See the 'R Internals' manual.
ASCII characters are not marked as Latin-1 nor UTF-8.
On Fri, 7 Nov 2008, Heinz Tuechler wrote:
Dear All,
Encoding() goes beyond my understanding. See the example. I would expect from
reading the help for Encoding() that strsplit preserves the encoding for each
To get output going into the Output tab, your either have to work
through the rkward menus, or you have to wrap up your commands in
rkward-specific code; you can get examples of the code by running any
statistical operation through the menus and checking the code produced
by rkward (it is shown
Thank you Phil and Bert!
I was sure there must be an efficient way using some kind of
indexing trick but totally did not see the as.matrix solution.
Thanks again
Philipp
On Wed, Nov 05, 2008 at 02:59:20PM -0800, Phil Spector wrote:
Philipp -
res = matrix(NA,5,3)
2008/11/7 tedzzx [EMAIL PROTECTED]:
The problem is that: There is some rounding problems, for example
Library(chron)
any(times(4:00:01)==times(4:00:00)+times(00:00:01)))
False
But,it should be true
FAQ 7.31 in disguise!
chron stores date-times as fractions, so you're comparing two
At 09:15 07.11.2008, Prof Brian Ripley wrote:
See the 'R Internals' manual.
Thank you, now I understand a little more.
My real problem, however is a data frame produced
by spss.get(). Is there a simple possibility to
mark all characters in that data.frame (except
ASCII characters),
Hi ,
can some body tell to me how to run a R method /function as a
background process from R interface
Thanks
K.Ravichandra
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
hello,
i am looking for a method to return rgb-values of predifined pixels of
jpg images.
can anybody help me?
thank you very much
best regards
hans-joachim klemmt
--
--
Dr. Hans-Joachim Klemmt
Forstoberrat
Organisationsprogrammierer IHK
Bayerische Landesanstalt
Are you looking for expand.grid?
expand.grid(1:2, letters[1:2])
Petr Pikal
[EMAIL PROTECTED]
724008364, 581252140, 581252257
[EMAIL PROTECTED] napsal dne 07.11.2008 01:53:56:
Dear R Gurus:
How do you put together a 2^2 (or even 2^k) factorial problem, please?
Since you have 2 levels
Heinz Tuechler wrote:
Dear Prof.Ripley!
Thank you very much for your attention. In the given example Encoding(),
or the encoding parameter of read.csv solve the problem. I hope your
patch will solve also the problem, when I read a spss file by
spss.get(), since this function has no encoding
Simone Gabbriellini wrote:
Tom,
I don't know if there are better ways, but this is the way I do:
I use Python for building the AB model, and RPy as an interface to R for
statistical analysis.
One of the best package for SNA in R is igraph, which has a nice Python
version.
But if you prefere
Dear all,
I am trying to bootstrap predictions from gnls models using the following code:
# a is the dataframe with which I am working; it contains the variables
# response.variable,LD,L,G,P and F
###
model=gnls(response.variable ~ a * LD/(b + LD),
params = list(a + b ~ L), start =
Colleagues,
I submitted this several days ago and no one responded, so I am trying
again, trying a different subject line:
I just encountered some unexpected behavior of difftime in
relationship to the change from daylight savings to standard time.
My understanding is that DST and ST take
On 11/7/2008 8:31 AM, roger koenker wrote:
Those of you with an interest in the US election and/or
statistical graphics may find the maps at:
http://www-personal.umich.edu/~mejn/election/2008/
interesting.
Nice stuff. Do you know if anyone has ported the cartogram code to R?
I see
Try this:
func - function(f, ...) f(...)
# e.g.
func(sin, 0) # same as sin(0)
func(max, 1, 2) # same as max(1, 2)
On Fri, Nov 7, 2008 at 5:21 AM, [EMAIL PROTECTED] wrote:
How can I apply function f, that I get as an argument as in
func - function(f, ...) {
.
.
.
}
to a list of
Tom,
I don't know if there are better ways, but this is the way I do:
I use Python for building the AB model, and RPy as an interface to R
for statistical analysis.
