Or with mapply
name - c(foo, bar, baz)
val - 1:3
mapply(assign, name, val, pos=1)
Cheers,
Simon.
On Tue, 2008-06-03 at 22:24 -0700, Moshe Olshansky wrote:
You can use either assign or eval.
Something like
name - c(foo, bar, baz)
for (i in 1:length(name)) assign(name[i],i)
---
Thanks for your reply.
In your last step If I create a duplicate ( two similar records )
# create a duplicate
vv[8,1] - 7
vv[8,2]-'g'
and then I merge vv with vv2 ,both duplicates are merged. Is there a way to
tell R to merge only the unique records.
--
I don't think merge will do what
I am troubled by what appears to be a glitch in the current
distribution, or in
its installation on our system. I've traced it, and found a work-
around. Is
this normal? Is there a cleaner solution?
The problem:
During a package installation, the warning message WARNING: ignoring
This indicates a serious problem with your build (a segfault).
Unfortunately as R is very well tested on Linux and we have never seen
this one reported, we have no clue as to why. You've told us very little
(what architecture, what compilers?) so although it seems to be something
specific to
On Tue, 3 Jun 2008, Duncan Murdoch wrote:
On 03/06/2008 6:53 PM, Chin-Cheng Su wrote:
Hi everyone,
I use R to plot my data set (x,y) in Windows XP, but keep getting error
message like follows,
...
plot(x, y)
Error in windows() : unable to start device devWindows
How can I solve this
Hello,
I have data at 10 locations, in each location there are time series
(T=56). Question is: when Im fitting variogram what happens with those
measures in each location? Are they taken as repeated measures? It's very
important for my to know this
Thanks a lot
ChCh jmo101 at student.canterbury.ac.nz writes:
Hello all,
I've become confused by the output produced by a call to
anova(model1,model2). First a brief background. My model used to predict
final tree height is summarised here:
Df Sum Sq Mean Sq F value
Dear all,
I would like to create an random time serie of POSIXct elements inside a
certain range of dates and time.
For example a sequences of date/time of 400 elements, randomly choosen,
with all dates between 2002-12-01 and 2002-12-31 and all time between
07:00:00 and 23:00:00.
Would anyone
Thomas
As far I can see eig (matlab) and eigen (R) both uses LAPACK as the engine, so
no differences apart from rounding errors should be expected.
You are not being particular friendly, since you have not told the whole story
about how you use eigs in matlab (you are giving some options to
My dataset contains missing data and I would like to do something like an EM
algorithm or a Markov Chain Monte Carlo approach to get rid of the missing
data.
Is there a function for imputation or simulation of missing data apart from
those in the randomForest library?
Thanks in advance
Birgit
Hi
the other possible option is to use split with suitable graphing technique
or with lapply and appropriate function
e.g.
boxplot(split(X, Y0))
histogram( ~ X|Y)
gives Marvin
[EMAIL PROTECTED] napsal dne 03.06.2008 21:11:29:
?by may be helpful here
eg if dat is your data.frame and yf is
Hi
use
str(object)
to see what is a structure of object
then use standard selecting by [..., ...]
In your case you shall probably use extracting function
coef(fit1)[1,]
for the first row and similarly for the second.
Regards
Petr
[EMAIL PROTECTED]
724008364, 581252140, 581252257
[EMAIL
I'm new to R so please forgive the newbie question; but i can't seem to
find a definitive answer to this.
I am wanting to do a PLS regression on some data. Which takes a formula
of the type
responseACC ~ dataACC
where dataACC is multivariate in nature and responseACC is a single value.
I
On 6/4/2008 5:32 AM, Birgitle wrote:
My dataset contains missing data and I would like to do something like an EM
algorithm or a Markov Chain Monte Carlo approach to get rid of the missing
data.
Is there a function for imputation or simulation of missing data apart from
those in the
Birgit,
not knowing your data, I would recommend R-package mice or function
aregImpute from R-package Hmisc as good multi-purpose tools.
