Hi
Can you be more specific? What else do you want?
x<-table(sample(letters[1:5], 100, replace=TRUE))
x
a b c d e
20 14 23 23 20
as.numeric(x)
[1] 20 14 23 23 20
y<-as.numeric(x)
Regards
Petr
>
> i want to retrieve count for each freq of given summary from table()
e.g. "A"
> cout,"B"
i want to retrieve count for each freq of given summary from table() e.g. "A"
cout,"B" count
& save it in different variables
> table(dil)
dil
A C D E F G H I K L M N
P Q
49433 10361 38490 44549 27182 44947 16310 39002 39829 58805 11228 29320
Hi
For such tasks I would use merge
merge(u, v, by.x="name", by.y = "enter", all.x=TRUE)
If you do not want coe column anymore you can easily get rid of it.
Regards
Petr
>
> Hi Valerie
>
> One way would be to use the match function.
>
> # Your Data
> u =data.frame(coe=c(0,0,0,0,0,0,0,0),
Try
round(1003, digits = -3)
HTH,
Jorge.-
On Mon, Feb 6, 2012 at 1:40 AM, arunkumar <> wrote:
> Hi
>
> I need to round integer
> \
> eg if my value is 1003
>
> i want to make it 1000
>
> Thanks in advance
>
>
>
> -
> Thanks in Advance
>Arun
> --
> View this message in context:
On Feb 6, 2012, at 1:40 AM, arunkumar wrote:
Hi
I need to round integer
\
eg if my value is 1003
i want to make it 1000
?round
--
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/li
Hi
I need to round integer
\
eg if my value is 1003
i want to make it 1000
Thanks in advance
-
Thanks in Advance
Arun
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Hi all,
I need your opinion about using R-Rapache to build a knowlegde base to
bioinformatics field.
I have copious amount of genotype information (some few million records)
and would like to store it in MySQL database. Then, using RMySQL connector
I would like to use it for further analysis usin
cat will wrap Josh's "overkill" approach in one line:
mylist<- list(1:3,3:9)
lapply(mylist , cat , sep=',' , fill=T , append=T , file='foo.csv')
Cheers
On Sun, Feb 5, 2012 at 7:42 PM, Joshua Wiley wrote:
> Hi Michael,
>
> Part of me imagines this is overkill, but this should be one option:
>
>
Peter,
Thanks for checking on this! We're having some difficulties forecasting
sales ... randomForest is doing a great job on days and periods where there
is high variability. However, such as in this case, we do get
dates/times/departments where it simply forecasts the most frequent value
(I un
Well, if you look at your data more carefully, you will see that the
histogram of y is heavily skewed towards 1 (small values). The 91/625
quantile is still 1 (there are 192 1s). It is therefore not surprising
that RF comes up with mostly 1s (in my attempt it came up with two 2s,
but that is a bit
On Feb 5, 2012, at 8:31 PM, Francis Keyes wrote:
With R and the hist function, is there a way to make a histogram in
which
the y axis denotes propotion with respect to a separate sample
dataset of
the same range instead of frequency?
hist() returns an object with both "counts" and "densit
This must be the same poster who reports on SO that _he_ figured out
the problem.
--
David.
On Feb 5, 2012, at 6:01 PM, dengyu19901102 wrote:
the structure of ssta_sst can be seen below.
dim(ssta_sst)
[1] 360 180 362
Also, the debugging for temp is right. The problem for that code is
t
indx <- match(u$nam, v$enter, nomatch = 0)
u$coe[indx != 0] <- v$coeff[indx]
Sent from my iPad
On Feb 5, 2012, at 15:45, Valerie Moore wrote:
> Hi,
>
> I have two data frames (u and v).
>
>> u
> coe nam
> 1 0 Time
> 2 0Poten
> 3 0 AdvExp
> 4 0Share
> 5 0 Chang
I am trying to use lm for a simple linear fit with weights. The results
I get from IDL (which I am more familiar with) seem correct and
intuitive, but the "lm" function in R gives outputs that seem strange to me.
Unweighted case:
> x<-1:4
> y<-(1:4)^2
> summary(lm(y~x))
Call:
lm(formula = y ~
michael,
what is the problem you are trying to solve? are you writing it out so you can
read it back in? if so, look at 'save/load' or 'dump/source'. is this the
format required by some other program?
