Hi All,I have a dataframe called 'means' as shown below:iris1.csv <- iris
iris2.csv <- iris
names <- c("iris1.csv", "iris2.csv")
dat <- mget(names)
lst4 <- lapply(dat, function(x) apply(x[,-5], 2, mean))
# Build the new data frame
means <- as.data.frame(do.call(rbind, lst4))
means$source <- names
Hi,
I hope that someone can provide a better way to implement it. This is my
implementation.
> data
X2 gbm_tcga lusc_tcga ucec_tcga_pub
1 gbm_tcga 1.0 0.14053719 -0.102847164
2 gbm_tcga 1.0 0.04413434 0.013568055
3 gbm_tcga 1.00
Hi,
I think you need not split the data.frame to get the desired result.
You can work with your list lst4 itself.
#Convert the vectors in the list to data.frames.
lst4 <- lapply(lst4, function(x) {as.data.frame(t(iris1.csv))})
#Get the data.frames in the list to the global environment
list2env(ls
On 17/02/15 12:59, smart hendsome wrote:
I'm very new to r-programming. I have rainfall data. I have tried to fit gamma
into my data but there is error. Anyone can help me please.
My rainfall data as I uploaded. When I try run the coding:
library(MASS)
KLT1<-read.csv('C:/Users/User/Dropbox/PhD
I'm very new to r-programming. I have rainfall data. I have tried to fit gamma
into my data but there is error. Anyone can help me please.
My rainfall data as I uploaded. When I try run the coding:
library(MASS)
KLT1<-read.csv('C:/Users/User/Dropbox/PhD
Materials/Coding_PhD_Thesis/Kelantan_Avera
Hi All,I have a dataframe called 'means' as shown below:iris1.csv <- iris
iris2.csv <- iris
names <- c("iris1.csv", "iris2.csv")
dat <- mget(names)
lst4 <- lapply(dat, function(x) apply(x[,-5], 2, mean))
# Build the new data frame
means <- as.data.frame(do.call(rbind, lst4))
means$source <- names(
DearGiovanni,Congratulationsfor the truly helpful plm package! Being new to R,
I have a problem with the plm function for financial markets timeseries data:
After having defined a large, unbalanced panel pdata.frame
(https://www.dropbox.com/s/2r9t1cu9v65gobk/Database_CN_2004.csv?dl=0)and
runn
Hi,
I'm trying to use method of moments estimation to estimate 3 unkown paramters
delta,k and alpha.
so I had system of 3 non linear equations:
1) [delta^(1/alpha) *gamma (k-(1/alpha)) ]/gamma(k) = xbar
2) [delta^(2/alpha) *gamma (k-(2/alpha)) ]/gamma(k) = 1/n *sum (x^2)
3) [delta^(3/alpha)
You can let lapply() do the preallocation and the looping for you with
ASL <- lapply(1:5, function(j) lapply(1:5, function(i) i^j))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Feb 16, 2015 at 9:46 AM, Jeff Newmiller
wrote:
> You have two named objects when your goal is to have one
I wonder if this might be better asked on a statistical list -- e.g.
stats.stackexchange.com
-- as this seems to involve complex statistical model comparison
issues, which are normally OT here.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Info
Thanks to Jim, Peter and Jeff who all saw the solution!
Best wishes
Troels
Den 16-02-2015 kl. 18:46 skrev Jeff Newmiller:
You have two named objects when your goal is to have one that contains five
others.
ASL <- vector( "list", 5 )
for (j in 1:5){
ASL[[j]] <- vector( "list", 5 )
for (i
Dear R-Help list,
I want to compare gam models including interaction with simpler models.
