Re: [R] R help on loops
Dear R users,
I am stuck here: My first function returns a vector of 5 values.
In my second function, I want to repeat this, a number of times, say 10
so that I have 10 rows and five columns but I keep on getting errors.
See the code and results below:
optm -function(perm, verbose = FALSE)
{
trace-c()
for (k in 1:perm){
trace[k]-Rspatswap(rhox=0.6,rhoy=0.6,sigmasqG=0.081,SsqR=1)[1]
perm[k]-k
mat-cbind(trace, perm = seq(perm))
}
if (verbose){
cat(***starting matrix\n)
print(mat)
}
# iterate till done
while(nrow(mat) 1){
high - diff(mat[, 'trace']) 0
if (!any(high)) break # done
# find which one to delete
delete - which.max(high) + 1L
mat - mat[-delete, ]
newmat-apply(mat,2,mean)[1]
sdm-sd(mat[,1])
sem-sdm/sqrt(nrow(mat))
maxv-mat[1,1]
minv-mat[nrow(mat),1]
}
stats-cbind(average=newmat,sd=sdm,se=sem,min=minv,max=maxv)
stats
}
test-optm(perm=20)
test
average sd se min max
trace 0.8880286 0.0009178193 0.0004589096 0.8870152 0.889241
itn-function(it){
siml-matrix(NA,ncol=5,nrow=length(it))
for(g in 1:it){
siml[g]-optm(perm=20)
}
siml-cbind(siml=siml)
siml
}
ans-itn(5)
Warning messages:
1: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
2: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
3: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
4: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
5: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
ans
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8874234 0.8861666 0.8880521 0.8870958 0.8876469
On 5/31/2013 9:53 PM, jim holtman wrote:
Is this what you want? I was not clear on your algorithm, but is
looks like you wanted descending values:
testx -
+ function(n, verbose = FALSE)
+ {
+ mat - cbind(optA = sample(n, n, TRUE), perm = seq(n))
+ if (verbose){
+ cat(***starting matrix\n)
+ print(mat)
+ }
+ # iterate till done
+ while(nrow(mat) 1){
+ high - diff(mat[, 'optA']) 0
+ if (!any(high)) break # done
+ # find which one to delete
+ delete - which.max(high) + 1L
+ mat - mat[-delete, ]
+ }
+ mat
+ }
testx(20, verbose = TRUE)
***starting matrix
optA perm
[1,] 131
[2,] 122
[3,]73
[4,] 104
[5,] 115
[6,]46
[7,] 117
[8,]28
[9,]69
[10,]5 10
[11,]6 11
[12,] 18 12
[13,]9 13
[14,] 16 14
[15,] 18 15
[16,]9 16
[17,]2 17
[18,]7 18
[19,] 15 19
[20,]7 20
optA perm
[1,] 131
[2,] 122
[3,]73
[4,]46
[5,]28
[6,]2 17
On Fri, May 31, 2013 at 2:02 PM, Laz lmra...@ufl.edu
mailto:lmra...@ufl.edu wrote:
Dear R Users,
I created a function which returns a value every time it's run (A
simplified toy example is attached on this mail).
For example spat(a,b,c,d) # run the first time and it gives you ans1=
10, perm=1 ;
run the second time and gives you ans2= 7, perm=2 etc
I store both the result and the iteration on a matrix called vector
with columns:1==ans, column2==permutation
The rule I want to implement is: compare between ans1 and ans2. If
ans1ans2, keep both ans1 and ans2. if ans1ans2, drop the whole
row of
the second permutation (that is drop both ans2 and perm2 but continue
counting all permutations).
Re-run the function for the 3rd time and repeat comparison
between the
value of the last run and the current value obtained.
Return matrix ans with the saved ans and their permutations and
another
full matrix with all results including the dropped ans and their
permutation numbers.
I need to repeat this process 1000 times but I am getting stuck below.
See attached R code.
The code below works only for the first 6 permutations. suppose I want
1000 permutations, where do I keep the loop?