One of the best package for SNA in R is igraph, which has a nice
Python version.
But if you prefere statnet (which is great
I know that there are two method to apply the Hosmer and Lemeshow’s. One of
them is calculated based on the fixed and pre-determined cut-off points of
the estimated probability of success. One of them is calculated based on
the percentiles of estimated probabilities.
In the previous post,i
At 13:34 07.11.2008, Peter Dalgaard wrote:
Heinz Tuechler wrote:
Dear Prof.Ripley!
Thank you very much for your attention. In the given example Encoding(),
or the encoding parameter of read.csv solve the problem. I hope your
patch will solve also the problem, when I read a spss file by
On Fri, Nov 7, 2008 at 7:53 AM, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 11/7/2008 8:31 AM, roger koenker wrote:
Those of you with an interest in the US election and/or
statistical graphics may find the maps at:
http://www-personal.umich.edu/~mejn/election/2008/
interesting.
Dear list,
I have to draw a simple plot. On y axe some numerical values that correspond
to various categories on axe x.
The table I am reading looks like:
cat Obj1 Obj2 Obj3
max 23 27 34
ave 21 25 32
min 19 23 30
In order to avoid that the first column is reordered alphabetically I used:
(found
Hello,
I am stuck when I want to add axes labels to my scatterplot with histograms.
I guess it must be something with par(mar=) or so, but could someone give
me a hint?
Here is what I got so far:
# adapted from
http://www.stat.ucl.ac.be/ISdidactique/comment/fichiers/r/scatterhist.rs and
from
On Fri, 7 Nov 2008, Peter Dalgaard wrote:
Heinz Tuechler wrote:
Dear Prof.Ripley!
Thank you very much for your attention. In the given example Encoding(),
or the encoding parameter of read.csv solve the problem. I hope your
patch will solve also the problem, when I read a spss file by
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Conditional logistic regression and small sample sizes: what to do?
Dear R-Gurus,
[I run R 2.8.0 on WinXP. I have no formal training in statistics.]
Please feel free to skip the 'Blah blah part' for questions below.
- ---Blah blah part
Why
On Fri, 7 Nov 2008, Dennis Fisher wrote:
Colleagues,
I submitted this several days ago and no one responded, so I am trying
again, trying a different subject line:
Well, you posted something that indicated you had not studied the relevant
help pages, without the information requested in the
On Fri, 7 Nov 2008, Christoph Scherber wrote:
Dear all,
I am trying to bootstrap predictions from gnls models using the following
code:
# a is the dataframe with which I am working; it contains the variables
# response.variable,LD,L,G,P and F
And without it your code is not reproducible.
On 11/7/2008 5:40 AM, baptiste auguie wrote:
perhaps something like,
func - function(f, ...) {
do.call(f, ...)
}
func(rnorm, list(n=3, mean=2, sd=3))
Alternatively, if the caller doesn't want to put the args in a list,
your func can do it:
func2 - function(f, ...) {
do.call(f,
Hello,
Two approaches depending when you want to trigger this background
calculation:
1) It is enough to trigger the background computation after each
top-level instruction entered at the command line: use ?addTaskCallback
2) You want to trigger the background calculation at a given time,
I'm having trouble running `updates.packages()' and installing into a
personal library.
Setup:
1. .Renviron file contains: R_LIBS_USER=~/lib/R/%p-library/%v
2. Hence personal library is: ~/lib/R/i486-pc-linux-gnu-library/2.6/
3. Library path upon starting R:
.libPaths()
[1]
Thanks a lot Keith : your function does the job (but I don't understand the
trick) !
Have a nice week-end (Thanks to you, I can have a good one :-)),
Ptit Bleu.
-
median.data.frame - function(x, ...)
sapply(x, median, ...)
I haven't tried it, but it might work
hth
Keith
Dear all,
Here comes a reproducible example, with the original data added.
The error message when running boot() is:
Error in gnls(response.variable ~ a * LD/(b + LD), params - list(a + :
Step halving factor reduced below minimum in NLS step
boot() in this case only seems to work for very
Unfortunately, I have the same error message.
lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not
with median.
