Hi,
Please dput() the example dataset. When I read from the one shown below, it
looks a bit altered.
library(zoo)
dat1<- read.zoo(text="2009-07-15,#N/A N/A,#N/A N/A,18.96858
2009-07-16,20.30685,20.40664,#N/A N/A
2009-07-17,20.78813,20.03991,20.40664
2009-07-20,21.41278,21.41278,20.03991
2009-07-
Euna Jeong writes:
> R> plot(gas1, ncomp=2, asp = 1, line = TRUE)
>
> This shows only the cross-validated predictions.
If you add the argument which = c("train", "validation") (see
?predplot.mvr), you will get both. However, you will get them in
separate panels in the plot.
If you wish to have
HI,
Try:
res<- lapply(seq_len(ncol(dat2)),function(i)
{
x1<-cbind(dat1New[,c(1:4)],dat2[,i]);
colnames(x1)[5]<- colnames(dat2)[i];
x2<-x1[x1[,5]!=0,];
x2$previoustripstore<-ave(x2$Store,x2$CUSTID,FUN=function(x)
c("",x[-length(x)]));
x2$Nexttripstore<- ave(x2$Store,x2$PANID,FUN=function(x) c(x[-1]
Hi,
I'd like to draw the trained predictions and the cross-validated
predictions in the same plot to compare two predictions.
In page 3, in the pls Package paper,
R> plot(gas1, ncomp=2, asp = 1, line = TRUE)
This shows only the cross-validated predictions.
Could you tell me how to do?
Thank y
HI Satish,
colnames(Output)[4]<- colnames(dat2)[i]; #guess this line should be:
colnames(x1)[4]<- colnames(dat2)[i]
Regarding the warning, I used
read.table(..., stringsAsFactors=FALSE). In your case, you might need to
either use that option while reading the data or convert the factor varia
Hi R Experts,
I was trying to develop model (bioclim, Domin) using a table data in 'dismo'
package. The data I have is at table format instead of images (stack of raster
images). I followed the procedures of "dismo" package to calculate the bioclim
but I could not figure it out how I can implem
On 31/08/13 22:35, prakashdevkumar wrote:
I have an Ubuntu Quantal 12.10 Server 64-bit instance. Trying to install R
libraries. Facing issue in installing library(qdap)
library(openNLP)
Can you suggest me how to go ahead.
No one should reply to you until you learn that what you are trying to
i
On Mon, Sep 2, 2013 at 5:01 PM, David Epstein
wrote:
> Dear Yihui
> Thanks very much for drawing my attention to knitr, which I had not heard of
> before. Also thanks for pointing out the bug in Sweave, which I don't fully
> understand, but I don't want to spend time and effort on understanding
Dear Yihui
Thanks very much for drawing my attention to knitr, which I had not heard of
before. Also thanks for pointing out the bug in Sweave, which I don't fully
understand, but I don't want to spend time and effort on understanding it. So I
hope you will find time to report the bug. I was pre
Sorry, there was a typo in my original message:
> df <- structure(list(var = c(1, 0, 0, 1, 0, 2, 2, 0, 2, 0),
+ cauc = c(6462.32876712329, 1585.27397260274,
2481.67808219178,
+ 344.178082191781, 8871.57534246575, 816.780821917808,
+ 6031.33561643836, 1013.52739726027, 4913.52739726027,
+
Hi,
You could try:
df2<- do.call(cbind,split(df[,-1],df[,1]))
res<-sapply(seq_len(ncol(df2)),function(i) {x<-df2[,i];x[duplicated(x)]<-NA;x})
dimnames(res)<- dimnames(df2)
res
# 0 1 2
#[1,] 1585.274 6462.3288 816.7808
#[2,] 2481.678 344.1781 6031.3356
#[3,] 8871.575
Thanks for the reproducible data set. The unstack() function
produces a list of three vectors, one for each value of var, but
it cannot combine them into a matrix since the number of entries
in each is not the same. To get that you need to pad each vector
with NAs:
> df <- structure(list(var = c(1