Example: Not a perfect code though it somehow works:
testx-function(perm){
optA-c()
#while(perm=2){
for (k in 1:perm){
#repeat {
optA[k]-sample(1:1000,1,replace=TRUE)
perm[k]-k
}
mat2-as.matrix(cbind(optA=optA,perm=perm))
result-mat2
lenm-nrow(result)
if(result[1,1]=result[2,1]) result-result[1,]
return(list(mat2=mat2,result=result))
#}
if(perm2){
mat2-as.matrix(cbind(optA=optA,perm=perm))
Re: [R] R help on loops
On 07-06-2013, at 10:59, Laz lmra...@ufl.edu wrote:
Dear R users,
I am stuck here: My first function returns a vector of 5 values.
In my second function, I want to repeat this, a number of times, say 10
so that I have 10 rows and five columns but I keep on getting errors.
See the code and results below:
optm -function(perm, verbose = FALSE)
{
trace-c()
for (k in 1:perm){
trace[k]-Rspatswap(rhox=0.6,rhoy=0.6,sigmasqG=0.081,SsqR=1)[1]
perm[k]-k
mat-cbind(trace, perm = seq(perm))
}
if (verbose){
cat(***starting matrix\n)
print(mat)
}
# iterate till done
while(nrow(mat) 1){
high - diff(mat[, 'trace']) 0
if (!any(high)) break # done
# find which one to delete
delete - which.max(high) + 1L
mat - mat[-delete, ]
newmat-apply(mat,2,mean)[1]
sdm-sd(mat[,1])
sem-sdm/sqrt(nrow(mat))
maxv-mat[1,1]
minv-mat[nrow(mat),1]
}
stats-cbind(average=newmat,sd=sdm,se=sem,min=minv,max=maxv)
stats
}
test-optm(perm=20)
test
average sd se min max
trace 0.8880286 0.0009178193 0.0004589096 0.8870152 0.889241
itn-function(it){
siml-matrix(NA,ncol=5,nrow=length(it))
for(g in 1:it){
siml[g]-optm(perm=20)
}
siml-cbind(siml=siml)
siml
}
ans-itn(5)
Warning messages:
1: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
2: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
3: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
4: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
5: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
ans
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8874234 0.8861666 0.8880521 0.8870958 0.8876469
1. Not reproducible code. Where does function Rspatswap come from?
2. You have several errors in function itn:
Argument it is a scalar: length(it) is 1. You need to do siml -
matrix(NA,ncol=5,nrow=it)
Next in the g-loop you want to fill row g so do: siml[g,] - …..
Finally why are you doing siml - cbind(siml=siml)?
Seems superfluous to me. Delete the line.
Berend
__
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] R help on loops
Dear Berend,
For reproducibility,
Rspatswap() is a function which originally returns a single value. For
example Rspatswap(...) and you get 0.8
So, run Rspatswap() 20 times and store all the values.
Then from these 20 values, calculate the calculate average,
sd,se,min,max to get something similar to:
average sd se min max
trace 0.8880286 0.0009178193 0.0004589096 0.8870152 0.889241
If we repeat the function 10 times, then I expect 10 rows with 5 columns but it
does not work ?
I hope I am now clearer...
On 6/7/2013 5:24 AM, Berend Hasselman wrote:
On 07-06-2013, at 10:59, Laz lmra...@ufl.edu wrote:
Dear R users,
I am stuck here: My first function returns a vector of 5 values.
In my second function, I want to repeat this, a number of times, say 10
so that I have 10 rows and five columns but I keep on getting errors.