For interaction models, I used gam(Y~ti(X1) + ti(X2) + ti(X1,X2))
removing the interaction, the models end as Y~ti(X1) + ti(X2)
How those models compare with models with the form Y ~ s(X1) + s(X2)
In my cas
Hi,
For the second one,
sqrt((sum(p[,1]^2*p[,2])-(sum(p[,1]*p[,2]))^2/sum(p[,2]))/(sum(p[,2])-1)),
please refer to the following link for an example to explain how it works.
http://www.lboro.ac.uk/media/wwwlboroacuk/content/mlsc/downloads/var_stand_deviat_group.pdf
For the first one:
sd(unli
Hi Peter,
That was my first port of call before I posted this thread. Unfortunately,
it does not seem to explicitly state which test is used or how the p-value
is calculated.
Thanks,
Rob.
--
View this message in context:
http://r.789695.n4.nabble.com/P-value-from-Matching-tp4703309p4703345.h
You have two named objects when your goal is to have one that contains five
others.
ASL <- vector( "list", 5 )
for (j in 1:5){
ASL[[j]] <- vector( "list", 5 )
for (i in 1:5) {
ASL[[j]][[i]] <- i^j
}
}
---
Jeff New
On 16 Feb 2015, at 17:43 , Troels Ring wrote:
> Dear friends - this is simple I know but I can figure it out without your
> help.
> I have for each of 2195 instances 10 variables measured at specific times
> from 6 to several hundred, so if I just take one of the instances, I can make
> a lis
Is this what you mean:
ASL <- list()
for (j in 1:5){
RES <- list()
for (i in 1:5) RES[[i]] <- i ^ j # create list
ASL[[j]] <- RES # store 'list of list'
}
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want t
Again, I come to think about violin plots which is more informative than
the error bars. But for some reason, the violin in the *west* became way
too slimmer than the *east* one, though the density plot tells me that is
not necessarily the case. I am still trying to figure that out, and that
would
Dear friends - this is simple I know but I can figure it out without
your help.
I have for each of 2195 instances 10 variables measured at specific
times from 6 to several hundred, so if I just take one of the instances,
I can make a list of the 10 variables together with their variable
times.
Or even easier is you use lmList() from the nlme package
library(nlme)
data(iris)
regression.list <- lmList(Sepal.Width ~ Petal.Width | Species, data = iris)
summary(regression.list)
coef(regression.list)
Best regards,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Resea
I named your data frame temp
> aggregate(temp[,-1], list(temp[,1]), function(x) sum(abs(x)>.2))
Group.1 gbm_tcga lusc_tcga ucec_tcga_pub
1 gbm_tcga4 1 0
2 lusc_tcga1 4 0
3 ucec_tcga_pub0 0 4
Cheer
Look at the function stepclass() in package klaR.
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David Moskowitz
Sent: S
Or for the slopes and t-values:
> do.call(rbind, lapply(mod, function(x) summary(x)[["coefficients"]][2,]))
Estimate Std. Error t value Pr(>|t|)
setosa 0.8371922 0.5049134 1.658091 1.038211e-01
versicolor 1.0536478 0.1712595 6.152348 1.41e-07
virginica 0.6314052 0.1428
Hi All,
how to compute only abs(value) > 0.2 of DF
X2 gbm_tcga lusc_tcga ucec_tcga_pub
gbm_tcga 1.0 0.1405371906 -0.1028471639
gbm_tcga 1.0 0.0441343447 0.0135680550
gbm_tcga 1.0 -0.2000397119 0.0389718175
gbm_tcga 1.0 0.
In R you would want to combine the results into a list. This could be done when
you create the regressions or afterwards. To repeat your example using a list:
data(iris)
taxon <- levels(iris$Species)
mod <- lapply(taxon, function (x) lm(Sepal.Width ~ Petal.Width,
data=iris, subset=Specie
Lovely, a much more elegant solution.
John Kane
Kingston ON Canada
-Original Message-
From: hyil...@gmail.com
Sent: Mon, 16 Feb 2015 02:30:09 +0800
To: jrkrid...@inbox.com, djmu...@gmail.com
Subject: Re: [R] ggplot2 shifting bars to only overlap in groups
Thanks so much, John and Denni
On 16 Feb 2015, at 00:31 , Rob Wood wrote:
> Hi all,
>
> When using the match command from the matching package, the output reports
> the treatment effect, standard error, t-statistic and a p-value. Which test
> is used to generate this p-value, or how us it generated?