Strange, isn't it?
Any other idea?
Thanks in advance,
Ptit Bleu.
Henrique Dallazuanna wrote:
Try this:
lapply(l, function(x)median(x[,sapply(x,
Hello,
sometimes it helps if I formulate my problem. I tried the following after I
posted my previous post.
scatterhist=function(x,y, xlab=, ylab=){
#oldpar = par(no.readonly = TRUE) # save default, for resetting...
zones=matrix(c(2,0,1,3),ncol=2,byrow=TRUE)
I am simulating sickness among a group of families. Part of the task is
to randomly draw who in the family will be sick, randomly drawing from
family ID's where Dad =1, Mom = 2, Kid1 = 3, Kid2 = 4., etc. My census of
Dads is of the form shown below.
Dad_ID Spouse (Y=1;N=0)
Maybe this?
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
At 11:23 07.11.2008, Shubha Vishwanath Karanth wrote:
Content-Type: text/plain
Content-Disposition: inline
Content-length: 569
Hi R,
I have certain checkings, which gives FALSE,
Dear R Users,
May be this message should be directy send to Douglas Bates ...
I just want to know if I can use the AIC value given in the output of an lmer
model to classify my logistic models.
I heard that the AIC value given in GLIMMIX output (SAS) is false because it
come from a calculation
Hi,
I'm trying to fit a proportional ordinal logistic model using function
polr() (package MASS).
Is there a way to fix certain betas in the regression (e.g. function
arima() allows this by defining fixed )
Maybe there is another function than polr() which allows that?
Thanks
Kazys
Test.
On Fri, Nov 7, 2008 at 2:40 AM, Alan Lue [EMAIL PROTECTED] wrote:
I'm having trouble running `updates.packages()' and installing into a
personal library.
Setup:
1. .Renviron file contains: R_LIBS_USER=~/lib/R/%p-library/%v
2. Hence personal library is:
Kurapati, Ravichandra \(Ravichandra\)
[EMAIL PROTECTED] napsal dne 07.11.2008 11:06:28:
Hi
I want NA counts also.
Well
barley[5:15, 1] - NA
barchart(yield ~ variety | site, data = barley,
groups = year, layout = c(1,6),
ylab = Barley Yield (bushels/acre),
Dear R users,
Beside lattice and vcd packages, what is my option if I want to create a
grouped and stacked categorical plots?
With lattice, what I usually do is breaking my data from a wide form to a
long form (using subset and rbind), such as:
TypeID - Var 1 - Var 2 - Var 3 - Val
to
On 07-Nov-08 00:53:56, Edna Bell wrote:
Dear R Gurus:
How do you put together a 2^2 (or even 2^k) factorial problem, please?
I'm not clear about what you mean here, especially about put together!
Can you describe what you want to end up with in more detail?
Since you have 2 levels for A and
Dear Martin,
You can use the same idea of concatenating the data in R. The following
reproduces your example:
spec - c(asfe, dias)
n1 - length(asfe)
n2 - length(dias)
CYT - c(28 * CYTF, 24.6 * 2 * CYTB6)
B559HP - c(B559HP, numeric(n2))
B559LP - c(numeric(n1), B559LP)
C550 - c(numeric(n1), C550)
Hi
I did not see any response and will not give you any answer too, just a
hint. Your code can not be reproduced and it is also difficult to say what
you really want.
Eg. this is rather confusing.
On a generated plot I observed that what ever records have counts as
NA, those are not
Hi,
I'm trying to plot a map of the pacific ocean, centered on the dateline, using
the maps package.
library(maps) # Basic library to draw maps
library(mapdata)# Library with specialized maps
library(mapproj)
map(database = world, fill = TRUE, col = 1, plot = TRUE,add=F,
xlim = c(120,300),
perhaps something like,
func - function(f, ...) {
do.call(f, ...)
}
func(rnorm, list(n=3, mean=2, sd=3))
baptiste
On 7 Nov 2008, at 10:21, [EMAIL PROTECTED] wrote:
How can I apply function f, that I get as an argument as in
func - function(f, ...) {
.
