[R] request: How to get column name
Dear R community I have a problem regarding which of the column in a matrix contains all of zero elements. e.g. x=c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8); x=matrix(x, nrow=4) the output is > x [,1] [,2] [,3] [,4] [1,]3058 [2,]3058 [3,]3058 [4,]3058 In this case the required column is second so the result should be "2". How can i get it? best regards Muhammad Azam [[alternative HTML version deleted]] __ 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.
[R] loop problem
Dear R members
I have a problem regarding storing the lists.
Let
L=number of distinct values of any predictor (say L=5)
P=number of predictors (say P=20)
g1 <- c()
for(i in 1:P){
if(L > 1){
for(j in 1:(L-1)){
g <-
g1[j] <- g
}
}
g2[]=sort.list(g1)
}
Now the question is: What should we use inside brackets of g2[], whether
"i" or some thing else? If L is not greater than 1 then there will be a "NULL"
for g2. We don't want to store it in g2, so how can we handle this problem.
Looking forward for some help. Thanks and
best regards
Muhammad Azam
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[R] Request: Optimum value of cost complexity parameter "k" in "tree" package
Dear R community
I have a question regarding the value of cost complexity parameter "k" used in
"tree" package for pruning purpose. Any help in finding the optimum value of
"k" is requested. Please give some suggestion in this regard. In the example
below i used k=0 but i don't know why? But if i use k=NULL, then it will not
plot the resultant tree.
library(tree)
ds=iris; iris=transform(iris, Species = factor(Species, labels = letters[1:3]))
miris <- tree(Species ~ ., data = iris, control=tree.control(nobs = 150,
minsize = 5, mincut = 2)); iris.prun=prune.tree(miris, method=c("misclass"),
best = NULL, k=0); iris.prun; summary(iris.prun); plot(iris.prun)
best regards
Muhammad Azam
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[R] request: cost complexity parameter "k"
Dear R community
I
have a question regarding the value of cost complexity parameter "k"
used in "tree" package for pruning purpose. In the example below i used k=0.
But if i take the value k=NULL, then it will not plot the resultant tree. Any
help in finding the
optimum value of "k" is requested. Please give some suggestion in this
regard. Thanks
library(tree)
ds=iris; iris=transform(iris, Species = factor(Species, labels = letters[1:3]))
miris
<- tree(Species ~ ., data = iris, control=tree.control(nobs = 150,
minsize = 5, mincut = 2)); iris.prun=prune.tree(miris,
method=c("misclass"), best = NULL, k=0); iris.prun; summary(iris.prun);
plot(iris.prun)
best regards
Muhammad Azam
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[R] request: maximum depth reached problem
Dear R community
Hope all of you are fine. I have a question regarding the an error message.
Actually, I am trying to generate classification trees using "tree" package. It
works well but for some datasets e.g., wine, yeast, boston housing etc. it
gives an error message.
Error in tree(V14 ~ ., data = training.data, method =
c("recursive.partitioning"), :
maximum depth reached
The structure of getting output is given below:
iris.tr = tree(Species ~., data=training.data,
method=c("recursive.partitioning"), split = c("gini"),
control=tree.control(nobs = 150, minsize = 5, mincut = 2))
Any suggestion will be appreciated to handle the above problem. Thanks and
best regards
Muhammad Azam
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[R] negative extents to matrix
s2 # vv=variable value
data=LN(data1,s1,s2); testdata=LN(data2,s1,s2); side <<- c("L")
funcp(data,testdata,levelp)
data=RN(data1,s1,s2); testdata=RN(data2,s1,s2); side <<- c("R")
funcp(data,testdata,levelp)
}
}
funcp(mm,ww,0)
clp=apply(vlp,3,function(vlp) which(rowSums(vlp)!=0))[1:n];
for(i in 1:nnodes){for(ii in 1:nc){if(any(vlp[,,i][ii,]!=0)){vlfp[[i]] <-
vlp[,,i][ii,]; vvlfp[[i]] <- vvlp[,,i][ii,]}}}; vlf1p <- do.call('rbind',vlfp);
vvlf1p <- do.call('rbind',vvlfp);
lp=vlf1p; indlp <- colSums(lp) != 0; lp=matrix(lp[,indlp],nrow=nrow(lp));
llp=vvlf1p; indllp <- colSums(llp) != 0;
llp=round(matrix(llp[,indllp],nrow=nrow(llp)),digits=4);
print("pruned tree"); matp <-
as.data.frame(matrix(paste(lp,"(",llp,")",sep=""),nrow=nrow(lp)));
deviance <- round(deviancep,digits=2); LorRnode <- nodesidep; class <- nodeclasp
resMatp <- cbind(matp,LorRnode,unit,misc.unit,deviance,class); print(resMatp)
eTrain <- round(mean(eTraining)/avelength*100,digits=2); eTest <-
round(tec2NF/nrow(ww)*100,digits=2); dev.sum <- sum(deviance);
print(cbind(eTrain,eTest,dev.sum))
}
CTPrundUnprund(iris,150,1,4,5,1,3,10,10)
best regards
Muhammad Azam
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[R] request: which integer in each column is in majority
Respected R helpers/ users I am one of the new R user. I have a problem regarding to know which of the integer in each column of the following matrix is in majority. I want to know that integer e.g. in the first column 1 is in majority. Similarly in the third column 4 is in majority. So what is the suitable way to get the desired integer for each column. I am looking for some kind reply. Thanks example: > x=matrix(c(1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,4,1,2,3,3),ncol=4) > x [,1] [,2] [,3] [,4] [1,]1234 [2,]1241 [3,]1342 [4,]2343 [5,]2343 best regards Muhammad Azam Ph.D. Student Department of Medical Statistics, Informatics and Health Economics University of Innsbruck, Austria [[alternative HTML version deleted]] __ 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.