>
I would assume this
? subset. You have most of the command already written
John Kane
Kingston ON Canada
> -Original Message-
> From: pnsinh...@gmail.com
> Sent: Mon, 16 Feb 2015 13:31:33 +0530
> To: r-help@r-project.org
> Subject: [R] Selection/filtering from Data
>
> >From a dataset , I want select age >
On 16/02/2015 6:37 AM, Ulrike Grömping wrote:
> Dear helpeRs,
>
> I have some png files in the inst/extdata directory of a package (e.g.,
> man.png), and I want to provide character strings containing the paths to
> these files; of course, these path strings have to depend on the individual
> inst
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Dear all,
I have a problem when trying to present the results of several
regression. Say I have run several regressions on a dataset and saved
the different results (as in the mini example below). I then want to
loop over the regression results in ord
You may have good reason to distrust the Excel solver :)
See below
John Kane
Kingston ON Canada
> -Original Message-
> From: rzw...@ets.org
> Sent: Sat, 14 Feb 2015 23:53:55 +
> To: r-help@r-project.org
> Subject: [R] Nonlinear integer programming (again)
>
> Oddly, Excel's Solver
Thanks for the clarification. The basic error on my side was that I had
misunderstood the "digits" option in the summary, I had not understood that
even integer numbers might end up being "rounded". Problem is solved now.
-Original Message-
From: Allen Bingham [mailto:aebingh...@gmail.co
Hi Arnab,
Actually, I don´t think so as in Bayesian Regression the estimation converges
towards the prior distribution and not a point estimate like in the frequentist
approach. But I’m not very sure honestly. I used a Bayesian Logit Model in R
(bayesm, rhierBinLogit by Rossi) to calculate indi
Dear helpeRs,
I have some png files in the inst/extdata directory of a package (e.g.,
man.png), and I want to provide character strings containing the paths to
these files; of course, these path strings have to depend on the individual
installation. So far, I have written a function that - if cal
On Mon, 16 Feb 2015, Rodica Coderie via R-help wrote:
Hello,
I've created a ctree model called fit using 15 input variables for a factor
predicted variable Response (YES/NO).
When I run the following :
table(predict(fit2), training_data$response)
I get the following result:
NO YES
NO 4
Zwick, Rebecca J ETS.ORG> writes:
> Oddly, Excel's Solver will produce a solution to such problems but
> (1) I don't trust it and
> (2) it cannot handle a large number of constraints.
> [...]
> My question is whether there is an R package that can handle this problem.
There are not many free in
Hello,
I've created a ctree model called fit using 15 input variables for a factor
predicted variable Response (YES/NO).
When I run the following :
table(predict(fit2), training_data$response)
I get the following result:
NO YES
NO 48694 480
YES 0 0
It appears that the NO re
Hi Parth,
Assume that your dataset is in the form of a data frame, named psdf.
psdf<-data.frame(age=sample(0:100,50),income=sample(8000:12000,50))
selectdf<-subset(psdf,age >= 36 & income > 1)
Jim
On Mon, Feb 16, 2015 at 7:01 PM, Partha Sinha wrote:
> >From a dataset , I want select age >=3
On 16/02/2015 07:49, PIKAL Petr wrote:
Hi Jeff
-Original Message-
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us]
Sent: Friday, February 13, 2015 3:56 PM
To: PIKAL Petr; r-help@r-project.org
Cc: Richard M. Heiberger
Subject: RE: [R] problems with packages installation
I agree th
On 16/02/15 13:36, Bingzhang Chen wrote:
Hi R users,
I have a problem on how to pass an extra argument to do. call:
The example codes are:
#
require(ggplot2)
plots = lapply(1:5, function(.x) qplot(1:10,rnorm(10), main=paste("plot",.x)))
require(gridE
>From a dataset , I want select age >=36 and income>1. How to do ?
parth
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