.
.
}
to a list of arguments
On 11/7/2008 12:00 PM, Stephen Collins wrote:
I am simulating sickness among a group of families. Part of the task is
to randomly draw who in the family will be sick, randomly drawing from
family ID's where Dad =1, Mom = 2, Kid1 = 3, Kid2 = 4., etc. My census of
Dads is of the form shown
On 11/7/2008 12:08 PM, [EMAIL PROTECTED] wrote:
Hi,
I'm trying to plot a map of the pacific ocean, centered on the dateline, using
the maps package.
library(maps) # Basic library to draw maps
library(mapdata)# Library with specialized maps
library(mapproj)
map(database = world, fill = TRUE,
On Fri, Nov 7, 2008 at 3:32 AM, Barry Rowlingson
[EMAIL PROTECTED] wrote:
2008/11/7 tedzzx [EMAIL PROTECTED]:
The problem is that: There is some rounding problems, for example
Library(chron)
any(times(4:00:01)==times(4:00:00)+times(00:00:01)))
False
But,it should be true
FAQ 7.31 in
How can I apply function f, that I get as an argument as in
func - function(f, ...) {
.
.
.
}
to a list of arguments list(a, b, c) (eg the ... argument of func above)
in order to obtain
f(a, b, c)
Thanks a lot,
Roberto
[[alternative HTML version deleted]]
You can provide a example of your data?
On Fri, Nov 7, 2008 at 9:14 AM, Ptit_Bleu [EMAIL PROTECTED] wrote:
Unfortunately, I have the same error message.
lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not
with median.
Strange, isn't it?
Any other idea?
Thanks in
I haven't looked at the detail, but I guess the answer is that mean works on
a data frame while median doesn't.
?mean
snip
For a data frame, a named vector with the appropriate method being applied
column by column.
-
I guess to use median you'll need nested '[l/s]apply's, the
Hi,
I'm not quite sure I understood everything but is this something close?
d - read.table(textConnection(Dad_ID SpouseYN NKids NSick
1 10 1
2 02 2
3 10 2
4 13 3), header=TRUE)
mapply(sample,
hi there
I have a vector with a set of data.I just wanna seperate them based on the
first p and q values metioned within the data.
[1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
[2] chr1q22-q24
[3] chr1q22-q24
[4]
Hi, I'm testing the efficiency of the Rmpi package regarding parallelization
using a cluster.
I've found and tried the task pull programming method, but even if it is
described as the best method, it seems to cause deadlock, anyone could help
me in using this method?
here is the code I've found
Rajasekaramya wrote:
hi there
I have a vector with a set of data.I just wanna seperate them based on the
first p and q values metioned within the data.
[1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
[2] chr1q22-q24
[3] chr1q22-q24
Hi: Is there a shorter way to color the abstract,sections subsections and page
headings of a document. I am using the code below in my preamble to accomplish
some of it:
[EMAIL PROTECTED],dvips]{color}
\else\usepackage[usenames,dvipsnames]{color}
% and fix pdf colour problems
Hello,
I have a list of data.frame
rowsplit : List of 15
$ (0,0.025] :'data.frame': 169 obs. of 7 variables:
$ (0.025,0.05]:'data.frame': 174 obs. of 7 variables:
$ (0.05,0.075]:'data.frame': 92 obs. of 7 variables:
$ (0.075,0.1] :'data.frame': 76 obs. of 7 variables:
$
Hi,
imagine the following two matrix:
Matrix A:
a b c
1 2 3
4 5 6
7 8 9
Matrix B:
a
4
7
I would like to remove those rows from matrix A which are present in both
matrices.
So after removing the corresponding rows the matrix A should look like this:
Matrix A:
a b c
1 2 3
Thanks in advance!
--
Dear all,
I would like to get standard errors (or confidence intervals) for *predicted*
values from an nls fit.