[R] request: To add an extra row in a matrix
Dear R users I have a problem regarding an addition of an extra "row" to a matrix. e.g. i have a matrix a <- matrix(1:6,2,3) > a [,1] [,2] [,3] [1,]135 [2,]246 I want to add a matrix having just one row. e.g. b <- matrix(7:9,1,3) > b [,1] [,2] [,3] [1,]789 Now i want to get result like this [,1] [,2] [,3] [1,]135 [2,]246 [3,]789 Can any body help to get the required result. Thanks and best regards Muhammad Azam Ph.D. Student Department of Medical Statistics, Informatics and Health Economics University of Innsbruck, Austria [[alternative HTML version deleted]] __ 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.
[R] request: An array declarion problem
Dear R users
I tried a lot to solve the following problem but could not. I have two arrays
having same order i.e 1 by 150.
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,
]-rr[ j, ]
}
Now i want to "rbind" the results of r1 and r2 for each time. Now the order of
r3[ j, ] will become 2 by 150. I used the following form in the loop
r3[ i,]=rbind(r1[ i, ], r2[ i, ])
But there is an error message
Error in r3[i, ] = rbind(r1[i, ], r2[i, ]) : number of items to replace is not
a multiple of replacement length
I am looking for some suggestion to solve the problem. Thanks and
best regards
Muhammad Azam
Ph.D. Student
Department of Medical Statistics,
Informatics and Health Economics
University of Innsbruck, Austria
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[R] request: a class having max frequency
Dear R users I have a very basic question. I tried but could not find the required result. using dat <- pima f <- table(dat[,9]) > f 0 1 500 268 i want to find that class say "0" having maximum frequency i.e 500. I used >which.max(f) which provide 0 1 How can i get only the "0". Thanks and best regards Muhammad Azam Ph.D. Student Department of Medical Statistics, Informatics and Health Economics University of Innsbruck, Austria [[alternative HTML version deleted]] __ 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.
[R] request: To access a particular list
Dear R community
I have a problem to access particular list. I have a code given below where
there is recursive process. It is not possible to run it because there are few
other functions involved inside like sv, LN, RN etc.
k=0; n=0; variable=c(); vr<-list()
func <- function(data,testdata)
{
.
.
if(..){
n<<-n+1; vr[[n]] <<- variable; print(vr)
}else
{
k <<- k + 1; data1 = data; data2 = testdata;
s=sv(data1)[[1]]; s1=s[1]; s2=s[2]; variable[k] <<- s1
data=LN(data1,s1,s2); testdata=LN(data2,s1,s2)
func(data,testdata);
data=RN(data1,s1,s2); testdata=RN(data2,s1,s2)
func(data,testdata)
}
}
func(m,w)
Actually i made an global array to store s1 named "variable". It contains "k"
values. When it enters into "if", i made a list of variable named "vr". So it
will make "n" lists. If i get the print out of "vr", i will get
[[1]]
[1] 4
[[1]]
[1] 4
[[2]]
[1] 4 4 3
[[1]]
[1] 4
[[2]]
[1] 4 4 3
[[3]]
[1] 4 4 3
[[1]]
[1] 4
[[2]]
[1] 4 4 3
[[3]]
[1] 4 4 3
[[4]]
[1] 4 4 3
But if i print(vr[[n]]), i get result like an array
[1] 4
[1] 4 4 3
[1] 4 4 3
[1] 4 4 3
I want to access a particular list vr[[n-1]], but it gives an error
"Error in print(vr[[n - 1]]) : attempt to select less than one element".
How can i tackle this problem to access vr[[n-1]]. Please give some suitable
suggestions.
Thanks and best regards
Muhammad Azam
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[R] request: most repeated sequnce
Dear R community Hope every one be in best of his/her health. I have a situation in which there are s-sectors. Each sector is further divided into r-rows and c-columns. All it makes an array having dimension (r,c,s). e.g. x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4, 0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0) x=array(x,dim=c(3,6,5)) > x , , 1 [,1] [,2] [,3] [,4] [,5] [,6] [1,]100000 [2,]100000 [3,]100000 , , 2 [,1] [,2] [,3] [,4] [,5] [,6] [1,]123000 [2,]123000 [3,]120000 , , 3 [,1] [,2] [,3] [,4] [,5] [,6] [1,]123400 [2,]123400 [3,]134000 , , 4 [,1] [,2] [,3] [,4] [,5] [,6] [1,]123400 [2,]123400 [3,]123400 , , 5 [,1] [,2] [,3] [,4] [,5] [,6] [1,]120000 [2,]223400 [3,]223400 I want to get the most repeated sequence (row-wise) of values in each sector. e.g. in sector 1 i.e. , , 1 the most repeated sequence is 1 (ignoring zeros). In , , 2 the most repeated sequence is 1 2 3. Similarly in last sector i.e. , , 5 such sequence is 2 2 3 4. Any body can help to solve this problem. Thanks best regards Muhammad Azam [[alternative HTML version deleted]] __ 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.
