
until eventually it will perform worse that [i,j] (because of copying I
suppose). So my point is just: It would just be nice if
On 12-06-26 2:48 PM, Frederico Mestre wrote:
Hello:
Sorry, this might look like a beginner question, but I'm just starting to
work on the C and R interface.
I'm trying to compile a C file (with a function) to load it to an R function
but, in the command line I keep getting a lot of errors,
On 12-06-26 3:28 PM, Jonas Michaelis wrote:
Dear list,
I am currently scraping some text data from several PDFs using the
readPDF() function in the tm package. This all works very well and in most
cases the encoding seems to be latin1 - in some, however, it is not. Is
there a good way in R to
Hello:
I just reinstalled R and Rtools.
It works perfectly now.
Thanks,
Frederico
-Mensagem original-
De: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Enviada em: quarta-feira, 27 de Junho de 2012 01:06
Para: Frederico Mestre
Cc: r-help@r-project.org
Assunto: Re: [R] Compile
This is what the addpanel argument to plot.Predict is for, something along
the lines of
ap - function(...) lpoints(age, weight)
plot(Predict(. . .), addpanel=ap)
Frank
David Winsemius wrote
On Jun 26, 2012, at 11:29 AM, Sarah Goslee wrote:
You could use points() instead of plot() for the
Dear R People:
I have dates as factors in the following:
poudel.df$DATE
[1] 1/2/2011 1/4/2011 1/4/2011 1/4/2011 1/6/2011 1/7/2011 1/8/2011
[8] 1/9/2011 1/10/2011
Levels: 1/10/2011 1/2/2011 1/4/2011 1/6/2011 1/7/2011 1/8/2011 1/9/2011
I want them to be regular dates which can be sorted,
Hi,
The error is due to less than 5 observations in some cells.
You can try,
fisher.test(tabele)
Fisher's Exact Test for Count Data
data: tabele
p-value = 0.0998
alternative hypothesis: two.sided
A.K.
- Original Message -
From: Omphalodes Verna omphalodes.ve...@yahoo.com
Hi,
Try:
do.call(c,do.call(c,x))
x1-do.call(c,do.call(c,x))
x2-flatlist(x)
identical(x1,x2)
[1] TRUE
A.K.
- Original Message -
From: Jeroen Ooms jeroen.o...@stat.ucla.edu
To: Neal Fultz nfu...@gmail.com
Cc: r-help@r-project.org
Sent: Tuesday, June 26, 2012 6:23 PM
Subject: Re:
Thank you both for your quick response and input. I will consider all of
your points and see what we are able to derive from there.
Thank you again for your time and expertise.
-Stephanie
---
Stephanie L. Simek
Carnivore Ecology Lab
Forest
Hi,
I hope this helps. Tested to some depth.
x1 - list(name=Jeroen, age=27, married=FALSE,
home=list(country=list(name=Netherlands, short=NL), city=Utrecht))
x2 - list(name=Jeroen, age=27, married=FALSE,
home=list(country=list(name=list(Country1=Netherlands,Country2=Spain),
On Tue, Jun 26, 2012 at 10:54 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Dear R People:
I have dates as factors in the following:
poudel.df$DATE
[1] 1/2/2011 1/4/2011 1/4/2011 1/4/2011 1/6/2011 1/7/2011 1/8/2011
[8] 1/9/2011 1/10/2011
Levels: 1/10/2011 1/2/2011 1/4/2011
Hello again:
Here is a solution to the dates without leading zeros:
pou1 - function(x) {
#Note: x is a data frame
#Assume that Column 1 has the date
#Column 2 has station
#Column 3 has min
#Column 4 has max
library(stringr)
w - character(length=nrow(x))
On 27/06/12 08:54, arun wrote:
Hi,
The error is due to less than 5 observations in some cells.
NO, NO, NO It's not the observations that matter, it is
the ***EXPECTED COUNTS***. These must all be at least
5 in order for the null distribution of the test statistic to be
Please don't change subject lines for follow-on comments. It messes up
threading in most readers: e.g.,
https://stat.ethz.ch/pipermail/r-help/2012-June/thread.html
Michael
On Tue, Jun 26, 2012 at 11:57 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Hello again:
Here is a solution to the
Thank you very much. The advice I followed (and which, for some reason, I do
not see here right now) was to use 'droplevels'. I needed the command for
several variables at the same time, so this was very convenient.
Hello,
Have you tried 'droplevels':
test
How could I select the rows of a dataset that have the maximum value in one
variable and to do this nested in another variable. It is a dataframe in long
format with repeated measures per subject.
I was not successful using aggregate, because one of the columns has character
values (and/or
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