Dear Friends.
Greetings!
I have asked the question of how to set up a blank file and write a list to
it as a row for many times, with the number of lists unknown.
I have received many beautiful solutions. Thanks go to Professor *Murdoch,
Professor *Menne, Professor Grothendieck and Dr.
Using R 2.5.1 on Windows XP Professional, and PBSmapping package version
2.51, I have encountered some behavior which puzzles me. I am including
the package's listed maintainer on this email but also seek the thoughts
of the R-help community.
I have a set of EventData, which I want to plot as
Hi
Emilio Gagliardi wrote:
Hi Paul,
I'm sorry for not posting code, I wasn't sure if it would be helpful without
the data...should I post the code and a sample of the data? I will remember
to do that next time!
It's important not only to post code, but also to make sure that other
I still have this problem. Does anybody know any solution?
Antje
Antje schrieb:
Hello,
I'm trying to plot a set of barplots like a matrix (2 rows, 10 columns
fromreduced_mat) to a pdf. It works with the following parameters:
pdf(test.pdf,width=ncol(reduced_mat)*2,
Thanks for all the comments,
The artificial dataset is as representative of my 440MB file as I could design.
I did my best to reduce the complexity of my problem to minimal
reproducible code as suggested in the posting guidelines. Having
searched the archives, I was happy to find that the topic
Hi
Daniel Brewer wrote:
Thanks for the replies, but I still cannot get what I want. I do not
want the label inside the plot area, but in the top left of the paper, I
suppose in the margins. When I try to use text to do this, it does not
seem to plot it outside the plot area. I have also
Daniel Brewer wrote:
Thanks for the replies, but I still cannot get what I want. I do not
want the label inside the plot area, but in the top left of the paper, I
suppose in the margins. When I try to use text to do this, it does not
seem to plot it outside the plot area. I have also tried
I am trying to run a GLMM on some binomial data. My fixed
factors include 2
dichotomous variables, day, and distance. When I run the model:
modelA-glmmPQL(Leaving~Trial*Day*Dist,random=~1|Indiv,family=
binomial)
I get the error:
iteration 1
Error in MEEM(object, conLin,
Dear R users;
I am modelling densities of some species of birds, so I have a problem with a
great ammount of zeros.
I have decided to try GLMs with the tweedie family, but in all the models I
have tried I got an NA for the AIC value.
Just to check the problem I've compared the a glm using
Hi,
I have 4 huge tables on which i want to do a PCA analysis and a kmean
clustering. If i run each table individually i have no problems, but if i want
to run it in a for loop i exceed the memory alocation after the second table,
even if i save the results as a csv table and i clean up all
Hi R-users,
I have to define a noise level function L and its energy in the various
moment of the day by:
if time is between 18:00:00 and 23:59:59 then L[j] - L[j]+5 and W -
10^((L+5)/10)
if time is between 22:00:00 and 05:59:59 == L - L+10 and W -
10^((L+10)/10)
else
L=L and W = W
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Nair,
Murlidharan T
Sent: Thursday, August 09, 2007 12:02 PM
To: Nordlund, Dan (DSHS/RDA); r-help@stat.math.ethz.ch
Subject: Re: [R] small sample techniques
n=300
30% taking A relief from pain
write.table(mydata.frame, mydata, col.names=NA, quote=F, sep=\t) will
solve the problem.
Deng
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Weiwei Shi
Sent: August 10, 2007 12:41 PM
To: r-help@stat.math.ethz.ch
Subject: [R] write.table
Hi,
I am always
Jim Lemon Wrote:
I also greatly enjoyed Ted's rebuttal of the Bar charts are
evil and must be banned argument. If bar charts are
appropriate for the audience, give 'em bar charts. One great
way to turn off your customers is to tell them what they can
and can't do with your product.
I
Daniel J. Nordlund
Research and Data Analysis
Washington State Department of Social and Health Services
Olympia, WA 98504-5204
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Welcome to the world of R.
I'm glad that you found the discussion enlightening, now that you have
thought about things a bit, here is some code to try out that shows some
of the alternatives to the original plots you provided (which is best
depends on your audience and what your main question of
I am having trouble with the rfImpute function in the randomForest package.