Re: [R] request: most repeated sequnce
Dear R community
Initially i thought my problem has been solved but one thing which i found e.g.
if
1. All the elements of a sector are zero e.g
, , 7
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]000000000 0
[2,]000000000 0
[3,]000000000 0
[4,]000000000 0
[5,]000000000 0
2. Majority of the rows consist of zeros e.g.
, , 5
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]440000000 0
[2,]440000000 0
[3,]000000000 0
[4,]000000000 0
[5,]000000000 0
Actually
zeros are not my values. I get values and fill the remaining parts with
zeros like "x=array(0,dim=c(3,6,5))". Now according to first strategy
"000000000 0" are most repeated
sequence of rows in both of above cases. But i don't want to consider
cases where all elements are zeros and interested to get "44
0000000 0" or just " 4 4 " in case 2.
Thanks and best regards
Muhammad Azam
- Original Message ----
From: jim holtman <[EMAIL PROTECTED]>
To: Muhammad Azam <[EMAIL PROTECTED]>
Cc: R Help ; R-help request <[EMAIL PROTECTED]>
Sent: Saturday, September 6, 2008 2:39:19 PM
Subject: Re: [R] request: most repeated sequnce
Here is a start. You can delete the zeros:
> x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,
+ 0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0)
> x=array(x,dim=c(3,6,5))
> apply(x,3,function(.mat){
+ rows <- table(apply(.mat,1,function(z){
+ paste(z,collapse=' ')
+ }))
+ names(rows[which.max(rows)])
+ })
[1] "1 0 0 0 0 0" "1 2 3 0 0 0" "1 2 3 4 0 0" "1 2 3 4 0 0" "2 2 3 4 0 0"
>
On Sat, Sep 6, 2008 at 4:54 AM, Muhammad Azam <[EMAIL PROTECTED]> wrote:
> Dear R community
> Hope every one be in best of his/her health. I have a situation in which
> there are s-sectors. Each sector is further divided into r-rows and
> c-columns. All it makes an array having dimension (r,c,s). e.g.
>
> x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,
> 0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0)
> x=array(x,dim=c(3,6,5))
>> x
> , , 1
>
> [,1] [,2] [,3] [,4] [,5] [,6]
> [1,]100000
> [2,]100000
> [3,]100000
>
> , , 2
>
> [,1] [,2] [,3] [,4] [,5] [,6]
> [1,]123000
> [2,]123000
> [3,]120000
>
> , , 3
>
> [,1] [,2] [,3] [,4] [,5] [,6]
> [1,]123400
> [2,]123400
> [3,]134000
>
> , , 4
>
> [,1] [,2] [,3] [,4] [,5] [,6]
> [1,]123400
> [2,]123400
> [3,]123400
>
> , , 5
>
> [,1] [,2] [,3] [,4] [,5] [,6]
> [1,]120000
> [2,]223400
> [3,]223400
>
> I want to get the most repeated sequence (row-wise) of values in each sector.
> e.g. in sector 1 i.e. , , 1
> the most repeated sequence is 1 (ignoring zeros). In , , 2 the most repeated
> sequence is 1 2 3. Similarly in last sector i.e.
> , , 5 such sequence is 2 2 3 4. Any body can help to solve this problem.
> Thanks
>
>
> best regards
> Muhammad Azam
>
>
>
>
>[[alternative HTML version deleted]]
>
> __
> 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.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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[R] request: table of sequences
Dear R community
The following code gives me the most repeated sequence of values. i.e.
x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,
0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
x=array(x,dim=c(3,6,7))
apply(x,3,function(.mat){
rows <- table(apply(.mat,1,function(z){
# remove the zeros
z <- z[z != 0]
paste(z,collapse=' ')
}))
# remove empty strings
rows <- rows[names(rows) != ""]
if (!is.null(rows)){
return(names(rows)[which.max(rows)])# return(table(names(rows)))
} else return(NULL)
})
I am trying to get frequencies of all the sequences instead of most repeated
using "return(table(names(rows)))". But i could not get the desired results (i
mean frequencies of all possible sequences)e.g. for first sector the result
should be
1
3
and so on. Please suggest me some suitable way. Thanks
Muhammad Azam
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[R] request: table of sequences
Dear R community Thanks. The problem has been solved by just using "return(rows)" Muhammad Azam [[alternative HTML version deleted]] __ 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.
[R] request: most repeated component of a list
Dear R community I have stored the results of arrays in a list consist of J-components (say 200 components). Each component containing same no of columns but may be different no of rows. e.g [[1]] [,1] [,2] [,3] [,4] [,5] [1,]40000 [2,]43400 [3,]43400 [4,]43000 [[2]] [,1] [,2] [,3] [,4] [,5] [1,]40000 [2,]44300 [3,]44300 [4,]44000 [5,]44000 [[3]] [,1] [,2] [,3] [,4] [,5] [1,]40000 [2,]44100 [3,]44100 [4,]44000 For 200 components i want to make a frequency table. How can i make a frequency table of these components or the most repeated component out of all? Any help in this regard will be appreciated. Muhammad Azam [[alternative HTML version deleted]] __ 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.