Hi,
On a bigger dataset:
#Speed:
set.seed(285)
dat1<-
as.data.frame(matrix(paste0(sample(1:10,69*3e5,replace=TRUE),sample(LETTERS[1:10],69*3e5,replace=TRUE)),ncol=69,nrow=3e5),stringsAsFactors=FALSE)
length(unique(unlist(dat1)))
#[1] 100
set.seed(3490)
dat2<-
data.frame(old=unique(unlist(dat1))
Hi,
You may try this:
set.seed(285)
dat1<-
as.data.frame(matrix(paste0(sample(1:10,100,replace=TRUE),sample(LETTERS[1:10],100,replace=TRUE)),10,10),stringsAsFactors=FALSE)
set.seed(3490)
dat2<-
data.frame(old=unique(unlist(dat1)),new=sample(1:100,63,replace=FALSE),stringsAsFactors=FALSE)
dat1
Gents:
The "eval(parse(...))" construction should almost always be avoided: it is
basically a misuse of R. There are exceptions, I suppose, but this does not
appear to be one of them.
Note that the use of numeric indexing does appear to be slightly faster
than logical indexing here, although I wo
HI,
You may try this:
dat1<- read.table(text="
CustID TripDate Store Bread Butter Milk Eggs
1 2-Jan-12 a 2 0 2 1
1 6-Jan-12 c 0 3 3 0
1 9-Jan-12 a 3 3 0 0
1 31-Mar-13 a 3 0 0 0
2 31-Aug-12 a 0 3 3 0
2 24-Sep-12 a 3 3 0 0
2 25-Sep-12 b 3 0 0 0
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- d
Dear all,
I think this is an easy task, but I don't know how to do it. Specifically, I
have 69 columns with 300.000 rows. In each cell there is a code like
"2E3", "4RR", etc.
I now have a list that replaces this with values, e.g.,
old new
2E3 5
4RR 3
etc.
The lis
Thank you A.K.
And do you have a solution without installing any package ?
Thank you in advance.
E.H.
Edouard Hardy
On Mon, Sep 2, 2013 at 5:56 PM, arun wrote:
>
>
> HI,
> In my first solutions:
> n<-3
>
>
> t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),n,nrow(Anew,colPro
tt <- function(x) {
obrien <- function(x) {
r <- rank(x)
(r - 0.5)/(0.5 + length(r) - r)
}
unlist(tapply(x, riskset, obrien))
}
hi, i am newer in R. when dealing with a survival data, i have fo
Thank you all for your responses.
The real problem is that all your answer work for products 2 by 2.
I now have to do the product n by n row.
Do you have a solution ?
Thank you in advance,
E.H.
Edouard Hardy
On Mon, Sep 2, 2013 at 5:43 PM, arun wrote:
> I guess in such situations,
>
>
> fun1<
Dear R users,
I'm now dealing with some big data set as big as 10GB, and my RAM got
16GB. Every time I run some models to this dataset using R, I have to
first load it into RAM via read.csv, which spends lots time.
I find the bigmemory package in the high performance task view of R, and
tried thi
My data is in this form: var has 3 conditions (0,1,2)
> df
var cauc
11 6462.3288
20 1585.2740
30 2481.6781
41 344.1781
50 8871.5753
62 816.7808
72 6031.3356
80 1013.5274
92 4913.5274
10 0 1517.2500
For the three conditions (0,1,2) I want the cauc-va
I think Thierry meant gsub("_", "_", version$platform); he just
typed too quickly. The point is to escape _ using \, but then people
are often trapped in the dreams of dreams of dreams of backslashes
like the movie Inception. And then due to a long-standing bug in
Sweave for \Sexpr{} (sorry I f
On 13-09-02 3:18 PM, David Epstein wrote:
Dear Thierry,
Your suggestion doesn't work on my version of R. Here's what I get
gsub("_", "\_", print(version$platform)
Error: '\_' is an unrecognized escape in character string starting ""\_"
print(gsub("_", "\_", version$platform))
Error: '\_' is
Hi,
Make sure you check the class of the columns. I forgot about that.
str(mat1)
int [1:10, 1:3] 1 2 10 18 2 16 1 18 12 19 ...