See the code and results below:
optm -function(perm, verbose = FALSE)
{
trace-c()
for (k in 1:perm){
trace[k]-Rspatswap(rhox=0.6,rhoy=0.6,sigmasqG=0.081,SsqR=1)[1]
perm[k]-k
mat-cbind(trace, perm = seq(perm))
}
if (verbose){
cat(***starting matrix\n)
print(mat)
}
# iterate till done
while(nrow(mat) 1){
high - diff(mat[, 'trace']) 0
if (!any(high)) break # done
# find which one to delete
delete - which.max(high) + 1L
mat - mat[-delete, ]
newmat-apply(mat,2,mean)[1]
sdm-sd(mat[,1])
sem-sdm/sqrt(nrow(mat))
maxv-mat[1,1]
minv-mat[nrow(mat),1]
}
stats-cbind(average=newmat,sd=sdm,se=sem,min=minv,max=maxv)
stats
}
test-optm(perm=20)
test
average sd se min max
trace 0.8880286 0.0009178193 0.0004589096 0.8870152 0.889241
itn-function(it){
siml-matrix(NA,ncol=5,nrow=length(it))
for(g in 1:it){
siml[g]-optm(perm=20)
}
siml-cbind(siml=siml)
siml
}
ans-itn(5)
Warning messages:
1: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
2: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
3: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
4: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
5: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
ans
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8874234 0.8861666 0.8880521 0.8870958 0.8876469
1. Not reproducible code. Where does function Rspatswap come from?
2. You have several errors in function itn:
Argument it is a scalar: length(it) is 1. You need to do siml -
matrix(NA,ncol=5,nrow=it)
Next in the g-loop you want to fill row g so do: siml[g,] - …..
Finally why are you doing siml - cbind(siml=siml)?
Seems superfluous to me. Delete the line.
Berend
__
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] R help on loops
Hi,
I am almost getting there, but still have errors. Thanks for your help.
I have tried improving but I get the following errors below:
itn-function(it){
+siml-matrix(NA,ncol=5,nrow=it)
+for(g in 1:it){
+siml[g]-optm(perm=20)[g]
+}
+siml
+}
itn(3)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8873775898 NA NA NA NA
[2,] 0.0015584824 NA NA NA NA
[3,] 0.0001414317 NA NA NA NA
itn-function(it){
+siml-matrix(NA,ncol=5,nrow=it)
+for(g in 1:it){
+siml[g]-optm(perm=20)
+}
+siml
+}
itn(3)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8880941 NA NA NA NA
[2,] 0.8869727 NA NA NA NA
[3,] 0.8877045 NA NA NA NA
Warning messages:
1: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
2: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
3: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
I expect something close to
average sd se min max
0.8881969 0.0008215379 0.000410769 0.8873842 0.8890167
0.884659 0.0004215379 0.000410769 0.2342 0.676307
0.8885839 0.0001215379 0.0002112 0.82752992 0.8836337
Thanks fpr you help.
On 6/7/2013 5:24 AM, Berend Hasselman wrote:
On 07-06-2013, at 10:59, Laz lmra...@ufl.edu wrote:
Dear R users,
I am stuck here: My first function returns a vector of 5 values.
In my second function, I want to repeat this, a number of times, say 10
so that I have 10 rows and five columns but I keep on getting errors.
See the code and results below:
optm -function(perm, verbose = FALSE)
{
trace-c()
for (k in 1:perm){
trace[k]-Rspatswap(rhox=0.6,rhoy=0.6,sigmasqG=0.081,SsqR=1)[1]
perm[k]-k
mat-cbind(trace, perm = seq(perm))
}
if (verbose){
cat(***starting matrix\n)
print(mat)
}
# iterate till done
while(nrow(mat) 1){
high - diff(mat[, 'trace']) 0
if (!any(high)) break # done
# find which one to delete
delete - which.max(high) + 1L
mat - mat[-delete, ]
newmat-apply(mat,2,mean)[1]
sdm-sd(mat[,1])
sem-sdm/sqrt(nrow(mat))
maxv-mat[1,1]
minv-mat[nrow(mat),1]
}
stats-cbind(average=newmat,sd=sdm,se=sem,min=minv,max=maxv)
stats
}
test-optm(perm=20)
test
average sd se min max
trace 0.8880286 0.0009178193 0.0004589096 0.8870152 0.889241
itn-function(it){
siml-matrix(NA,ncol=5,nrow=length(it))
for(g in 1:it){
siml[g]-optm(perm=20)
}
siml-cbind(siml=siml)
siml
}
ans-itn(5)
Warning messages:
1: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
2: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
3: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
4: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
5: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
ans
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8874234 0.8861666 0.8880521 0.8870958 0.8876469
1. Not reproducible code. Where does function Rspatswap come from?
2. You have several errors in function itn:
Argument it is a scalar: length(it) is 1. You need to do siml -
matrix(NA,ncol=5,nrow=it)
Next in the g-loop you want to fill row g so do: siml[g,] -
..