Re: [R] request: most repeated component of a list
May be i could not explain properly. Actually there are components of list i.e. [[1]] to [[500]]. Each component containing r-rows (may be different for each [[ k ]] and c-columns same for all). I have to compare all the [[ k ]] components of that list and found the one appearing maximum no of times. e.g. from three components [[1]] to [[3]] given below. The most repeated is [,1] [,2] [,3] [,4] [,5] [1,]40000 [2,]43400 [3,]43400 [4,]43000Please help to find it. Thanks and best regards Muhammad Azam - Original Message From: jim holtman <[EMAIL PROTECTED]> To: Muhammad Azam <[EMAIL PROTECTED]> Cc: R-help request <[EMAIL PROTECTED]>; R Help Sent: Wednesday, September 10, 2008 5:59:28 PM Subject: Re: [R] request: most repeated component of a list If want you want is the summary from all of them, then 'rbind' the data together into one matrix and analyze it: totalMat <- do.call(rbind, listOfMatrices) On Wed, Sep 10, 2008 at 11:49 AM, Muhammad Azam <[EMAIL PROTECTED]> wrote: > Dear R community > I have stored the results of arrays in a list consist of J-components (say 200 components). Each component containing same no of columns but may be different no of rows. e.g > [[1]] > [,1] [,2] [,3] [,4] [,5] > [1,]40000 > [2,]43400 > [3,]43400 > [4,]43000 > > [[2]] > [,1] [,2] [,3] [,4] [,5] > [1,]40000 > [2,]43400 > [3,]43400 > [4,]43000 > > [[3]] > [,1] [,2] [,3] [,4] [,5] > [1,]40000 > [2,]44100 > [3,]44100 > [4,]44000 > [5,]44000 > > For 200 components i want to make a frequency table. How can i make a frequency table of these components or the most repeated component out of all? Any help in this regard will be appreciated. > > > Muhammad Azam > > > >[[alternative HTML version deleted]] > > __ > 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. > -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[alternative HTML version deleted]] __ 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.
Re: [R] request: most repeated component of a list
Thanks for the effort but still we are far from the desired result. May be this
example will help you to understand the situation. Example
a1=c(1:12); a1=array(a1,dim=c(3,4)); a2=c(1:12); a2=array(a2,dim=c(3,4));
a3=c(1:16)
a3=array(a3,dim=c(4,4));
a=list(a1,a2,a3);
a
[[1]]
[,1] [,2] [,3] [,4]
[1,]147 10
[2,]258 11
[3,]369 12
[[2]]
[,1] [,2] [,3] [,4]
[1,]147 10
[2,]258 11
[3,]369 12
[[3]]
[,1] [,2] [,3] [,4]
[1,]159 13
[2,]26 10 14
[3,]37 11 15
[4,]48 12 16
Here [[1]] and [[2]] are same out of three (internal values wise). The whole
array [[1]] or [[2]] is in majority. So i want to get the whole array or
component of list which is in majority. The result should be like this
[,1] [,2] [,3] [,4]
[1,]147 10
[2,]258 11
[3,]369 12
Hope it is much more clear as before.
best regards
Muhammad Azam
- Original Message
From: Adam D. I. Kramer <[EMAIL PROTECTED]>
To: Muhammad Azam <[EMAIL PROTECTED]>
Cc: R Help
Sent: Thursday, September 11, 2008 9:53:40 AM
Subject: Re: [R] request: most repeated component of a list
That is indeed different from what I thought the first time.
x <- sapply(1:length(l), function(x) {
sum(sapply(l, function(y) {
if ( nrow(l[[x]]) != nrow(y) | ncol(l[[x]]) != ncol(y) ) FALSE
else sum(y != l[[x]]) == 0
}))
} )
names(x) <- names(l)
Then, x has the same names as l, and x[i] is the number of matches that
l[[i]] has...so you want the index or indices of max(x).
--Adam
On Thu, 11 Sep 2008, Muhammad Azam wrote:
> May be i could not explain properly. Actually there are components of
> list i.e. [[1]] to [[500]]. Each component containing r-rows (may be
> different for each [[ k ]] and c-columns same for all). I have to
> compare all the [[ k ]] components of that list and found the one
> appearing maximum no of times. e.g. from three components [[1]] to
> [[3]] given below. The most repeated is
>
> [,1] [,2] [,3] [,4] [,5]
> [1,]40000
> [2,]43400
> [3,]43400
> [4,]43000Please help to find it. Thanks and
>
> best regards
> Muhammad Azam
>
>
> ----- Original Message
> From: jim holtman <[EMAIL PROTECTED]>
> To: Muhammad Azam <[EMAIL PROTECTED]>
> Cc: R-help request <[EMAIL PROTECTED]>; R Help
> Sent: Wednesday, September 10, 2008 5:59:28 PM
> Subject: Re: [R] request: most repeated component of a list
>
> If want you want is the summary from all of them, then 'rbind' the
> data together into one matrix and analyze it:
>
> totalMat <- do.call(rbind, listOfMatrices)
>
> On Wed, Sep 10, 2008 at 11:49 AM, Muhammad Azam <[EMAIL PROTECTED]> wrote:
>> Dear R community
>>
> I have stored the results of arrays in a list consist of J-components
> (say 200 components). Each component containing same no of columns but
> may be different no of rows. e.g
>> [[1]]
>> [,1] [,2] [,3] [,4] [,5]
>> [1,]40000
>> [2,]43400
>> [3,]43400
>> [4,]43000
>>
>> [[2]]
>> [,1] [,2] [,3] [,4] [,5]
>> [1,]40000
>> [2,]43400
>> [3,]43400
>> [4,]43000
>>
>> [[3]]
>> [,1] [,2] [,3] [,4] [,5]
>> [1,]40000
>> [2,]44100
>> [3,]44100
>> [4,] 44000
>> [5,]44000
>>
>>
> For 200 components i want to make a frequency table. How can i make a
> frequency table of these components or the most repeated component out
> of all? Any help in this regard will be appreciated.