Regards, Ulrike
--
View this message in context:
http://www.nabble.com/missing-data-imputation---simulation-tp17642736p17643601.html
Sent from the R help
Sorry for the noise -- but I want to see what happens
to this message when it gets to tyhe archives.
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 04-Jun-08
Many thenks to both of you:
Will have a look.
Birgit
Chuck Cleland wrote:
On 6/4/2008 5:32 AM, Birgitle wrote:
My dataset contains missing data and I would like to do something like an
EM
algorithm or a Markov Chain Monte Carlo approach to get rid of the
missing
data.
Is there a
Dear R users,
Suppose I have two different response variables y1, y2 that I regress separately on the same
explanatory variable, x; sample sizes are n1=n2.
Is it legitimate to compare the regression slopes (equal variances assumed) by
using
lm(y~x*FACTOR),
where FACTOR gets y1 if y1 is the
Hi
[EMAIL PROTECTED] napsal dne 04.06.2008 10:57:26:
I'm new to R so please forgive the newbie question; but i can't seem to
find a definitive answer to this.
I am wanting to do a PLS regression on some data. Which takes a formula
of the type
responseACC ~ dataACC
where dataACC is
François Aucoin:
I have tried several R's functions for optimization but the results I
yield are not correct.
Is there anybody who can help me?
I couldn't get it to estimate the correct values either, so I guess either
your random number generator 'rkappa' is wrong, or your 'Neg.Log.Like'
Hi all,
I've been trying to include a nice example in the Rd file for a function
and cannot work out how to get it through R check. If I use:
if(require(maps,quietly=TRUE)) {
...
Warning message:
'Sys.putenv' is deprecated.
Use 'Sys.setenv' instead.
See help(Deprecated)
The warning message
I am trying to define groupings from levels of factor variables and this the
warning message that R give
not meaningful for factors.
The nature of my task is this. I have a variable stage which has the levels
(1B, 2A, 2B) - these are the AJCC TNM stages of cancer, and another variable
When I run your code, I get a different error:
j=10; ss=150; r=array(0 , c( j , ss )); rr=array(0 , c( j , ss ));
r1=array(0 , c( j-1 , ss )); r2=array(0 , c( j-1 , ss ));
r3=array(0 , c( 2 , ss ))
for(i in 1:j-1){
+r1[ i , ] - r[ j+1, ]-r[ j, ]; r2[ i , ] - rr[
j+1,
What exactly are you trying to do? In the first case you are making a
logical comparison and that is legal for . In the second you are trying
to do a logical operation () between two factors and that operation is
not defined. This is what the error message is saying.
Also you first attempt is
On 6/4/2008 8:09 AM, Jim Lemon wrote:
Hi all,
I've been trying to include a nice example in the Rd file for a function
and cannot work out how to get it through R check. If I use:
if(require(maps,quietly=TRUE)) {
...
Warning message:
'Sys.putenv' is deprecated.
Use 'Sys.setenv' instead.
It works. Thanks Jim. I guess this will be a lots of coffee morning kinda day.
jim holtman [EMAIL PROTECTED] wrote: What exactly are you trying to do? In
the first case you are making a logical comparison and that is legal for .
In the second you are trying to do a logical operation ()
Laura Saltyte wrote:
Hello,
I have data at 10 locations, in each location there are time series
(T=56). Question is: when I’m fitting variogram what happens with those
measures in each location? Are they taken as repeated measures? It's very
important for my to know this
Thanks a lot
Hi
Dear Christoph,
If I've understood properly what you intend to do, no, it doesn't make
sense. I assume from your description that you don't have two independent
samples, but rather you really have n observations and that the variables x,
y1, and y2, are measured on the same individuals. If that's
Hi R,
I have a list of matrices. I need to get the sum of all the matrices in
the list.
Example:
a=b=c=d=matrix(1:4,2,2)
l=list(a,b,c,d)
I need:
a+b+c+d
[,1] [,2]
[1,]4 12
[2,]8 16
Something like do.call(+,l) is not working...why is this?
I may not
Thanks Greg.
A useful solution.