Here is a sample...
clunk.roughfix-na.roughfix(clunk)
clunk.impute-rfImpute(CONVERT~.,data=clunk)
ntree OOB 1 2
300: 26.80% 3.83% 85.37%
ntree OOB 1 2
300: 18.56% 5.74% 51.22%
Error
[Gabor Grothendieck]
table(col(mat), mat)
Clever, simple, and elegant! :-)
--
François Pinard http://pinard.progiciels-bpi.ca
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
You've spotted it!
table(df$area)
0 1 2 3 4 5 7
21 27 71 46 19 3 1
There are no values in area 6.
Thank you very much.
Jabez
- Original Message
From: Petr PIKAL [EMAIL PROTECTED]
To: Jabez Wilson [EMAIL PROTECTED]
Cc: R-Help r-help@stat.math.ethz.ch
Sent: Friday, 10
An even simpler solution is:
mat2 - 1 * (mat1 0.25)
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
On Fri, 10 Aug 2007, Monica Pisica wrote:
Thanks! I will look into ...
I have 4 GB RAM, and i was monitoring the memory with Windows task
manager so i was looking how R gets more and more memory allocation
from less than 100Mb to 1500Mb .
Then you are almost certainly fragmenting
Will something like this help?
mm - matrix(rnorm(100),nrow=10)
mm
nn - mm .5
nn
--- Lanre Okusanya [EMAIL PROTECTED] wrote:
Hello all,
I am working with a 1000x1000 matrix, and I would
like to return a
1000x1000 matrix that tells me which value in the
matrix is greater
than a
I hope you don't mind that I offer also two solutions. No.1 is really
bad. No.2 should be on par with the other ones.
Best,
Roland
mydata - matrix(rnorm(10*10), ncol=10)
threshold.value - 1.5
mydata2 - matrix(0, nrow=nrow(mydata), ncol=ncol(mydata))
mydata3 - matrix(0, nrow=nrow(mydata),
I know this has been dealt with before on this list, but the previous
messages lacked detail, and I haven't figured it out yet.
The model is:
\x_{ij} = \mu + \alpha_i + \beta_j
\alpha is a random effect (subjects), and \beta is a fixed effect
(condition).
I have a link function:
p_{ij} = .5
I did not read ?write.table in details about CSV section.
Thanks.
On 8/10/07, Yinghai Deng [EMAIL PROTECTED] wrote:
write.table(mydata.frame, mydata, col.names=NA, quote=F, sep=\t) will
solve the problem.
Deng
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]
If X or Y contains missing values, _you_ supplied missing values as the
'lims' argument and it will be those missing values that are reported.
I do not see how you expect to be able to do density estimation with
missing values: they are unknown and so no part of the answer is known. If
you are
Hi,
I am always with this qustion when I tried to write a data.frame with
row.names and col.names. I have to re-make the data frame to let its
first column be the rownames and let row.names=F so that I can align
the colnames correctly.
Is there a way or option in write.table to automatically do
haha Paul,
It's important not only to post code, but also to make sure that other
people can run it (i.e., include real data or have the code generate
data or use one of R's predefined data sets).
Oh, I hadn't thought of using the predefined datasets, thats a good idea!
Also, isn't this
It is not entirely clear what you are using for y values in
smooth.spline,
but it would appear that it is just the point estimates. I would
suggest
using instead -- at each x value -- a few equally spaced quantiles of
the estimated proportions. Implicitly, smooth.spline expects to be
hi,
maybe I should re-phrase my question a bit:
is there a way to get explicit formulae like Y ~ sum of CiXi from the
model build by lda{MASS} to calculate $x (value) ?
I assume scaling is the coeff and Xi is from test data and Y is $x
called LD1. But I want to confirm this.
Thanks.
Weiwei
Sorry, forgot to attach the graph.
On 8/10/07, Rose Hoberman [EMAIL PROTECTED] wrote:
I am looking for a function that can fit a smooth function to a vector
of estimated proportions, such that the smoothed value is within
specified confidence bounds of each proportion. In other words, given
Is this what you want:
x - matrix(runif(100), 10)
round(x, 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0.268 0.961 0.262 0.347 0.306 0.762 0.524 0.062 0.028 0.226
[2,] 0.219 0.100 0.165 0.131 0.578 0.933 0.317 0.109 0.527 0.131
[3,] 0.517 0.763 0.322 0.374 0.910
Try this where we have constructed the example to illustrate that
it does handle the case where not all values are in each column:
mat - matrix(rep(1:6, each = 4), 6)
table(col(mat), mat)
On 8/10/07, Tom Cohen [EMAIL PROTECTED] wrote:
Dear list,
I have the following dataset and want to
Jim Lemon wrote:
Daniel Brewer wrote:
Thanks for the replies, but I still cannot get what I want. I do not
want the label inside the plot area, but in the top left of the paper, I
suppose in the margins. When I try to use text to do this, it does not
seem to plot it outside the plot area.