Finally why are you doing siml - cbind(siml=siml)?
Seems superfluous to me. Delete the line.
Berend
[[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] R help on loops
On 07-06-2013, at 11:57, Laz lmra...@ufl.edu wrote:
Dear Berend,
I have made some changes but I still get errors:
For reproducibility,
Rspatswap() is a function which originally returns a single value. For
example Rspatswap(...) and you get 0.8
So, run Rspatswap() 20 times and store all the values.
Then from these 20 values, calculate the calculate average, sd,se,min,max to
get something similar to:
average sd se min max
trace 0.8880286 0.0009178193 0.0004589096 0.8870152 0.889241
If we repeat the function 10 times, then I expect 10 rows with 5 columns but
it does not work ?
Replacing Rspatswap(…)[1] with runif(1) (you could/should have done that to
provide reproducible code) and the modifications I suggested you make in your
function itn() like this
itn-function(it){
siml-matrix(NA,ncol=5,nrow=it)
for(g in 1:it){
siml[g,]-optm(perm=20)
}
siml
}
running itn(10) I get a matrix with 10 rows and 5 columns.
So what do you mean by it does not work?
You sometimes get an error message from apply due to a bad dim() like this
Error in apply(mat, 2, mean) : dim(X) must have a positive length
When the result of mat[-delete,] is a matrix with a single row the result is
simplified to a vector. See ?[ in particular the drop argument.
So this should work in function optm()
mat - mat[-delete,, drop=FALSE]
Berend
__
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] R help on loops
Please, please do not use html formatted mail.
It is clearly requested by the mailing list
The code of your last mail is a mess and when replying it becomes even more of
a mess.
I told you to do
siml[g,] - optm(perm=20)
See the comma after g.
and not what you are now doing with siml[g]-optm(perm=20)[g]
I'm giving up.
Berend
On 07-06-2013, at 12:15, Laz lmra...@ufl.edu wrote:
Hi,
I am almost getting there, but still have errors. Thanks for your help. I
have tried improving but I get the following errors below:
itn-function(it){
+
siml-matrix(NA,ncol=5,nrow=it)
+
for(g in 1:it){
+
siml[g]-optm(perm=20)[g]
+
}
+
siml
+
}
itn(3)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8873775898 NA NA NA NA
[2,] 0.0015584824 NA NA NA NA
[3,] 0.0001414317 NA NA NA NA
itn-function(it){
+
siml-matrix(NA,ncol=5,nrow=it)
+
for(g in 1:it){
+
siml[g]-optm(perm=20)
+
}
+
siml
+
}
itn(3)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8880941 NA NA NA NA
[2,] 0.8869727 NA NA NA NA
[3,] 0.8877045 NA NA NA NA
Warning messages:
1: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
2: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
3: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
I expect something close to
average sd se min max
0.8881969 0.0008215379 0.000410769 0.8873842 0.8890167
0.884659 0.0004215379 0.000410769 0.2342 0.676307
0.8885839 0.0001215379 0.0002112 0.82752992 0.8836337
Thanks fpr you help.
On 6/7/2013 5:24 AM, Berend Hasselman wrote:
On 07-06-2013, at 10:59, Laz lmra...@ufl.edu
wrote:
Dear R users,
I am stuck here: My first function returns a vector of 5 values.
In my second function, I want to repeat this, a number of times, say 10
so that I have 10 rows and five columns but I keep on getting errors.