Sent from my iPad
On Feb 5, 2012, at 21:54, baptiste auguie
wrote:
> Hi,
>
> If you d
With R and the hist function, is there a way to make a histogram in which
the y axis denotes propotion with respect to a separate sample dataset of
the same range instead of frequency?
[[alternative HTML version deleted]]
__
R-help@r-project.org
Don't worry, i have figured it out. I forgot the most basic method by adding
another increasing loop counter.
k=0
for (yr in 1982:2010) {
temp <- ssta_sst[,,year_sst==yr & (month_sst>=6 & month_sst<=8)]
k=k+1
for (i in 1:360) {
for (j in 1:180) {
the structure of ssta_sst can be seen below.
> dim(ssta_sst)
[1] 360 180 362
Also, the debugging for temp is right. The problem for that code is the temp
doesn't pass down as it suppose to be, it only gives the last value in 2010.
> for (yr in 1982:2010) {
+ temp <- ssta_sst[,,year_sst==yr & (mo
How do you calculate the percentage of data within 2SD, 3SD, 4SD, 5SD, and
6SD of the mean? I used the following link as the data I'm working with:
nb10 <- read.table("http://www.adjoint-functors.net/su/web/314/R/NB10";) if
this helps answer my question. Can you please explain how to calculate t
New to R - rookie question.
I'm a mechanical engineer and enjoying using R to make high quality graphs.
I've searched.
I want to put text notation on graph plot areas and have the text background
"box" white to cover over the grid lines.
my command so far
text(15,5200,"Air Flow",cex=.8,
Hi All,
Currently I am using R 2.14.0 version in which RMySQL package is available
to establish connection with MySql database. I want to use'quantmod' package
which is use for stock charts. Could anyone tell how to establish connection
to MySQL DB in quantmod package? I have tried but got the fol
You can first simulate Sigma from the inverse Wishart distribution and
then simulate from the matrix normal given the realized Sigma.
As per a DuckDuckGo search, InvWishart function in the contributed R
package MCMCpack may be what you need for the first step.
HTH,
Ranjan
On Sun, 5 Feb 2012 15:
Hi,
I have two data frames (u and v).
> u
coe nam
1 0 Time
2 0 Poten
3 0 AdvExp
4 0 Share
5 0 Change
6 0 Accounts
7 0 Work
8 0 Rating
> v
coeff enter
1 0.7272727 Accounts
2 0.322 Time
3 0.0500123 Poten
I want to update the values
We have a pretty simple R forecasting problem, using randomForest. Our
script is below. We have data with about 6 columns, and 625 rows to do the
training on, and then we're trying to forecast 90 or so rows.
The x matrix has 3 years worth of sales data, starting at 6:15am. However,
for some rea
Hi Tom,
I don't believe there is a standard location. You can set variables or have
functions run (say to source other functions in) in your .Rprofile file. That
way, you always have them. Once you get more than a handful of functions, it
is probably time to consider putting them into your o
Dear All,
I'm now looking for R-code on how to find the Goodness of Fits for
Archimedean Copulas. If anyone have a guide for this problems please lets
me know.
Your prompt action is much appreciated.
Regards,
Ummul
[[alternative HTML version deleted]]
__
I'm new but am already starting to accumulate utility functions for
the fairly specialized kinds of data files with which I work. I'd like
to keep those functions in a single folder, or filetree rooted at a
specific folder under my $HOME (I'm running linux), for ease of
* source()-ing
* git commi
Hi,
If you don't mind having NAs for missing values, try the following,
mylist = list(1:3, 4:7)
library(plyr)
write.csv(do.call(rbind.fill.matrix, lapply(mylist, matrix, nrow=1)), file="")
HTH,
b.
On 6 February 2012 15:01, Michael wrote:
> Hi all,
>
> I have a list of vector of numbers - the
Hi Michael,
Part of me imagines this is overkill, but this should be one option:
## your data
mylist <- list(1:3, 3:6)
## open a writeable connection to a file
con <- file("test.csv", "w")
## first collapse each element of the list to be a comma separated
string, then write each
## element of new
Hi Valerie
One way would be to use the match function.