Daniel Brewer wrote:
Hi all,
Another plotting question I am afraid. Is there anyway of putting a
legend for a plot in a margin rather than within the figure. I am
trying to plot a 3x2 plot and I want to have:
1) One key along the bottom for all the plots
2) A label (a,b,c) for each plot
Thanks. That got me onto the right track. Because it is a multiplot
and I wanted it along the bottom, I found that I had to use par(xpd=NA)
and then position it relative to the last of the multiplots. After a
bit of trial and error I got there.
Thanks
Lauri Nikkinen wrote:
Very simple
Hi all,
Another plotting question I am afraid. Is there anyway of putting a
legend for a plot in a margin rather than within the figure. I am
trying to plot a 3x2 plot and I want to have:
1) One key along the bottom for all the plots
2) A label (a,b,c) for each plot (see previous emails)
Is
How can we get variance/covariance components in a linear model with
random effects when the response is multivariate?
e.g. variance components estimates are obtained through lme or lmer in
the univariate case but these functions do not seem to extend to the
multivariate case. I'd like to
Gabor Grothendieck wrote:
You could put the numbers inside the bars in which
case it would not add to the height of the bar:
x - 1:5
names(x) - letters[1:5]
bp - barplot(x)
text(bp, x - .02 * diff(par(usr)[3:4]), x)
Indeed, the boxed.labels function makes this pretty easy.
Dear R users,
I have been trying to combine two anova outputs into one single table
(for later publication). The outputs are of different length, and share
only some common explanatory variables.
Using merge() or melt() (from the reshape package) did not work out.
Here are the model outputs
Hello Alberto, hello Felix,
aside of monthplot() and stl(), there is the possibility to use Census
X-12-ARIMA. The program can be downloaded from:
http://www.census.gov/srd/www/x12a/
It should be mentioned that this is *not* a pure R solution, but one can set up
the relevant scripts and
Hi
I have run into a problem and i wonder if anyone has a smart way of doing
this.
For example i have this data frame for 5 different test groups:
Res1 - c(1,5,4,-0.5,3)
Res2 - c(-1,8,2,0,3)
Mean - c(0.5,1,1.5,-.5,2)
MyFrame - data.frame(Res1,Res2,Mean,row.names=c(G1,G2,G3,G4,G5))
where the
I don't understand why one would run a 64-bit version of R on a 2GB
server, especially if one were worried about object size. You can run
32-bit versions of R on x86_64 Linux (see the R-admin manual for a
comprehensive discussion), and most other 64-bit OSes default to 32-bit
executables.
Quoting Greg Snow [EMAIL PROTECTED]:
My original intent was to get the original posters out of the mode of
thinking they want to match what the spreadsheet does and into thinking
about what message they are trying to get across. To get them (and
possibly others) thinking I made the
Hi,
Try whit:
if(time[j] = 18:00:00 23:59:59)
...
...
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 10/08/07, KOITA Lassana - STAC/ACE [EMAIL PROTECTED]
wrote:
Hi R-users,
I have to define a noise level function L and its energy in the various
moment of
On Fri, 10 Aug 2007, Monica Pisica wrote:
Hi,
I have 4 huge tables on which i want to do a PCA analysis and a kmean
clustering. If i run each table individually i have no problems, but if
i want to run it in a for loop i exceed the memory alocation after the
second table, even if i save
[Apologies to Gabor, who I sent a personal copy of the reply
erroneously instead of posting to List directly]
[...]
Perhaps what you really intend is to
take the average over those elements in each row of the first matrix
which correspond to 1's in the second in the corresponding
row of the
Thanks for the replies, but I still cannot get what I want. I do not
want the label inside the plot area, but in the top left of the paper, I
suppose in the margins. When I try to use text to do this, it does not
seem to plot it outside the plot area. I have also tried to use mtext,
but that
Hi,
I am using xcms library to read mass spectrum data. I generate objects from
CDF files using the command line
SME10 - xcmsRaw(SME_10.CDF)
I have 50 CDF files with different name and I don't want to repeat the command
for each one. Is there any option to read all the files and generate
Christoph Scherber wrote:
Dear R users,
I have been trying to combine two anova outputs into one single table
(for later publication). The outputs are of different length, and share
only some common explanatory variables.