See the code and results below:
optm -function(perm, verbose = FALSE)
{
trace-c()
for (k in 1:perm){
trace[k]-Rspatswap(rhox=0.6,rhoy=0.6,sigmasqG=0.081,SsqR=1)[1]
perm[k]-k
mat-cbind(trace, perm = seq(perm))
}
if (verbose){
cat(***starting matrix\n)
print(mat)
}
# iterate till done
while(nrow(mat) 1){
high - diff(mat[, 'trace']) 0
if (!any(high)) break # done
# find which one to delete
delete - which.max(high) + 1L
mat - mat[-delete, ]
newmat-apply(mat,2,mean)[1]
sdm-sd(mat[,1])
sem-sdm/sqrt(nrow(mat))
maxv-mat[1,1]
minv-mat[nrow(mat),1]
}
stats-cbind(average=newmat,sd=sdm,se=sem,min=minv,max=maxv)
stats
}
test-optm(perm=20)
test
average sd se min max
trace 0.8880286 0.0009178193 0.0004589096 0.8870152 0.889241
itn-function(it){
siml-matrix(NA,ncol=5,nrow=length(it))
for(g in 1:it){
siml[g]-optm(perm=20)
}
siml-cbind(siml=siml)
siml
}
ans-itn(5)
Warning messages:
1: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
2: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
3: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
4: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
5: In siml[g] - optm(perm = 20) :
number of items to replace is not a multiple of replacement length
ans
[,1] [,2] [,3] [,4] [,5]
[1,] 0.8874234 0.8861666 0.8880521 0.8870958 0.8876469
1. Not reproducible code. Where does function Rspatswap come from?
2. You have several errors in function itn:
Argument it is a scalar: length(it) is 1. You need to do siml -
matrix(NA,ncol=5,nrow=it)
Next in the g-loop you want to fill row g so do: siml[g,] - …..
Finally why are you doing siml - cbind(siml=siml)?
Seems superfluous to me. Delete the line.
Berend
__
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] R help on loops
Dear R Users,
I created a function which returns a value every time it's run (A
simplified toy example is attached on this mail).
For example spat(a,b,c,d) # run the first time and it gives you ans1=
10, perm=1 ;
run the second time and gives you ans2= 7, perm=2 etc
I store both the result and the iteration on a matrix called vector
with columns:1==ans, column2==permutation
The rule I want to implement is: compare between ans1 and ans2. If
ans1ans2, keep both ans1 and ans2. if ans1ans2, drop the whole row of
the second permutation (that is drop both ans2 and perm2 but continue
counting all permutations).
Re-run the function for the 3rd time and repeat comparison between the
value of the last run and the current value obtained.
Return matrix ans with the saved ans and their permutations and another
full matrix with all results including the dropped ans and their
permutation numbers.
I need to repeat this process 1000 times but I am getting stuck below.
See attached R code.
The code below works only for the first 6 permutations. suppose I want
1000 permutations, where do I keep the loop?
Example: Not a perfect code though it somehow works:
testx-function(perm){
optA-c()
#while(perm=2){
for (k in 1:perm){
#repeat {
optA[k]-sample(1:1000,1,replace=TRUE)
perm[k]-k
}
mat2-as.matrix(cbind(optA=optA,perm=perm))
result-mat2
lenm-nrow(result)
if(result[1,1]=result[2,1]) result-result[1,]
return(list(mat2=mat2,result=result))
#}
if(perm2){
mat2-as.matrix(cbind(optA=optA,perm=perm))
result-mat2
lenm-nrow(result)
if(result[1,1]=result[2,1]) result-result[1,]
if(result[lenm-1,1]=result[lenm,1]) result-result[-lenm,]
if(result[(lenm-2),1]=result[(lenm-1),1])
result-result[-(lenm-1),]
if(result[(lenm-3),1]=result[(lenm-2),1])
result-result[-(lenm-2),]
if(result[(lenm-4),1]=result[(lenm-3),1])
result-result[-(lenm-3),]
if(result[(lenm-5),1]=result[(lenm-4),1])
result-result[-(lenm-4),]
return(list(mat2=mat2,result=result))
}
}
## Now calling my function below:
testx(perm=6)
$mat2
optA perm
[1,] 2721
[2,] 8582
[3,] 8343
[4,] 8624
[5,] 6505
[6,] 4056
$result
optA perm
2721
testx(perm=2)
$mat2
optA perm
[1,] 3981
[2,] 8162
$result
optA perm
3981
[[alternative HTML version deleted]]
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Re: [R] R help on loops
Is this what you want? I was not clear on your algorithm, but is looks
like you wanted descending values:
testx -
+ function(n, verbose = FALSE)
+ {
+ mat - cbind(optA = sample(n, n, TRUE), perm = seq(n))
+ if (verbose){
+ cat(***starting matrix\n)
+ print(mat)
+ }
+ # iterate till done
+ while(nrow(mat) 1){
+ high - diff(mat[, 'optA']) 0
+ if (!any(high)) break # done
+ # find which one to delete
+ delete - which.max(high) + 1L
+ mat - mat[-delete, ]
+ }
+ mat
+ }
testx(20, verbose = TRUE)
***starting matrix
optA perm
[1,] 131
[2,] 122
[3,]73
[4,] 104
[5,] 115
[6,]46
[7,] 117
[8,]28
[9,]69
[10,]5 10
[11,]6 11
[12,] 18 12
[13,]9 13
[14,] 16 14
[15,] 18 15
[16,]9 16
[17,]2 17
[18,]7 18
[19,] 15 19
[20,]7 20
optA perm
[1,] 131
[2,] 122
[3,]73
[4,]46
[5,]28
[6,]2 17
On Fri, May 31, 2013 at 2:02 PM, Laz lmra...@ufl.edu wrote:
Dear R Users,
I created a function which returns a value every time it's run (A
simplified toy example is attached on this mail).