# Your Data
u =data.frame(coe=c(0,0,0,0,0,0,0,0),
name=c("Time","Poten","AdvExp","Share","Change","Accounts","Work","Rating"))
v = data.frame(coeff=c(0.7272727,0.322,0.0500123),
enter=c("Accounts","Time","Pot
Hi all,
I have a list of vector of numbers - the reason I used list of vector was
that I each list have different numbers of numbers which I don't know
before run-time.
mylist[[1]] = c(1, 2, 3)
mylist[[2]] = c (3, 4, 5, 6)
...
...
etc.
Could you please tell me if there is a way to dump all thes
On Sun, 5 Feb 2012, Matyas Sustik wrote:
Thanks for the quick response! That is exactly what I was looking for.
You're welcome.
On Sun, Feb 5, 2012 at 4:34 PM, Achim Zeileis
There is an introduction in the following report:
http://epub.wu.ac.at/3269/
[...]
Authors@R: c(person(given = "Jo
Hi Philip,
You have gotten several good responses. For a generalization of this,
I would suggest _not_ using nested if or ifelse statements. They
quickly become difficult to read.
d <- c(1, 1, 2, 2, 2, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0)
require(car)
dnew <- recode(d, "2 = 0; 1 = 1; 0
Thanks for the quick response! That is exactly what I was looking for.
On Sun, Feb 5, 2012 at 4:34 PM, Achim Zeileis
> There is an introduction in the following report:
> http://epub.wu.ac.at/3269/
[...]
> Authors@R: c(person(given = "John", family = "Doe", role = "aut"),
> person(given = "Jane"
Much easier: use ifelse (the vectorized function) instead as follows:
ifelse(A < 2, ifelse(A < 1, 2, 1), 0)
But you could probably just do 2 - A in this case which would be easiest.
Michael
On Sun, Feb 5, 2012 at 5:30 PM, Philip Robinson
wrote:
> I have a vector of 2,1,0 I want to change to 0,
On Feb 5, 2012, at 5:30 PM, Philip Robinson wrote:
I have a vector of 2,1,0 I want to change to 0,1,2 respectively (the
data
is allele dosages)
I have tried multiple nested if/else statements and looked at the ?
if help
and cannot work out what is wrong, other people have posted code
whic
On Feb 5, 2012, at 5:30 PM, Philip Robinson wrote:
I have a vector of 2,1,0 I want to change to 0,1,2 respectively (the
data
is allele dosages)
I have tried multiple nested if/else statements and looked at the ?
if help
and cannot work out what is wrong, other people have posted code
whic
On Sun, 5 Feb 2012, Matyas Sustik wrote:
Hi All,
Could someone point me to an example on how to use Author@R in a package
DESCRIPTION file?
There is an introduction in the following report:
http://epub.wu.ac.at/3269/
Say with 2 authors: John Doe and Jane Doe, the latter is a maintainer
whi
I have a vector of 2,1,0 I want to change to 0,1,2 respectively (the data
is allele dosages)
I have tried multiple nested if/else statements and looked at the ?if help
and cannot work out what is wrong, other people have posted code which is
identical and they state works.
Any help would be great
Hi All,
I am building my first package. I have the files under revision
control. Normally, I have a $Id$ tag in my files
which expand by my editor and so the current revision etc. I tried to
add it to the DESCRIPTION file by having
the first line to be:
# $Id$
R CMD build complained. I hoped
Hi Uwe,
> 1. I don't believe you really need all the flags from above. If so, within a
> package, use a Makevars file.
>
> 2. R CMD SHLIB QUIC.cpp shoudl do the trick already, perhaps some linker
> flags are required for blas that can be specified in the same line, see R
> CMD SHLIB --help
Thanks
Hi All,
Could someone point me to an example on how to use Author@R in a package
DESCRIPTION file? Say with 2 authors: John Doe and Jane Doe, the
latter is a maintainer which could replace:
Author: John Doe, Jane Doe
Maintainer: Jane Doe
"Both ‘Author’ and ‘Maintainer’ can be omitted (as from R
How about using the legend function ...
plot(rnorm(100))
legend(60,2,"100 Random Normal Draws",cex=.8,text.col="blue",
box.col="red",bg="yellow")
You can customize my effort to fit your specific needs
HTH
Pete
Henry wrote
>
> New to R - rookie question.