Using merge() or melt() (from the reshape package) did not work
see ROCR or accuracy package.
Justin BEM
BP 1917 Yaoundé
Tél (237) 99597295
(237) 22040246
- Message d'origine
De : gallon li [EMAIL PROTECTED]
À : r-help r-help@stat.math.ethz.ch
Envoyé le : Vendredi, 10 Août 2007, 4h15mn 36s
Objet : [R] compute ROC curve?
Hello,
i have
On Fri, 10 Aug 2007, Roberto Olivares Hernandez wrote:
Hi,
I am using xcms library to read mass spectrum data. I generate objects
from CDF files using the command line
SME10 - xcmsRaw(SME_10.CDF)
I have 50 CDF files with different name and I don't want to repeat the
command for each
Henrique Dallazuanna wrote:
Hi,
Try whit:
if(time[j] = 18:00:00 23:59:59)
This code is obviously wrong and does not help for the next few lines in
the questioner's message, please do not post unsensible stuff.
Uwe Ligges
...
...
KOITA Lassana - STAC/ACE wrote:
Hi R-users,
I have to define a noise level function L and its energy in the various
moment of the day by:
if time is between 18:00:00 and 23:59:59 then L[j] - L[j]+5 and W -
10^((L+5)/10)
What kind of object is time? Just a character or some
I am looking for a function that can fit a smooth function to a vector
of estimated proportions, such that the smoothed value is within
specified confidence bounds of each proportion. In other words, given
a small number of trials and large confidence intervals, I would
prefer the function to
mat2-matrix(as.numeric(mat10.25), ncol=1000)
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 10/08/07, Lanre Okusanya [EMAIL PROTECTED] wrote:
Hello all,
I am working with a 1000x1000 matrix, and I would like to return a
1000x1000 matrix that tells me which value
Another couple of things to think about:
You could use the layout function to set up your multiple plots and
include an extra plotting area at the bottom to place the legend in.
If you stick with the solution below then the cnvrt.coords function from
the TeachingDemos package may be useful (will
I have a data.frame with rownames taken from a database. Unfortunately, one
of the rownames (automatically obtained from the DB) is an empty string. I
often do computations on the DB s.t. the answers (rows) are indexed with
respect to the rownames so a computation on a DB record might
[Tom Cohen]
I have the following dataset and want to know how many times each value
occur in each column.
data
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] -100 -100 -100000000 -100
[2,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100
[3,]
Hello!
I am trying to do a smooth with the kde2d function, and I'm getting an error
message about NAs. Does anyone have any suggestions? Does this function
not do well with NAs in general?
fit - kde2d(X, Y, n=100,lims=c(range(X),range(Y)))
Error in if (from == to || length.out 2) by - 1 :
1. matrices are stored columnwise so R is better at column-wise operations
than row-wise.
I am seeing this by my code which contains more t() than
what seems healthy. However, the summaries are patient-wise
over repeated measurements. Out of convention, I am storing
patients in rows and
I am afraid the above example will not work. In original dataset of
Jabez Wilson numerical range is from 0..7.
So try this one:
df-as.factor(c(0,I,II,III,IV,V,VI,VII)[df$area+1])
Hope this is what you want,
Rainer
Henrique Dallazuanna schrieb:
Hi,
df
ht area
1 3203
2 410
Hi,
df
ht area
1 3203
2 4104
3 2302
4 3603
5 1261
6 2802
7 2602
8 2802
9 2802
10 2602
df$area - as.factor(df$area)
levels(df$area) - c(I, II, III, IV)
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 10/08/07,
Dear R Help,
I have a set of data of heights of trees described by area that they are in.
The areas are numerical (0 to 7).
htarea
1 320 3
2 410 4
3 230 2
4 360 3
5 126 1
6 280 2
7 260 2
8 280 2
9 280 2
10 260 2
...
180 450 4
181 90 1
182
Try this:
positions - order(ranks)
On 8/10/07, Tom.O [EMAIL PROTECTED] wrote:
Hi
I have run into a problem and i wonder if anyone has a smart way of doing
this.