>>
>>
>> Muhammad Azam
>>
>>
>>
>>[[alternative HTML version deleted]]
>>
>> __
>> 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.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
>
>
> [[alternative HTML version deleted]]
>
> __
> 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.
>
[[alternative HTML version deleted]]
__
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.
Re: [R] request: most repeated component of a list
Thanks a lot for this effort.
- Original Message
From: Dimitris Rizopoulos <[EMAIL PROTECTED]>
To: Muhammad Azam <[EMAIL PROTECTED]>
Cc: [EMAIL PROTECTED]; R Help
Sent: Thursday, September 11, 2008 10:52:16 AM
Subject: Re: [R] request: most repeated component of a list
try the following:
ff <- function (x) {
do.call("paste", c(as.data.frame(x), sep = "\r", collapse = ""))
}
pats <- sapply(a, ff)
ind <- which.max(table(pats))
a[[ind]]
I hope it helps.
Best,
Dimitris
> Thanks for the effort but still we are far from the desired result. May be
> this example will help you to understand the situation. Example
> a1=c(1:12); a1=array(a1,dim=c(3,4)); a2=c(1:12);
> a2=array(a2,dim=c(3,4)); a3=c(1:16)
> a3=array(a3,dim=c(4,4));
> a=list(a1,a2,a3);
> a
> [[1]]
> [,1] [,2] [,3] [,4]
> [1,]147 10
> [2,]258 11
> [3,]369 12
>
> [[2]]
> [,1] [,2] [,3] [,4]
> [1,]147 10
> [2,]258 11
> [3,]369 12
>
> [[3]]
> [,1] [,2] [,3] [,4]
> [1,]159 13
> [2,]26 10 14
> [3,]37 11 15
> [4,]48 12 16
>
> Here [[1]] and [[2]] are same out of three (internal values wise). The
> whole array [[1]] or [[2]] is in majority. So i want to get the whole
> array or component of list which is in majority. The result should be like
> this
> [,1] [,2] [,3] [,4]
> [1,] 147 10
> [2,]258 11
> [3,]369 12
>
> Hope it is much more clear as before.
>
> best regards
> Muhammad Azam
>
>
> - Original Message
> From: Adam D. I. Kramer <[EMAIL PROTECTED]>
> To: Muhammad Azam <[EMAIL PROTECTED]>
> Cc: R Help
> Sent: Thursday, September 11, 2008 9:53:40 AM
> Subject: Re: [R] request: most repeated component of a list
>
> That is indeed different from what I thought the first time.
>
> x <- sapply(1:length(l), function(x) {
>sum(sapply(l, function(y) {
> if ( nrow(l[[x]]) != nrow(y) | ncol(l[[x]]) != ncol(y) ) FALSE
> else sum(y != l[[x]]) == 0
>}))
> } )
>
> names(x) <- names(l)
>
> Then, x has the same names as l, and x[i] is the number of matches that
> l[[i]] has...so you want the index or indices of max(x).
>
> --Adam
>
> On Thu, 11 Sep 2008, Muhammad Azam wrote:
>
>> May be i could not explain properly. Actually there are components of
>> list i.e. [[1]] to [[500]]. Each component containing r-rows (may be
>> different for each [[ k ]] and c-columns same for all). I have to
>> compare all the [[ k ]] components of that list and found the one
>> appearing maximum no of times. e.g. from three components [[1]] to
>> [[3]] given below. The most repeated is
>>
>> [,1] [,2] [,3] [,4] [,5]
>> [1,]40000
>> [2,]43400
>> [3,]43400
>> [4,]43000Please help to find it. Thanks and
>>
>> best regards
>> Muhammad Azam
>>
>>
>> - Original Message
>> From: jim holtman <[EMAIL PROTECTED]>
>> To: Muhammad Azam <[EMAIL PROTECTED]>
>> Cc: R-help request <[EMAIL PROTECTED]>; R Help
>>
>> Sent: Wednesday, September 10, 2008 5:59:28 PM
>> Subject: Re: [R] request: most repeated component of a list
>>
>> If want you want is the summary from all of them, then 'rbind' the
>> data together into one matrix and analyze it:
>>
>> totalMat <- do.call(rbind, listOfMatrices)
>>
>> On Wed, Sep 10, 2008 at 11:49 AM, Muhammad Azam <[EMAIL PROTECTED]>
>> wrote:
>>> Dear R community
>>>
>> I have stored the results of arrays in a list consist of J-components
>> (say 200 components). Each component containing same no of columns but
>> may be different no of rows. e.g
>>> [[1]]
>>> [,1] [,2] [,3] [,4] [,5]
>>> [1,]40000
>>> [2,]43400
>>> [3,]43400
>>> [4,]43000
>>>
>>> [[2]]
>>> [,1] [,2] [,3] [,4] [,5]
>>> [1,]40000
>>> [2,]43400
>>> [3,]43400
>>> [4,]43000
>>>
>>> [[3]]
>>> [,1] [,2] [,3] [,4] [,5]
>>> [1,]40000
>>> [2,]44100
>>> [3,]44100
>>> [4,]44000
>>> [5,]44000
[R] request: How to ignore columns having zero sums
Dear friends I have an array consist of r-rows and c-columns e.g. x=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,0,0,0,0,0,0,0,0); x1=array(x, dim=c(4,6)) output is > x1 [,1] [,2] [,3] [,4] [,5] [,6] [1,]123400 [2,]123400 [3,]123400 [4,]123400 How can i ignore columns having zero sums? Help in this regard is needed. Thanks M.Azam [[alternative HTML version deleted]] __ 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.