> I'm a mechanical engineer and enjoy
How about
# Read Data
nb10 <- read.table("http://www.adjoint-functors.net/su/web/314/R/NB10";)
# Calculate Stats
total = length(nb10[,1])
mean = mean(nb10[,1])
sd = sd(nb10[,1])
# Function ... nSD is the number of SD you are looking at
pData <- function(nSD){
lo = mean - nSD/2*sd
hi =
On Feb 5, 2012, at 1:20 PM, Christopher Kelvin wrote:
Dear All,
Can you help me, with the code below how do I obtain the fisher
information from it.
A modest amount of searching the archives (which is really your
responsibility):
search.r-project.org/cgi-bin/namazu.cgi?query="fisher
+
Am 05.02.2012 15:54, schrieb Alaios:
Dear all
I am using lapply (actually mclapply that share the same syntax).
I want to call the same function that takes as input a vector. My initial data
structure is a matrix that I want to cut it to multiple vectors (one vector for
every row of the matrix
--
Luis Solís
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Dear All,
Can you help me, with the code below how do I obtain the fisher information
from it.
Is my q<-replicate(1000,x) the right way to do simulation.
thank you.
x<-rweibull(100,0.8,1.5)
q<-replicate(1000,x)
z<-function(p){
beta<-p[1]
eta<-p[2]
log1<-(n*log(beta)-n*beta*log(eta)+(beta-1)*sum(l
On Feb 5, 2012, at 9:54 AM, jim holtman wrote:
Is this what you are after:
x <- c(1327211358, 1327221999, 1327527296, 1327555433, 1327701042,
+ 1327761389, 1327780993, 1327815670, 1327822964, 1327897497,
1327897527,
+ 1327937072, 1327938300, 1327957589, 1328044466, 1328127921,
1328157588,
On Feb 5, 2012, at 2:53 AM, sagarnikam123 wrote:
how to avoid writing index of variable in write.table command,
e.g. output--->
It's time to "get with the program", sagarnikam123. You are asked in
the Posting Guide to provide a reproductible example. At the very
least you should provide s
This was indeed very helpful - thanks a lot ! :-))
On 4 February 2012 14:56, Ranjan Maitra wrote:
> I think this should be easy to write a function doing this.
>
> Assume that Y is matrix normal with mean matrix mu and row and column
> dispersion matrices Sigma and Gamma, respectively.
>
> Isn'
It is not homework. My problem is this: I used nonlinear regression to fit a
lognormal cumulative distribution to the probability of a buildings' collapse
against a measure of seismic intensity. I am able to obtain the asymptotic
confidence intervals. I am not , however, certain that they are
On Feb 4, 2012, at 8:29 PM, dengyu19901102 wrote:
i want to select June, July and August data from the 3D array
(`ssta_sst`,
360*180*362).
the loop works but the output of ssta_winter has the identical
values for
`ssta_winter[,,i]`.seen below. I have set it up as an array of
(360,180,29)
Hi Brian,
that explains it. I will update R to 2.14.x tomorrow.
Thanks for your post.
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_
Hello everyone
I was wondering how would one simulate from a Normal Wishart Distribution
in R.
A normal inverted Wishart distribution is denoted by NIW (M,C,d,S), where
X/(Sigma) ~ N( M,C,(Sigma) ) -> a matrix normal distribution, (Sigma) ->
column dispersion matrix
(Sigma) ~ IW (d,S) -> inver
It's jsut a vocabulry trouble. The good name is TermDocumentMatrix.
Certainly it's depend of the version of TM.
best regards
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R
Dear R users,
I am using nlme to fit a pharmacokinetic model. The base model is
parameterized in terms of CL, V1, V2 and Q.
basemodel<-nlme(Conc ~TwoCompModel(CL,Q,V1,V2,Time,ID),
data = data2, fixed=list(CL+Q+V1+V2~1),
random = pdDiag(CL+V1+V2~1),
start=c(CL=log(20),Q=log(252),V1=log(24.9),V2=lo
Hi there,
can somebody give me a guide as to how to generate data from weibull
distribution with censoring
for example, the code below generates only failure data, what do i add to
get the censored data, either right or interval censoring
q<-rweibull(100,2,10).