Birgit
Am 02.06.2008 um 23:13 schrieb Greg Snow:
An alternative way to draw the symbols (or some approximation of
them) is to use the my.symbols function from the TeachingDemos
package along with ms.male and ms.female (or your improvement of
these, also
try a simple for loop, it will be fast enough in this case, e.g.,
matSums - function (lis) {
out - array(data = 0, dim = dim(lis[[1]]))
for (i in seq(along = lis))
out - out + lis[[i]]
out
}
a - b - c - d - matrix(1:4, 2, 2)
l - list(a, b ,c, d)
matSums(l)
I hope it helps.
Shubha Vishwanath Karanth wrote:
I need:
a+b+c+d
[,1] [,2]
[1,]4 12
[2,]8 16
Something like do.call(+,l) is not working...why is this?
Because do.call constructs a function call with the elements of l as
arguments, so you end up with:
+(1:4, 1:4, 1:4, 1:4)
but
Dear Shubha,
This problem was coincidentally used as an illustration in the Help Desk
column in the current R News.
Actually, the brute-force method of using a loop to accumulate the sum works
quite well; a more elegant alternative, recently brought to my attention by
Kurt Hornik, uses the
Thanks all...Reduce() is the new function I learnt today... Thanks...
BR, Shubha
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
-Original Message-
From: Barry Rowlingson
G'day Shubha,
On Wed, 4 Jun 2008 20:23:35 +0530
Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote:
Something like do.call(+,l) is not working...why is this?
Well, as the error message says, + is either a unary or a binary
operator, i.e. it takes either one or two arguments, but not more.
I
Dear R-help,
I am trying to create groupedData objects using the nlme library. I'm
missing something basic, I know:
Here is the first example in ch.1 of Pinheiro Bates (2000):
library(nlme)
x2=Rail$travel;x1=Rail$Rail;eg1=data.frame(x1,x2);eg1gd=Rail
print(eg1gd)
x11();print(plot(eg1gd))
Bill Venables responded and got me what i wanted with this:
[EMAIL PROTECTED] wrote:
cv - scan(
textConnection(CW-W730 CW-W720 CW-W710 CW-W700 CW-W690 CW-W680
CW-W670 CW-W660 CE-W997 CE-W987 CE-W977 CE-W967
CE-W956 CE-W944 CE-W934 CE-W924 7W-W760
Dear R forum,
I have a concern regarding a mathematical expression of the probability
function (see below). I know how to write it for only one index i, but we have
two : i (country) and j (year). We have a set of N observations of country year
ij (or ith country in jth year).
Basically, I
HI Toby,
I think that the problem is that you have not specified using Rail and
travel in the model statement, but rather are still using x1 and x2.
It also does not realize that you want x1 to be a factor. Try
x1=factor(c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6))
Mark,
Others have given answers to the question that you asked, in the spirit of
fortune(108) I am going to answer some of the questions that you should have
asked:
Q1: Should I ever do this?
Short answer: No
Less short answer: probably not
Longer answer: You should only do this once you fully
Toby Marthews Toby.Marthews at lsce.ipsl.fr writes:
Dear R-help,
I am trying to create groupedData objects using the nlme library. I'm
missing something basic, I know:
Here is the first example in ch.1 of Pinheiro Bates (2000):
...
...but, as you can see, the coefficients I get at
Hi Dieter,
I never really used groupedData objects at all. I can imagine that if
you become even more immersed in nlme and lattice, groupedData could
reduce your typing, and sometimes simplify your life, but it was not a
hurdle I chose willingly.
Hank
On Jun 4, 2008, at 1:15 PM, Dieter
Hi,
Are there any functions in R that could be used to estimate the phase-shift
between two semi-sinusoidal vectors? Here is what I have tried so far, using
the spectrum() function -- possibly incorrectly:
# generate some fake data, normalized to unit circle
x - jitter(seq(-2*pi, 2*pi,
On Wednesday 04 June 2008, Dylan Beaudette wrote:
Hi,
Are there any functions in R that could be used to estimate the phase-shift
between two semi-sinusoidal vectors? Here is what I have tried so far,
using the spectrum() function -- possibly incorrectly:
# generate some fake data,
Hi,
Are there any functions in R that could be used to estimate the phase-shift
between two semi-sinusoidal vectors? Here is what I have tried so far, using
the spectrum() function -- possibly incorrectly:
# generate some fake data, normalized to unit circle
x - jitter(seq(-2*pi, 2*pi,
I'm new to R - and trying to create a plot similiar to the spider plot at
http://www.statgraphics.com/eda.htm#radar .