Convert it to numeric.
mat1New<- sapply(split(mat1,col(mat1)),as.numeric)
n<- 40
nrow(mat1New)%%40
#[1] 0
system.time({
j40<- n*seq_len(nrow(mat1New)/n)
vec1<
On Aug 28, 2013, at 11:53 AM, Gabriel Wajnberg wrote:
>
> Good Afternoon,
> My name is Gabriel, I'm doing an analysis if there is increase or decrease in
> dependence on the mutated genes, using 3 or more genes using the fisher exact
> test.I performed with success an analysis for two genes usi
Hi,
You can try this:
n<- 3
j3<-n*seq_len(nrow(A)/n)
vec1<- rep("j3",n)
eval(parse(text=paste0("A","[",paste0(vec1,"-",seq(n)-1),",]",collapse="*")))
# [,1] [,2] [,3]
#[1,] 28 80 162
#[2,] 162 80 28
Just saw Bert's new solution:
n<-3
j <- seq_len(nrow(A))%%n
b <- A[j==0,]
for(i in
Dear Thierry,
Your suggestion doesn't work on my version of R. Here's what I get
> gsub("_", "\_", print(version$platform)
Error: '\_' is an unrecognized escape in character string starting ""\_"
> print(gsub("_", "\_", version$platform))
Error: '\_' is an unrecognized escape in character string s
## For arbitrary n, just loop over the lag!
Here's an alternative version using logical instead of numerical indexing
(assumes nrow(A) %% n = 0).
You could also use the numerical indexing as before, of course. Doubt that
it would make much of a speed difference, but you can try it and see.
j <- s
Dear John,
thank you very much for your answer. I take a look at these packages (Rvmmin
and Rcgmin). That sounds very interesting.
For the example: The method relies on data which I always try to avoid to send
on the r-help list - not that my data is confidential - but it becomes even
more cu
HI,
You could modify Bert's solution:
n<-3
j3<-n*seq_len(nrow(A)/n)
A[j3,]*A[j3-1,]*A[j3-2,] ##assuming that nrow(dataset)%%n==0
# [,1] [,2] [,3]
#[1,] 28 80 162
#[2,] 162 80 28
#Speed comparison
set.seed(28)
mat1<- matrix(sample(1:20,1e5*3,replace=TRUE),ncol=3)
n<-4
system.tim
I use microbenchmark to time various of my code segments and find it
very useful. However, by accident I called it with the expression I
wanted to time quoted. This simply measured the time to evaluate the
quote. The following illustrates the difference. When explained, the
issue is obvious, bu
Hi,
If you check ?str()
str(zseq)
#List of 6
# $ : int [1:19] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:20] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:21] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:22] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:23] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:24] 1 2 3 4 5 6 7 8 9 10 ...
#
Hi,
No problem.
n<- 4
t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),n,nrow(Anew,function(x)
apply(x,2,prod)))
# V1 V2 V3
#1 252 640 1134
#2 18 30 20
This could be a bit slow if you have big dataset.
A.K.