Thank you
Grace Kam
student, Univer
On Sun, Feb 05, 2012 at 06:54:02AM -0800, Alaios wrote:
> Dear all
> I am using lapply (actually mclapply that share the same syntax).
>
> I want to call the same function that takes as input a vector. My initial
> data structure is a matrix that I want to cut it to multiple vectors (one
> vecto
Hi there,
can somebody give me a guide as to how to generate data from weibull
distribution with censoring
for example, the code below generates only failure data, what do i add to
get the censored data, either right or interval censoring
q<-rweibull(100,2,10).
Thank you
Grace Kam
student, Univer
Hi Tal,
Untested, but it should just be a matter of giving tabularx style formatting
in the align option to xtable(), and then using "tabularx" as the
tabular.environment argument to print.xtable()
see ?xtable and ?print.xtable for details.
Best,
Ista
On Sunday, February 05, 2012 01:16:38 PM
Is this what you are after:
> x <- c(1327211358, 1327221999, 1327527296, 1327555433, 1327701042,
+ 1327761389, 1327780993, 1327815670, 1327822964, 1327897497, 1327897527,
+ 1327937072, 1327938300, 1327957589, 1328044466, 1328127921, 1328157588,
+ 1328213951, 1328236836, 1328300276, 1328335936, 132
Dear all
I am using lapply (actually mclapply that share the same syntax).
I want to call the same function that takes as input a vector. My initial data
structure is a matrix that I want to cut it to multiple vectors (one vector for
every row of the matrix) and then feed that to the function by
On 12-02-05 3:01 AM, Alaios wrote:
Dear all,
I am using browse() to debug my R applications. My problem is that I need a bit
more control of the debugging process.
For example 'I want R to stop at the 7th iteration and check. One can do think
that I can put the browse inside an if statement bu
On 05/02/2012 08:10, ioanna wrote:
How do you use bootstrap to estimate the confidence as well as the prediction
intervals in nonlinear regression ?
With difficulty!
There is far too little here to go on, and this seems an odd question
unless it is homework (why dictate a problem-strewn metho
write.table(..., row.names = FALSE)
(not row.name )
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
-
Hi Alex,
some hints and advices on debugging can be found in the 'Writing R Extensions
Manual':
http://cran.r-project.org/doc/manuals/R-exts.html#Debugging-R-code
Depending on the IDE/editor you are using, there may be additional
debugging tools/options, but without further informations about yo
How do you use bootstrap to estimate the confidence as well as the prediction
intervals in nonlinear regression ?
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Dear Ista,
Do you happen to know of a way to use the {tabularx} LaTeX pacakge, so to
have it work with xtables?
I would rather have my columns adjust themselves then to rotate my table.
Thank you in advance,
Tal
Contact
Details:
how to avoid writing index of variable in write.table command,
e.g. output--->
index,character,state
"1" "M" "2"
"2" "K" "3"
"3" "R" "1"
"4" "E" "1"
"5" "S" "1"
"6" "H" "1"
"7" "K" "1"
"8" "H" "1"
"9" "A" "1"
"10" "E" "3"
"11" "Q" "1"
i tried col.name=FALSE & row.name=FALSE but it can't
--
View
Thanks for your answer.
You're right, when I'm looking at the files with IrfanView everything is ok.
> you need to learn how to shrink the full image into
> the display window.
guess, that's what I'm lacking...
I tried to make the graphics smaller in R, but it doesn't have any effect, when
I have a list of numbers corresponding to timestamps, a sample of which follows:
c(1327211358, 1327221999, 1327527296, 1327555433, 1327701042,
1327761389, 1327780993, 1327815670, 1327822964, 1327897497, 1327897527,
1327937072, 1327938300, 1327957589, 1328044466, 1328127921, 1328157588,
1328213951,
On Sun, Feb 5, 2012 at 3:01 AM, Alaios wrote:
> Dear all,
> I am using browse() to debug my R applications. My problem is that I need a
> bit more control of the debugging process.
>
> For example 'I want R to stop at the 7th iteration and check. One can do
> think that I can put the browse insi
Dear all,
I am using browse() to debug my R applications. My problem is that I need a bit
more control of the debugging process.
For example 'I want R to stop at the 7th iteration and check. One can do think
that I can put the browse inside an if statement but this limits by lot the
flexibili
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