I have tried to implement code from p.225 in MASS (bootstrapping a nls fit), but this gives only the
confidence intervals of the parameter estimates, but not an overall confidence
Berwin A Turlach wrote:
rant on
That's the problem with introductory textbook whose author think they
do the students a favour by using notation as z_alpha, z_0.01,
z_(alpha/2) instead of z_(1-alpha), z_0.99, z_(1-alpha/2),
respectively. In my opinion this produces in the long run only
Do anyone know anything about the use of R for agent-based social
simulation? It should be possible, and would be convenient for the
simple reason that there are several nice packages containing useful
stuff for SNA (Social Network Analysis). Information about packages,
web sites,
On Fri, 7 Nov 2008, [EMAIL PROTECTED] wrote:
Hi,
I'm trying to fit a proportional ordinal logistic model using function
polr() (package MASS).
Is there a way to fix certain betas in the regression (e.g. function
arima() allows this by defining fixed )
Maybe there is another function than
Martin Elff wrote:
Hi Tom,
my package 'memisc' contains a sort of an infrastructure for doing
simulations. As a fun exercise I also used it to create a 'toy' agent based
simulation of Schelling's neighbourhood model. Although it is not a serious
application, at least it shows that agent
Hi R,
I have certain checkings, which gives FALSE, but actually it is true. Why does
this happen? Note that the equations that I am checking below are not even the
case of recurring decimals...
1.4^2 == 1.96
[1] FALSE
1.2^3==1.728
[1] FALSE
Thanks in advance, Shubha
Those of you with an interest in the US election and/or
statistical graphics may find the maps at:
http://www-personal.umich.edu/~mejn/election/2008/
interesting.
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
On 11/7/2008 3:24 AM, Hans-Joachim Klemmt wrote:
hello,
i am looking for a method to return rgb-values of predifined pixels of
jpg images.
can anybody help me?
See the rimage package, with function read.jpeg. It will read the whole
file into a large array, with separate red, green, blue
I'm trying to snow working with openmpi and sge. Everything appears
to work, except, the program only runs on 2 node. If i tell snow to
run on 9 nodes, it spawns 9 processes on a single node. I'm at a loss
of ideas. Any help would be appreciated.
Setup: Rocks Cluster, OpenMPI, R 2.8.0, Snow,
Is the following intended or not?
func- function(y) match.call()
z - func(y =2)
z
func(y = 2)
z[[a]] - 5
z
func(y = 2, 5) ## Note that the second argument **is not** named
## BUT...
z - func(y =2)
z$a - 5
z
func(y = 2, a = 5) ## The second argument **is** named
### End of
On Fri, Nov 7, 2008 at 4:02 PM, hadley wickham [EMAIL PROTECTED] wrote:
The source code (in C) for this type of cartogram (Diffusion-based
method for producing density equalizing maps) is available from here:
http://www-personal.umich.edu/~mejn/cart/download/
From the documentation [1]:
If you
Forgive the spam. Let me try that e-mail again.
I'm trying to get snow working with openmpi and sge. Everything
appears to work, except the program only runs on 1 node. If i tell
snow to run on 5 nodes, it spawns 5 processes on a single node, 6
nodes = 6 procs on 1 node and so on. I'm at a
Hi,
I'm dealing with a lattice plot inserted into a tk widget and would like
to know when a user has clicked on the plot area of a plot (i.e. inside
the axes). For example,
library(tkrplot)
library(lattice)
tt - tktoplevel()
makePlot - function() print(xyplot(1 ~ 1))
printCoords -
I am trying to simplify my code by adding a for loop that will load and
compute a sequence of code 10 time. They way i run it now is that the same
8 lines of code are basically reproduced 10 times. I would like to replace
the numeric value in the code (e.g. Bin1, Bin2Bin10) each time the
Dear R-users
Thanks to Jose Pinheiro, Douglas Bates and coworkes for providing R with the
nlme package.
Could someone help me, please, to specify a correct random formula for a
mixed model, that specifies no random effect on a higher level?