I can't figure out several things... most of which I would think would be
straightforward
How can I change the lines for each series plotted instead of creating a
filled
Hi - Writing to see if someone can suggest whether a problem warrants a bug
report. It concerns the use of stepFlexclust in the flexclust package. The
problem concerns the size of clusters returned.
Versions: R-2.7.0 on Windows XP; RODBC_1.2-3
code snippet:
r8 -
My apriori apologies if this is felt to be the wrong list for this issue -
although it starts with R, it's a combination of programs that creates the
problem.
Currently we are using windows metafile format for in-text tables for
reports created in Word. However, we've discovered some artifactual
I am trying to plot the following data using dotchart
intersect.data-structure(list(X = structure(c(1L, 3L, 8L, 9L, 10L, 11L, 12L,
13L, 14L, 15L, 2L, 4L, 5L, 6L, 7L), .Label = c(1-100, 1001-1100,
101-200, 1101-1200, 1201-1300, 1301-1400, 1401-1500,
201-300, 301-400, 401-500, 501-600, 601-700,
since they are not in data1 , I do not want them to be in the merge result.
Moshe Olshansky-2 wrote:
Where do you want to place ID's which are in data2 but NOT in data1?
--- On Wed, 4/6/08, kayj [EMAIL PROTECTED] wrote:
From: kayj [EMAIL PROTECTED]
Subject: [R] merge two data sets
Thanks a lot for your help, I appreciate it.
David Winsemius wrote:
Thanks for your reply.
In your last step If I create a duplicate ( two similar records )
# create a duplicate
vv[8,1] - 7
vv[8,2]-'g'
and then I merge vv with vv2 ,both duplicates are merged. Is there a way
Wrong list, so re-directing to r-help. Consider r-sig-debian for Debian
questions too, but subscribe or else your posts bounce.
On 4 June 2008 at 07:18, Rongrong wrote:
|
| I am a new R user and I have a question of embedding R to generate png
| On Debian, I installed R by source code.
|
|
Thanks! I put the labels as follows
dotchart(as.matrix(intersect.data[-1]), labels=intersect.data[,1], cex=0.8)
Murli
From: Henrique Dallazuanna [EMAIL PROTECTED]
Sent: Wednesday, June 04, 2008 1:49 PM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Hello,
A few questions about the following examples:
1. Why do the two plotting versions not produce the same result?
2. Is the 'scale_x_continuous' (or *_y_* or *_*_discrete) geom the best
way to setup grids (as in visual guide-lines) in polar (or for that
matter, any) coordinate system?
3. Why
Jorge,
You can use the package BB to try and solve this problem.
I have re-written your functions a little bit.
# --
# Constants
# --
l=1
m=0.4795
s=0.4795
# --
# Functions to estimate f_i-k_i
#
can someone show me how to convert a table to a data.frame or a matrix ?
I tried below and as.data.frame rearranges the columns
similarly to a melt from reshape and as.matrix didn't change it. I
actually would prefer to change it to a dataframe but if someone
can show me how to convert it to a
Hi Mark,
Try this:
as.data.frame.matrix(temp)
or
as.data.frame(unclass(temp))
On Wed, Jun 4, 2008 at 4:20 PM, [EMAIL PROTECTED] wrote:
can someone show me how to convert a table to a data.frame or a matrix ? I
tried below and as.data.frame rearranges the columns
similarly to a melt from
Either of these will do it:
head(temp, Inf)
class(temp) - matrix
On Wed, Jun 4, 2008 at 3:20 PM, [EMAIL PROTECTED] wrote:
can someone show me how to convert a table to a data.frame or a matrix ? I
tried below and as.data.frame rearranges the columns
similarly to a melt from reshape and
Dear friend,
In an R program running permanently on a server I would like to read hour by
hour the temperature in *C and the humidity from a site like this (actually,
from many of such sites):
http://www.wunderground.com/global/stations/16239.html
How can I read the content of the site and
Il Thursday 01 May 2008 01:07:13 Charilaos Skiadas ha scritto:
Actually it's been out for a couple of weeks now at least. I just
finished my first reading of it, and I must say it was spectacular.