Re: [R] request: How to ignore columns having zero sums
Dear Dimitris Thanks a lot. - Original Message From: Dimitris Rizopoulos <[EMAIL PROTECTED]> To: Muhammad Azam <[EMAIL PROTECTED]> Cc: R Help Sent: Tuesday, October 14, 2008 10:10:54 AM Subject: Re: [R] request: How to ignore columns having zero sums try this: x <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,0,0,0,0,0,0,0,0); x1 <- array(x, dim=c(4,6)) ind <- colSums(x1) != 0 x1[, ind] I hope it helps. Best, Dimitris Muhammad Azam wrote: > Dear friends > I have an array consist of r-rows and c-columns e.g. > x=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,0,0,0,0,0,0,0,0); > x1=array(x, dim=c(4,6)) > output is >> x1 > [,1] [,2] [,3] [,4] [,5] [,6] > [1,]123400 > [2,]123400 > [3,]123400 > [4,]123400 > How can i ignore columns having zero sums? Help in this regard is needed. > Thanks > > > M.Azam > > > > [[alternative HTML version deleted]] > > __ > 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. > -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 [[alternative HTML version deleted]] __ 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.
Re: [R] request: How to ignore columns having zero sums
Dear Jorge Thanks a lot. - Original Message From: Jorge Ivan Velez <[EMAIL PROTECTED]> To: Muhammad Azam <[EMAIL PROTECTED]> Cc: R mailing list Sent: Tuesday, October 14, 2008 1:52:15 PM Subject: Re: [R] request: How to ignore columns having zero sums Dear Muhammad, Try also x <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,0,0,0,0,0,0,0,0); x1 <- array(x, dim=c(4,6)) x1[,apply(x1,2,function(x) !all(x==0))] HTH, Jorge On Tue, Oct 14, 2008 at 4:02 AM, Muhammad Azam <[EMAIL PROTECTED]> wrote: Dear friends I have an array consist of r-rows and c-columns e.g. x=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,0,0,0,0,0,0,0,0); x1=array(x, dim=c(4,6)) output is > x1 [,1] [,2] [,3] [,4] [,5] [,6] [1,]123400 [2,]123400 [3,]123400 [4,]123400 How can i ignore columns having zero sums? Help in this regard is needed. Thanks M.Azam [[alternative HTML version deleted]] __ 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. [[alternative HTML version deleted]] __ 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.
[R] request: How can we ignore a component of list having no element
Dear friends There is a list of arrays comprising different no of rows and columns even sometimes NULL, such as [[2]] given below. How can we ignore [[2]] or others like this in the complete list. Any help in this regard is needed. Thanks [[1]] [,1] [,2] [1,]31 [2,]31 [3,]31 [[2]] NULL [[3]] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,]3100000 [2,]3100000 [3,]3100000 [4,]3131321 [5,]3131321 [6,]3131320 [[4]] [,1] [,2] [,3] [,4] [1,]3000 [2,]3133 [3,]3133 [4,]3130 OR x1=c(1,2,3); x2=c(1,2,3,4,6); x3=c(); x=list(x1,x2,x3) M.Azam [[alternative HTML version deleted]] __ 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.
Re: [R] request: How can we ignore a component of list having no element
Dear Mahbub Thanks a lot for the help. - Original Message From: Mahbub Latif <[EMAIL PROTECTED]> To: Muhammad Azam <[EMAIL PROTECTED]> Sent: Wednesday, October 15, 2008 12:30:13 PM Subject: Re: [R] request: How can we ignore a component of list having no element try this, junk <- sapply(x,function(i) !is.null(i)) y <- x[junk] On Wed, Oct 15, 2008 at 11:23 AM, Muhammad Azam <[EMAIL PROTECTED]> wrote: Dear friends There is a list of arrays comprising different no of rows and columns even sometimes NULL, such as [[2]] given below. How can we ignore [[2]] or others like this in the complete list. Any help in this regard is needed. Thanks [[1]] [,1] [,2] [1,]31 [2,]31 [3,]31 [[2]] NULL [[3]] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,]3100000 [2,]3100000 [3,]3100000 [4,]3131321 [5,]3131321 [6,]3131320 [[4]] [,1] [,2] [,3] [,4] [1,]3000 [2,]3133 [3,]3133 [4,]3130 OR x1=c(1,2,3); x2=c(1,2,3,4,6); x3=c(); x=list(x1,x2,x3) M.Azam [[alternative HTML version deleted]] __ 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. -- A.H.M. Mahbub Latif Assistant Professor Applied Statistics Institute of Statistical Research and Training University of Dhaka, Dhaka 1000, Bangladesh web : http://www.isrt.ac.bd/mlatif [[alternative HTML version deleted]] __ 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.