I have the following dataset of timeseries of
So, I have training data, and testing data
however, when I try to predict values for the testing data, it gives me
values for the training data. what gives?
the following is my code:
train = read.table(train.txt, header = TRUE, sep = )
test = read.table(test.txt, header = TRUE, sep = )
Hi guys,
I am truing to draw a circle and choose 10 points on the circle and put a
number from 1 to 10 above each point. I don't have a problem with drawing a
circle and choosing 10 points. my problem is with numbering . is it even
possible to put a number above each chosen point? if so any idea
Hi,
there is a new mailing list for R and HPC: [EMAIL PROTECTED]
This is probably a better list for your question. Do not forgett, first
of all you have to register: https://stat.ethz.ch/mailman/listinfo/r-sig-hpc
Did you tried to run openmpi and R without SGE?
Like this: orterun -n 1 R
?text
On Fri, Nov 7, 2008 at 12:42 PM, Alex99 [EMAIL PROTECTED] wrote:
Hi guys,
I am truing to draw a circle and choose 10 points on the circle and put a
number from 1 to 10 above each point. I don't have a problem with drawing a
circle and choosing 10 points. my problem is with numbering .
Hi,
there is a new mailing list for R and HPC: [EMAIL PROTECTED]
This is probably a better list for your question. Do not forget, first
of all you have to register: https://stat.ethz.ch/mailman/listinfo/r-sig-hpc
I tried your code and it is working!
Please send us your sessionInfo() output.
# I would put this in a list in the following manner
Bin - lapply(1:10, function(.file){
#---
#Loads bin data frame from csv files with acres and TAZ data
fileName -
paste(I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin,
.file,
On Friday 07 November 2008 (19:45:04), you wrote:
Martin Elff wrote:
Hi Tom,
my package 'memisc' contains a sort of an infrastructure for doing
simulations. As a fun exercise I also used it to create a 'toy' agent
based simulation of Schelling's neighbourhood model. Although it is not a
I estimated a negative binomial model using zelig.
z.out- zelig(NEWBHC~ PW80 + CHNGBLK + XBLK,data=data, model=negbin)
How do I calculate predicted probabilities for this model? Is it the same
process as a poisson regression?
Thanks in advance
Joe
[[alternative HTML version deleted]]
Hello,
I have some rather large matrices. Is there a way (without having to
loop) to cap all the values of a data frame to a given ceiling?
E.g.
junk - cbind(c(1,2,3,4,5),c(2,4,6,8,10))
junk
[,1] [,2]
[1,]12
[2,]24
[3,]36
[4,]48
[5,]5 10
replace
Assuming the data frame is all numeric:
DF[] - pmax(10, unlist(DF))
On Fri, Nov 7, 2008 at 4:16 PM, Grey Moran [EMAIL PROTECTED] wrote:
Hello,
I have some rather large matrices. Is there a way (without having to
loop) to cap all the values of a data frame to a given ceiling?
E.g.
junk -
Are they objects of class matrix or data.frame? You seem to make
reference to both, but know there is a difference.
Try:
junk[junk 5] - 5
Grey Moran wrote:
Hello,
I have some rather large matrices. Is there a way (without having to
loop) to cap all the values of a data frame to a given
Dear Heather!
Thank you very, very much for your solution which also reproduces exactly
the results I get from the SigmaPlot analysises. :)
Martin
Heather Turner schrieb:
Dear Martin,
You can use the same idea of concatenating the data in R. The following
reproduces your example:
That should be pmin:
On Fri, Nov 7, 2008 at 4:25 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Assuming the data frame is all numeric:
DF[] - pmax(10, unlist(DF))
On Fri, Nov 7, 2008 at 4:16 PM, Grey Moran [EMAIL PROTECTED] wrote:
Hello,
I have some rather large matrices. Is there a
Thanks to all who replied - lots of good ideas.
The one I prefered at the end was:
junk[junk 5] - 5
Grey
On Fri, Nov 7, 2008 at 4:38 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
That should be pmin:
On Fri, Nov 7, 2008 at 4:25 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Assuming the
On Fri, Nov 7, 2008 at 4:46 PM, Grey Moran [EMAIL PROTECTED] wrote:
Thanks to all who replied - lots of good ideas.
