On Dec 25, 2012, at 9:52 AM, Swagath wrote:
Dear all...Merry Christmas
I would like to split a long dataframe. The dataframe looks like this
x<-c('0:00:00', '0:30:00', '1:00:00', '1:30:00', '2:00:00',
'2:30:00', '3:00:00', '0:00:00', '0:30:00', '1:00:00', '1:30:00',
'2:00:00', '2:30:00',
First, hello,
Second, http://r.789695.n4.nabble.com/path-analysis-td2528558.html#a2530207
Last, Regards
Le 26/12/2012 04:11, Ali Mahmoudi a écrit :
What's the function of 'path analysis ' to do it with R?
Please help me.Thanks.
[[alternative HTML version deleted]]
What's the function of 'path analysis ' to do it with R?
Please help me.Thanks.
[[alternative HTML version deleted]]
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Dear all...Merry Christmas
I would like to split a long dataframe. The dataframe looks like this
x<-c('0:00:00', '0:30:00', '1:00:00', '1:30:00', '2:00:00', '2:30:00',
'3:00:00', '0:00:00', '0:30:00', '1:00:00', '1:30:00', '2:00:00',
'2:30:00', '3:00:00', '3:30:00', '4:00:00','0:00:00', '0:30:
Hi,
You could use library(data.table)
x <- data.frame(A=rep(letters,2), B=rnorm(52), C=rnorm(52), D=rnorm(52))
res<- with(x,aggregate(cbind(B,C,D),by=list(A),mean))
colnames(res)[1]<-"A"
x1<-data.table(x)
res2<- x1[,list(B=mean(B),C=mean(C),D=mean(D)),by=A]
identical(res,data.frame(res2))
#[1]
Hi,
Jim's method was found to be faster than data.table()
n <- 1
nLevels <- 10
nRows <- 120
Cols <- list(rep(list(sample(nRows)), n))
df <- data.frame(levels = sample(nLevels, nRows, TRUE), Cols)
colnames(df)[-1] <- paste0('col', 1:n)
# convert to matrix for faster processing
df.m <-
HI Eliza,
Try this:
set.seed(15)
mat1<-matrix(sample(1:2000,1776,replace=TRUE),ncol=444)
colnames(mat1)<-paste("Col",1:444,sep="")
res1<-lapply(1:37,function(i) mat1[,seq(i,444,37)])
res2<-lapply(1:37,function(i) {a<-mat1[,i:444];a[,c(TRUE,rep(FALSE,36))]})
#your code
identical(res1,res2)
#[1] TR
aggregate() is not efficient. try by().
On Tue, Dec 25, 2012 at 11:34 AM, Martin Batholdy
wrote:
> Hi,
>
>
> I need to aggregate rows of a data.frame by computing the mean for rows
> with the same factor-level on one factor-variable;
>
> here is the sample code:
>
>
> x <- data.frame(rep(letters
Hello,
Did you contact the package maintainer?
Mark M. Fredrickson
There is also a webpage:
https://github.com/markmfredrickson/optmatch
Regards,
Pascal
Le 18/12/2012 21:37, MA a écrit :
Hello
My optmatch package is loaded and otherwise running fine.
I get an error after lcds successfully c
On Dec 24, 2012, at 11:21 AM, jenna wrote:
I am trying to get the means of each participants avg saccade
amplitude as a
function of the group they were in (designated by "shape_"), but
there are
missing values in the datasetthis is what i tried...
with(data055,tapply(AVERAGE_SACCADE_AM
According to the way that you have used 'aggregate', you are taking
the column means. Couple of suggestions for faster processing:
1. use matrices instead of data.frames ( i converted your example just
before using it)
2, use the 'colMeans'
I created a 120 x 10 matrix with 10 levels and its
I'd suggest the 'data.table' package. That is
one of the prime uses it was created for.
Pat
On 25/12/2012 16:34, Martin Batholdy wrote:
Hi,
I need to aggregate rows of a data.frame by computing the mean for rows with
the same factor-level on one factor-variable;
here is the sample code:
You might consider using the sqldf package.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. D
Perhaps you should read the help file for rnorm more carefully.
?rnorm
Keep in mind that the normal probability distribution is a density function, so
the smaller the standard deviation is, the greater the magnitude of the density
function is.
--
Dear Arun,
thank-you very much. i m extremely sorry for spoiling your Christmas.
eliza
> Date: Tue, 25 Dec 2012 08:51:05 -0800
> From: smartpink...@yahoo.com
> Subject: Re: [R] for loop not working
> To: eliza_bo...@hotmail.com
> CC: r-help@r-project.org; kri...@ymail.com
>
> HI Eliza,
>
> Try
Hi,
You should have provided a reproducible example.
To convert in general, ?as.POSIXct()
A.K.
- Original Message -
From: Antonio Rodriges
To: r-help@r-project.org
Cc:
Sent: Tuesday, December 25, 2012 7:28 AM
Subject: [R] as.Date to as.POSIXct
Hello,
I have converted some UNIX tim
Hi,
I need to aggregate rows of a data.frame by computing the mean for rows with
the same factor-level on one factor-variable;
here is the sample code:
x <- data.frame(rep(letters,2), rnorm(52), rnorm(52), rnorm(52))
aggregate(x, list(x[,1]), mean)
Now my problem is, that the actual data-s
Hello R-helpers..
I want to ask about how I can sample data sets without having the infinite
numbers coming out. For example,
set.seed(1234)
a<-rnorm(15,0,1)
b<-rnorm(15,0,1)
c<-rnorm(15,0,1)
d<-rnorm(15,0,36)
After come out with the sample, I need to do a transformation (by Hoaglin,
1985) f
Simplest way is the call a system command, using R CMD.
See :http://stackoverflow.com/questions/6695105/call-r-scripts-in-matlab
But there are more complicated solutions are proposed:
http://www.mathworks.co.uk/matlabcentral/fileexchange/5051
This is uses R-(D)-COM
In my opinion most robust int
Hello,
I have converted some UNIX time stamps with as.Date as follows
dates_unix <- seq(
as.Date(convertTimeToString(timeStart)),
length = length(data),
by="1 mon")
and now I would like to convert "dates_unix" from type D
A bit of data might be useful. Make a small example and post the data with
dput().
On 24.12.2012, at 20:21, jenna wrote:
> I am trying to get the means of each participants avg saccade amplitude as a
> function of the group they were in (designated by "shape_"), but there are
> missing values in
Dear Rui
If you impose the homogeneity (adding-up) restrictions, your system
becomes singular, because the error terms of the share equations
always sum up to zero. Therefore, you can arbitrarily delete one of
the share equations and obtain the coefficients that were not directly
estimated by the
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