Congratulations Deepayan, the book gave me exactly the kind of
lattice knowledge I needed, and
Jubilo,
I believe we have the same problem. For a hacky solution, please see
my post
http://groups.google.com/group/r-help-archive/browse_thread/thread/bbd4ef5857dc99ef
- Stu
On May 22, 11:05 am, Jubilo [EMAIL PROTECTED] wrote:
Hello,
I'm a new user to R, and I was trying to install
To Whom it May Concern:
I have been using R version 2.6.2 for awhile now, installed on a
PowerMac G5 running OS 10.4.11 I typically get Quartz graphics
output from R into Adobe Illustrator CS2 simply by copying and
pasting. Upon upgrading to R version 2.7.0, I now receive an error
from
Dear R users,
I want to calculate the bias and variance of an estimator for several values
of two parameters a and b.
For example :
b1 b2
a1 bias bias
variance variance
a2 bias bias
variance variance
Can one do array of arrays ? I have tried and it did not
I have been trying to figure out how to get superscript/subscript in the
main title for a plot. I have tried various approaches and suggestions but
none of them work. I am trying to get the following as the main title of my
plot:
Emission of CO2 with time
(but note that 2 is subscript.)
I have
Try this:
plot(1:10, main = expression(Emission~of~CO[2]~with~time))
On Wed, Jun 4, 2008 at 5:31 PM, Tariq Perwez [EMAIL PROTECTED] wrote:
I have been trying to figure out how to get superscript/subscript in the
main title for a plot. I have tried various approaches and suggestions but
none
Hi Tariq,
try:
plot(x,y,main=expression(Emission of CO[2]* with time))
Cheers,
Christoph
Wednesday, June 4, 2008, 10:31:08 PM, you wrote:
I have been trying to figure out how to get superscript/subscript in the
main title for a plot. I have tried various approaches and suggestions but
Perhaps something about like this:
merge(vv[-c(which(duplicated(vv))-1, which(duplicated(vv))),], vv2, by=1)
On Wed, Jun 4, 2008 at 3:08 PM, kayj [EMAIL PROTECTED] wrote:
Thanks a lot for your help, I appreciate it.
David Winsemius wrote:
Thanks for your reply.
In your last
here is the data:
y-c(5,2,3,7,9,0,1,4,5)
id-c(1,1,6,6,7,8,15,15,19)
t-c(50,56,50,56,50,50,50,60,50)
table1-data.frame(y,id,t)//longitudinal data
the above is only part of data.
what I want to do is to use the linear model for each id ,then get the
estimate value,like:
Try this:
f - function(x)any(is.na(coefficients(x)))
models - by(table1[c(y, t)], table1$id, FUN=lm)
models[!unlist(lapply(models, f))]
On Wed, Jun 4, 2008 at 6:20 PM, Manli Yan [EMAIL PROTECTED] wrote:
here is the data:
y-c(5,2,3,7,9,0,1,4,5)
id-c(1,1,6,6,7,8,15,15,19)
Dear Courtney,
Are you exporting the graphs as postscript files? This is the usual way I
do it when moving graphs between R and Illustrator CS2. I´m afraid I do
not have a Mac, but I suppose CS2 runs similarly on both systems.