For example i have this data frame for 5 different test groups:
Res1 - c(1,5,4,-0.5,3)
Res2 - c(-1,8,2,0,3)
Mean -
This works fine for one plot, but if it is a multiple plot (mfrow=c(2,2)
say) then each individual label is placed in the same position i.e.
absolute top left on the canvas. I would like it top left of each
individual plot.
Thanks anyway. Got any idea how to fix this?
Dan
Paul Murrell wrote:
Thanks. That works if it is only a single plot, but if there are
multiple plots (e.g. par(mfrow=c(2,2))) it confusingly puts the label in
the absolute top left always i.e. the top left of plot one.
Dan
S Ellison wrote:
Try something like
mtext(side=3, line=-1, text=Here again?, adj=0,
Dear Antje
I cannot see that you have got any replies yet, so I will make and
attempt. However, I am sure other have more formally correct solutions.
When you call the pdf(), you can set paper=a4 (or a4r for
landscape). However, the width and the height of your plot should then
not exceed the
Dear list,
I have the following dataset and want to know how many times each value occur
in each column.
data
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] -100 -100 -100000000 -100
[2,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100
[3,]
Hi
[EMAIL PROTECTED] napsal dne 10.08.2007 13:41:53:
Dear R Help,
I have a set of data of heights of trees described by area that they are
in.
The areas are numerical (0 to 7).
htarea
1 320 3
2 410 4
3 230 2
4 360 3
5 126 1
6 280 2
7 260 2
8
Hi,
This is a general statistics question that I believe occurs often so may have
some R functions/packages dedicated to it.
Suppose you want to check the accuracy of a classifier using a large training
data-set where each row represents an observation. Is there a simple approach
for removing
1. matrices are stored columnwise so R is better at column-wise operations
than row-wise.
2. Here is one way to do it (although I am not sure its better than the
index approach):
row.apply - function(f, a, b)
t(mapply(f, as.data.frame(t(a)), as.data.frame(t(b
3. The code for the
Hi
mat-sample(c(-50,0,-100), 100,replace=T)
dim(mat)-c(10,10)
mat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]0000 -500000 0
[2,] -100 -100 -50 -5000 -100 -50 -100 -50
[3,]0 -50 -100 -1000 -50 -10000 -100
Thanks! I will look into ...
I have 4 GB RAM, and i was monitoring the memory with Windows task manager so i
was looking how R gets more and more memory allocation from less than 100Mb
to 1500Mb .
My initial tables are between 30 to 80 Mb and the resulting tables that
incorporate
that was ridiculously simple. duh.
THanks
Lanre
On 8/10/07, jim holtman [EMAIL PROTECTED] wrote:
Is this what you want:
x - matrix(runif(100), 10)
round(x, 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0.268 0.961 0.262 0.347 0.306 0.762 0.524 0.062 0.028
Hello all,
I am working with a 1000x1000 matrix, and I would like to return a
1000x1000 matrix that tells me which value in the matrix is greater
than a theshold value (1 or 0 indicator).
i have tried
mat2-as.matrix(as.numeric(mat10.25))
but that returns a 1:10 matrix.
I have also tried for
Alberto Monteiro wrote:
I have a time series x = f(t), where t is taken for each
month. What is the best function to detect if _x_ has a seasonal
variation? If there is such seasonal effect, what is the
best function to estimate it?
From my own experience, I had the impression that there is
On 10-Aug-07 18:05:50, Lanre Okusanya wrote:
Hello all,
I am working with a 1000x1000 matrix, and I would like to return a
1000x1000 matrix that tells me which value in the matrix is greater
than a theshold value (1 or 0 indicator).
i have tried
mat2-as.matrix(as.numeric(mat10.25))
but
Tom,
If all values (-100,0,-50) would be in every column then simple
apply(data,2,table)
would work. Even if there aren0t all values in every column you could
correct that and insert additional lines with all values for all columns
like
data -
Good day. I am employed at a public entity that handles million
information and records of several variables and distributed in several
topics. For the statistical analyses we use a statistical package, which
allows us to call directly of the database (ORACLE) the information and
to realize
I am running R 2.5.1 using Mac OSX 10.4.10. xlsReadWrite is a Windows
binary. Instead, install and load packages: (1) gtools:(2) gdata. These
are both Windows and Mac binaries. gdata depends on gtools, so be sure
to load gtools first or set the installation depends parameters. Then you
can use
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