From: Edouard Hardy
T
HI,
In my first solutions:
n<-3
t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),n,nrow(Anew,colProds))
# [,1] [,2] [,3]
#1 28 80 162
#2 162 80 28
#3 1 3 5
n<-4
t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),n,nrow(Anew,colProds))
# [,1]
You have to escape the underscore
\Sexpr{gsub("_", "\_", print(version$platform))}
Best regards,
Thierry
Van: r-help-boun...@r-project.org [r-help-boun...@r-project.org] namens David
Epstein [david.epst...@warwick.ac.uk]
Verzonden: maandag 2 september 2
I guess in such situations,
fun1<- function(mat){
if(nrow(mat)%%2==0){
j<- 2*seq_len(nrow(mat)/2)
b<- mat[j,]* mat[j-1,]
}
else {mat1<- mat[-nrow(mat),]
j<- 2*seq_len(nrow(mat1)/2)
b<- rbind(mat1[j,]*mat1[j-1,],mat[nrow(mat),])
}
b
}
fun1(A)
# [,1] [,2] [,3]
#[1,] 4 10 18
#[2,
I tried example('apply'). Among the various examples, there was the following:
apply> z <- array(1:24, dim = 2:4)
apply> zseq <- apply(z, 1:2, function(x) seq_len(max(x)))
apply> zseq ## a 2 x 3 matrix
[,1] [,2] [,3]
[1,] Integer,19 Integer,21 Integer,23
[2,] Integer,20 In
I am working with Sweave and would like to print out into my latex document the
result of the R command
version$platform
So what I first tried in my .Rnw document was \Sexpr{print(version$platform)}.
However, the output from this command is the string "x86_64-apple-darwin10.8.0"
(without the quo
Hi Bert,
Thanks. It is a better solution.
If nrow() is not even.
Anew<- rbind(A,c(1,3,5))
j<-seq_len(nrow(Anew)/2)###
Anew[j,]*Anew[j-1,]
#Error in Anew[j, ] * Anew[j - 1, ] : non-conformable arrays
t(sapply(split(as.data.frame(Anew),as.numeric(gl(nrow(Anew),2,7))),colProds))
[,1] [,2] [,3]
Hi r-help,
I have been using your RHmm package for some time and have recently
had to try using the package for a new dataset.
Basically I have a dataset with a number of discrete observation
variables that change over time, and I would love to try modeling them
using a HMM.
Basically I was wond
These elaborate manipulations are unnecessary and inefficient. Use indexing
instead:
j <- 2*seq_len(nrow(A)/2)
b <- A[j,]*A[j-1,]
b
[,1] [,2] [,3]
[1,]4 10 18
[2,] 63 64 63
[3,] 18 104
[,1] [,2] [,3]
[1,]4 10 18
[2,] 63 64 63
[3,] 18 104
[,1
This may be one of the many mysteries of the internals of L-BFGS-B,
which I have found fails from time to time. That is one of the reasons
for Rvmmin and Rcgmin (and hopefully sooner rather than later Rtn - a
truncated Newton method, currently working for unconstrained problems,
but still glitc
Dear R-helpers,
I'm trying to combine two box plots having two dependent variables:
var1- Orange.area and
var2- Iridescent.area;
two independent categorical factors (each has two levels - 'High' & 'Low'):
fact1- Quantity
fact2- Quality
the data frame (df) is:
Quantity Quality Orange.are
Hi,
You could try:
A<- matrix(unlist(read.table(text="
1 2 3
4 5 6
7 8 9
9 8 7
6 5 4
3 2 1
",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL)
library(matrixStats)
res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds))
res1
# [,1] [,2] [,3]
#1 4 10 18
#2 6
Hi Ramón,
It is for the column index.
For ex:
tags_totals[order(tags_totals[,1],decreasing=TRUE),1,drop=FALSE] #same as
previous solution as there is only one column.
# [,1]
#Grupos 23
#Wikis 15
#Glosarios 11
#Bases de datos 7
#Taller 5
Hello R experts,
I am trying to use hmap from package seriation or bertinplot.
I have two questions:
How can I specify smaller font? I tried with
pushViewport(viewport(layout=grid.layout(nrow = 1, ncol = 2),
+ gp = gpar(fontsize = 8)))
but didn't work for the font with bertinplot.
Also for hmap I
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