Best wishes
Christoph
To Whom it May Concern:
I have been using
If I want to convert one column to date and time format that R can
understand and manipulate then I would do this
initial$Started-as.POSIXlt(initial$Started)
But what happens if I have many such columns and I do not want to have
my code be stuffed with 10 similar lines (one for each variable aka
On 03-Jun-08 16:16:13, chaogai wrote:
Hi,
I noticed the following fortune in R 2.7 and 2.6.2:
fortune('Spreads')
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
Is it possible to have one array for bias and one for variance? such as..
biasAr=
b1 b2
a1 biasbias
a2 biasbias
and
varAr=
b1 b2
a1 var var
a2 var var
then you can combine the two in a data frame?
thanks
y
Michael Prince wrote:
Dear R users,
I
I need to create 100 variable ,whose name is id.1,id.2id.100
then I need to let a vector say id-c(id.1,id.2id.100)
any easy way to do this?
thanks a lot~
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Try this:
for(i in 1:100)assign(sprintf(id.%d, i), value = sample(1))
id - ls(patt = id.[0-9])
On Wed, Jun 4, 2008 at 7:52 PM, Manli Yan [EMAIL PROTECTED] wrote:
I need to create 100 variable ,whose name is id.1,id.2id.100
then I need to let a vector say id-c(id.1,id.2id.100)
any
Hi
I think my question might be answered before but i can't find the correct
one.
Here is the question.
I have a data frame from a csv file which contains about 30 variables.
lit is like this. (the values are just number I arbitrary created here)
id group calorie weight height bmi
The problem here is that the compiler cannot find the include files
for mpi. Notice that the first checks that fail are:
checking mpi.h usability... no
checking mpi.h presence... no
checking for mpi.h... no
One solution is to create a file named ~/.R/Makevars with the
following line:
This bug has been resolved, after our administrator updated the path
of R_HOME
in our scripts. I believe that the cause of trouble was the parallel
nature
of the installation, as mentioned in the following thread:
http://groups.google.com/group/r-help-archive/browse_thread/thread/632175125a7c4
I am having problem calling ars function in ARS package. My parameter domain
does not have upper bound or lower bound, so I did not givevalues to
parameters ubx,lbx. But it keeps telling me the starting values I provided
didn't have value to the left or to the right of the mode of the target
Hi All,
I am looking into merging 3 data sets I know how to do that by merging data1
with data2 and then merging the result with data 3. I was wondering if it
can be done all at once so I tried,
M-merge(data1,data2,data3, by=”ID”)
It does not work!
Any ideas?
--
View this message in
Hello Everyone,
In preparation for an upcoming talk, I would like to assemble a list
of companies that provide consulting, services, products, or training
for R.
I am already aware of a number of such companies including (in
alphabetical order):
BlueReference http://inference.us
Hi everyone:
I have been struggling with this repeated data type for whole afternoon,I
sent two emails to server for help,many people kindly responded , hereby
thank you so much,but since I dont want to write to much in email,so I
divide the problem in parts,so far this seem did not work out
I'm sure this must be a nls() newbie question, but I'm stumped.
I'm trying to do the example from Draper
and Yang (1997). They give this snippet of S-Plus code:
Specify the weight function:
weight - function(y,x1,x2,b0,b1,b2)
{
pred - b0+b1*x1 + b2*x2
parms - abs(b1*b2)^(1/3)
(y-pred)/parms
}
I have the data structure below and I'm attempting to send it into
barchart using the R code below it. I don't get an error but I don't get
any output either. Deepyan's new Lattice book is amazing and there are
some examples sort of similar to what i'm doing but I couldn't see a
way of using
Hello,
I'm trying to run an lmer model with family poisson but receive the
following
warning message:
model-lmer(count~tem+dat+alt+year+tem:dat+tem:alt+tem:year+dat:alt+dat:year+alt:year+(1|id),family=poisson)
Error in objective(.par, ...) :
Leading minor of order 2 in downdated X'X is not
I could suggest merge(merge(data1,data2), data3).
However, one problem I notice is that is assigns age=12 and gender=M to
everyone with id=1, and so on.
How are we to know that person with id=1 in data1 is the same person with id=1
in data2 and data3?
Bill Date: Wed, 4 Jun 2008 16:24:04
By the way, I used age and gender as examples. Plus, I've seen something very
similar posted a while ago. If the problem I listed isn't an issue, then that
code will work.