[R] Request: How to draw a tree
Dear friends As a result I get an array containing certain no of rows and columns. In the resultant array first row represents first node of a tree starting from left side, second row represents second node of that tree and so on. In the example below, the resultant tree contains 6 nodes. We get the first node on L.H.S by splitting top node using variable 10, node 2 by taking start from the R.H.S node then splitting it using variables 7 and 11 respectively [ Note: 0's are not the variables ]. All the nodes are connected on this way to get the whole tree. We can easily draw it manually. Now the question is, is it possible to draw it using R. Any help in this regard is needed. Thanks [,1] [,2] [,3] [,4] [1,] 10000 [2,] 107 110 [3,] 107 110 [4,] 107 13 11 [5,] 107 13 11 [6,] 107 130 regards M. Azam [[alternative HTML version deleted]] __ 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.
[R] Request: Most repeated sequence considering combinations at each row
Dear friends
Hope you all are fine. Suppose we have a list of arrays.
a1=c(4,4,4,4,0,4,4,4,0,3,3,0,0,0,0,0); a1=array(a1,dim=c(4,4));
a2=c(4,4,4,4,0,4,4,4,0,3,3,0,0,0,0,0); a2=array(a2,dim=c(4,4));
a3=c(4,4,4,4,0,3,3,4,0,4,4,0,0,0,0,0); a3=array(a3,dim=c(4,4));
a4=c(4,4,4,4,4,0,3,3,3,3,0,4,4,4,0,0,0,0,0,0); a4=array(a4,dim=c(5,4));
a5=c(4,4,4,4,4,0,4,4,4,4,0,3,3,3,0,0,1,1,0,0); a5=array(a5,dim=c(5,4));
a6=c(4,4,4,4,4,0,1,1,1,1,0,4,4,4,0,0,3,3,0,0); a6=array(a6,dim=c(5,4));
a7=c(1,1,1,1,1,0,4,4,4,4,0,3,3,3,0,0,4,4,0,0); a7=array(a7,dim=c(5,4));
a8=c(4,4,4,4,4,0,3,3,3,3,0,1,1,1,0,0,4,4,0,0); a8=array(a8,dim=c(5,4));
l=list(a1,a2,a3,a4,a5,a6,a7,a8);
x <- sapply(1:length(l), function(x) {
sum(sapply(l, function(y) {
if ( nrow(l[[x]]) != nrow(y) | ncol(l[[x]]) != ncol(y) ) FALSE
else sum(y != l[[x]]) == 0
}))
} ); l; x
Using the above function, we are able to get frequency of each most repeated
similar components of the list. For example, [[1]] and [[2]] are most repeated
similar out of all. But if we consider the "combinations" at each row of each
array. Then [[3]] will be included with [[1]] and [[2]]. Also [[5]], [[6]] and
[[8]] will be similar. How can we modify the above function to get the desired
most repeated sequence in this case? Any help in this regard is needed.
best regards
M.Azam
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: Request: Most repeated sequence considering combinations at each row
Dear Petr
Thanks for the response. Hope it will now help me to proceed.
best regards
M.Azam
From: Petr PIKAL <[EMAIL PROTECTED]>
Cc: R Help ; [EMAIL PROTECTED]
Sent: Monday, October 27, 2008 8:43:27 AM
Subject: Odp: [R] Request: Most repeated sequence considering combinations at
each row
Hi
not sure if this is what you want. It does not do fuzzy matching but make
a exact evaluation equal row sums of arrays.
rle(do.call("c",lapply(lapply(l, rowSums), function(x) paste(x,
collapse=""
Maybe something similar can be done without conversion to character.
Regards
Petr
[EMAIL PROTECTED] napsal dne 24.10.2008 12:34:15:
> Dear friends
> Hope you all are fine. Suppose we have a list of arrays.