The one I prefered at the end was:
junk[junk 5] - 5
Grey
On Fri, Nov 7, 2008 at 4:38 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
That should be pmin:
On Fri, Nov 7,
A few more. These each have the advantage of not
destroying the original data frame:
# based on Erik's
DF2 - replace(DF, DF 10, 10)
# based on my previous one
DF2 - replace(DF, TRUE, pmin(10, unlist(DF)))
DF2 - (DF + 10)/2 - abs(DF - 10)/2
On Fri, Nov 7, 2008 at 4:38 PM, Gabor Grothendieck
__
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leo_wa wrote:
I know that there are two method to apply the Hosmer and Lemeshow’s. One of
them is calculated based on the fixed and pre-determined cut-off points of
the estimated probability of success. One of them is calculated based on
the percentiles of estimated probabilities.
Both of
Hi,
I am trying to get R 2.8.0 for Mac OS from CRAN, but I thing I am
doing something wrong because when R starts I have annoying error
messages:
During startup - Warning messages:
1: Setting LC_CTYPE failed, using C
2: Setting LC_COLLATE failed, using C
3: Setting LC_TIME failed, using C
4:
Peter Dalgaard wrote:
Rajasekaramya wrote:
hi there
I have a vector with a set of data.I just wanna seperate them based on the
first p and q values metioned within the data.
[1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
[2] chr1q22-q24
[3]
Hi.
i have a data, and there is 3 columns, Month, Year and Total. and there is
over 1000 rows for them because there is 87 years data for every month, so
there is month from Jan-Dec, and year from 1900-1987,
so i was wondering if i would want to make 12 groups (Jan,Feb...,Dec),
and put each
I am trying to combine two data sets, one with daily values and one with weekly
values. SurveyData conatins environmental data collected on a daily basis.
sat.data contains satellite sea surface temperature that is an average of
satellite measurements over a six day period. I would like to
Wacek Kusnierczyk wrote:
Peter Dalgaard wrote:
Rajasekaramya wrote:
hi there
I have a vector with a set of data.I just wanna seperate them based on the
first p and q values metioned within the data.
[1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
[2] chr1q22-q24
Wacek Kusnierczyk wrote:
Rajasekaramya wrote:
hi there
I have a vector with a set of data.I just wanna seperate them based on the
first p and q values metioned within the data.
[1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
[2] chr1q22-q24
Here are a few more solutions. x is the input vector
of character strings.
The first is a slightly shorter version of one of Wacek's.
The next three all create an anonymous grouping variable
(using sub, substr/gsub and strapply respectively)
whose components are p and q and then tapply
is used
Hello dear R people,
for my MSc thesis I need to program some functions, and some of them
simply do not work. In the following example, I made sure both vectors
have the same length (10), but R gives me the following error:
Error in if (vector1[i] == vector2[j]) { :
missing value where
2008/11/8 Bert Gunter [EMAIL PROTECTED]:
Is the following intended or not?
func- function(y) match.call()
z - func(y =2)
z
func(y = 2)
z[[a]] - 5
z
func(y = 2, 5) ## Note that the second argument **is not** named
## BUT...
z - func(y =2)
z$a - 5
z
func(y = 2, a = 5) ## The
is this what you want?
vector1
[1] 65 1 34 100 42 20 79 43 89 10
vector2
[1] 34 65 47 91 48 32 23 74 92 86
for (i in 1:10) {
+ for (j in 1:10) {
+ if (vector1[i] == vector2[j])
+ show(c(i,j))
+ }
+ }
[1] 1 2
[1] 3 1
On Sat, Nov 8, 2008 at
hi friend,
this is from your previous posts on Kruskal-Wallis test:)
i came up with this one:
A5 - read.table('kew.dat' ,header=TRUE)
plot(factor(A5$Month, levels=month.abb), A5$Rain)
is that what you want?
On Sat, Nov 8, 2008 at 7:03 AM, Swanton0822 [EMAIL PROTECTED] wrote:
Hi.
i have a
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