Best,
Bill
From: [EMAIL PROTECTED]: [EMAIL PROTECTED]; [EMAIL PROTECTED]: RE: [R] merging
3 data sets at onceDate:
dydata
x1 x2 y
1 9 27 248
2 9 22 213
3 4 23 190
4 11 16 183
5 -6 25 144
6 11 14 169
7 -4 13 72
8 2 8 73
9 10 13 156
10 8 30 263
11 12 10 147
12 -7 5 -2
13 0 10 75
14 12 0 77
15 9 8 115
16 12 24 245
17 34 23 370
18 12 1 84
19 10 37 324
20 26 30 371
weight -
id-c(1,1,1,1,3,3,3,7,7,7,7,11,11,11)
how to sort this kind of data to
id:(1,1,1,1,2,2,2,3,3,3,3,4,4,4.)
thanks~
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Not following this very carefully, but today I did something similar
with
Reduce(merge,list(d1,d2,d3)) ...
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Are these the ranks of the data?
help.search(rank)
Manli Yan wrote:
id-c(1,1,1,1,3,3,3,7,7,7,7,11,11,11)
how to sort this kind of data to
id:(1,1,1,1,2,2,2,3,3,3,3,4,4,4.)
thanks~
[[alternative HTML version deleted]]
no,the id is variable of a table,such as:
treatment id age response
low 1 50 20
low 1 60 30
high5 50 30
high5 60 40
...
I want to rearranage the table according the id (increasing),since id is not
strictly from 1~n,it is in
Try also,
id- paste('id.',1:100,sep=)
HTH,
Jorge
On Wed, Jun 4, 2008 at 6:52 PM, Manli Yan [EMAIL PROTECTED] wrote:
I need to create 100 variable ,whose name is id.1,id.2id.100
then I need to let a vector say id-c(id.1,id.2id.100)
any easy way to do this?
thanks a lot~
Jorge Ivan Velez wrote:
Dear John,
Assuming that your information is in the list x, does
substr(x,1,2)
work for you?
HTH,
Jorge
Jorge,
i tried
split(rtt, substr(rtt,1,3))
and that worked also. (i didn't think to nest it when you first
suggested it.) i just have to clean up the levels
An easy and good way to make a bunch of variables is to store them in a list.
For example:
mylist - replicate(100, sample(1:100, 10), simplify=FALSE)
id - paste('id.',1:100, sep='')
names(mylist) - id
sapply(mylist, median)
id.1 id.2 id.3 id.4 id.5 id.6 id.7 id.8 id.9
For the first question, what do you want to do with all the subsets? There are
tools like split, by, lapply, lmList, etc. that make working with all the
subsets easy. If you tell us what your final goal is, we may be able to help
with a simple solution that does not need the intermediate
Hi Mark,
I get output, after a while,
and reams of it. Very likely
not what you wanted.
Can you describe what you are trying to
see in barcharts for these data? It's
not obvious to me from the code below.
Best
Steve McKinney
-Original Message-
From: [EMAIL PROTECTED] on behalf of
Looks like you have some independent
variables in your model that correlate
perfectly so your design matrix is
not of full rank, probably because
your independent variable data is not
balanced.
Does a simpler fit such as
model-lmer(count~tem+(1|id),family=poisson)
give you a result?
You might
id-c(1,1,1,1,3,3,3,7,7,7,7,11,11,11,2,2,2,4,4,4,4,8,8,8)
sort(id)
[1] 1 1 1 1 2 2 2 3 3 3 4 4 4 4 7 7 7 7 8 8 8 11 11 11
Quoting Manli Yan [EMAIL PROTECTED]:
no,the id is variable of a table,such as:
treatment id age response
low 1 50 20
low
hi:lot thanks,how to use list to extract,I type allFit$coefficents,it came
to nothing,
such as I need to extract the estimates,how to do it by using list
2008/6/3 Austin, Matt [EMAIL PROTECTED]:
How about
library(nlme)
allFits - lmList(y ~ t|id, data=table1, pool=FALSE)
or
allFits -
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