> a1=c(4,4,4,4,0,4,4,4,0,3,3,0,0,0,0,0); a1=array(a1,dim=c(4,4));
a2=c(4,4,4,
> 4,0,4,4,4,0,3,3,0,0,0,0,0); a2=array(a2,dim=c(4,4));
> a3=c(4,4,4,4,0,3,3,4,0,4,4,0,0,0,0,0); a3=array(a3,dim=c(4,4));
a4=c(4,4,4,4,
> 4,0,3,3,3,3,0,4,4,4,0,0,0,0,0,0); a4=array(a4,dim=c(5,4));
a5=c(4,4,4,4,4,0,4,
> 4,4,4,0,3,3,3,0,0,1,1,0,0); a5=array(a5,dim=c(5,4));
a6=c(4,4,4,4,4,0,1,1,1,1,
> 0,4,4,4,0,0,3,3,0,0); a6=array(a6,dim=c(5,4));
a7=c(1,1,1,1,1,0,4,4,4,4,0,3,3,
> 3,0,0,4,4,0,0); a7=array(a7,dim=c(5,4));
a8=c(4,4,4,4,4,0,3,3,3,3,0,1,1,1,0,0,
> 4,4,0,0); a8=array(a8,dim=c(5,4));
> l=list(a1,a2,a3,a4,a5,a6,a7,a8);
>
> x <- sapply(1:length(l), function(x) {
> sum(sapply(l, function(y) {
> if ( nrow(l[[x]]) != nrow(y) | ncol(l[[x]]) != ncol(y) ) FALSE
> else sum(y != l[[x]]) == 0
> }))
> } ); l; x
>
> Using the above function, we are able to get frequency of each most
repeated
> similar components of the list. For example, [[1]] and [[2]] are most
repeated
> similar out of all. But if we consider the "combinations" at each row of
each
> array. Then [[3]] will be included with [[1]] and [[2]]. Also [[5]],
[[6]] and
> [[8]] will be similar. How can we modify the above function to get the
desired
> most repeated sequence in this case? Any help in this regard is needed.
>
> best regards
> M.Azam
>
>
>
>
>[[alternative HTML version deleted]]
>
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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[R] request: How to combine three matrices in the desired form
Dear R-friends I have three matrices e.g. var <- matrix(c(4,4,4,4,0,4,4,4,0,3,3,0),nrow=4); val <- matrix(c(0.6,0.6,0.6,0.6,0,1.6,1.6,1.6,0,4.9,4.9,0),nrow=4); nod <- matrix(c(-1,-1,1,1),ncol=1) > var [,1] [,2] [,3] [1,]400 [2,]443 [3,]443 [4,]440 > val [,1] [,2] [,3] [1,] 0.6 0.0 0.0 [2,] 0.6 1.6 4.9 [3,] 0.6 1.6 4.9 [4,] 0.6 1.6 0.0 > nod [,1] [1,] -1 [2,] -1 [3,]1 [4,]1 Always there is same number of rows and columns for "var" and "val". Also no of rows of all three are always same. Combining them we get > cbind(var,val,nod) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,]400 0.6 0.0 0.0 -1 [2,]443 0.6 1.6 4.9 -1 [3,]443 0.6 1.6 4.91 [4,]440 0.6 1.6 0.01 Is there any way to write them in the following form? [,1] [,2] [,3] [,4] [1,]4(0.6)0 0 -1 [2,]4(0.6)4(1.6)3(4.9)-1 [3,]4(0.6)4(1.6)3(4.9) 1 [4,]4(0.6)4(1.6)01 Any help in this regard will be appreciated. best regards M.Azam [[alternative HTML version deleted]] __ 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.
Re: [R] request: How to combine three matrices in the desired form
Dear Jim
Thanks a lot for such a nice solution.
best regards
M.Azam
From: Jim Lemon <[EMAIL PROTECTED]>
Cc: R Help
Sent: Thursday, October 30, 2008 10:06:57 AM
Subject: Re: [R] request: How to combine three matrices in the desired form
Hi Muhammad,
Try this:
cbind(matrix(paste(var,"(",val,")",sep=""),nrow=4),as.character(nod))
Jim
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[R] request: how to assign alphabets to integer values
Dear R community
I am trying to assign alphabets to integer values 1, 2, 3 etc. in y given
below. Can any body suggest some simple way to do the same job?
ds=iris; dl=nrow(ds)
c1=ds[,1]; c2=ds[,2]; c3=ds[,3]; c4=ds[,4]; c5=ds[,5];
iris=cbind(c1,c2,c3,c4,c5)
y=iris[,5]
y1=which(y==1); y[y1] <- c("a"); y2=which(y==2); y[y2] <- c("b");
y3=which(y==3); y[y3] <- c("c");
iris=cbind(c1,c2,c3,c4,y)
Thnks and best regards
M. Azam
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Re: [R] request: how to assign alphabets to integer values
Dear Gabor Grothendieck
Thanks a lot for the help.
M. Azam
From: Gabor Grothendieck <[EMAIL PROTECTED]>
Cc: R-help request <[EMAIL PROTECTED]>; R Help
Sent: Monday, December 1, 2008 12:46:06 PM
Subject: Re: [R] request: how to assign alphabets to integer values
Try:
transform(iris, Species = factor(Species, labels = letters[1:3]))
> Dear R community
> I am trying to assign alphabets to integer values 1, 2, 3 etc. in y given
> below. Can any body suggest some simple way to do the same job?
>
> ds=iris; dl=nrow(ds)
> c1=ds[,1]; c2=ds[,2]; c3=ds[,3]; c4=ds[,4]; c5=ds[,5];
> iris=cbind(c1,c2,c3,c4,c5)
> y=iris[,5]
> y1=which(y==1); y[y1] <- c("a"); y2=which(y==2); y[y2] <- c("b");
> y3=which(y==3); y[y3] <- c("c");
> iris=cbind(c1,c2,c3,c4,y)
>
> Thnks and best regards
> M. Azam
>
>
>
> [[alternative HTML version deleted]]
>
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
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