For example spat(a,b,c,d) # run the first time and it gives you ans1=
10, perm=1 ;
run the second time and gives you ans2= 7, perm=2 etc
I store both the result and the iteration on a matrix called vector
with columns:1==ans, column2==permutation
The rule I want to implement is: compare between ans1 and ans2. If
ans1ans2, keep both ans1 and ans2. if ans1ans2, drop the whole row of
the second permutation (that is drop both ans2 and perm2 but continue
counting all permutations).
Re-run the function for the 3rd time and repeat comparison between the
value of the last run and the current value obtained.
Return matrix ans with the saved ans and their permutations and another
full matrix with all results including the dropped ans and their
permutation numbers.
I need to repeat this process 1000 times but I am getting stuck below.
See attached R code.
The code below works only for the first 6 permutations. suppose I want
1000 permutations, where do I keep the loop?
Example: Not a perfect code though it somehow works:
testx-function(perm){
optA-c()
#while(perm=2){
for (k in 1:perm){
#repeat {
optA[k]-sample(1:1000,1,replace=TRUE)
perm[k]-k
}
mat2-as.matrix(cbind(optA=optA,perm=perm))
result-mat2
lenm-nrow(result)
if(result[1,1]=result[2,1]) result-result[1,]
return(list(mat2=mat2,result=result))
#}
if(perm2){
mat2-as.matrix(cbind(optA=optA,perm=perm))
result-mat2
lenm-nrow(result)
if(result[1,1]=result[2,1]) result-result[1,]
if(result[lenm-1,1]=result[lenm,1]) result-result[-lenm,]
if(result[(lenm-2),1]=result[(lenm-1),1])
result-result[-(lenm-1),]
if(result[(lenm-3),1]=result[(lenm-2),1])
result-result[-(lenm-2),]
if(result[(lenm-4),1]=result[(lenm-3),1])
result-result[-(lenm-3),]
if(result[(lenm-5),1]=result[(lenm-4),1])
result-result[-(lenm-4),]
return(list(mat2=mat2,result=result))
}
}
## Now calling my function below:
testx(perm=6)
$mat2
optA perm
[1,] 2721
[2,] 8582
[3,] 8343
[4,] 8624
[5,] 6505
[6,] 4056
$result
optA perm
2721
testx(perm=2)
$mat2
optA perm
[1,] 3981
[2,] 8162
$result
optA perm
3981
[[alternative HTML version deleted]]
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--
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 to do it.
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Re: [R] help with loops
On 08/11/2011 07:51 PM, William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of R. Michael
Weylandt
Sent: Thursday, August 11, 2011 10:09 AM
To: Srinivas Iyyer
Cc: r-help@r-project.org
Subject: Re: [R] help with loops
No problem,
By the way, you can't (or at least shouldn't) use return() outside of a
function -- that was the source of your old error message.
If you, for whatever reason, couldn't use unlist() you would write:
OurUnlist - function(c, unique = F) {
if (!is.list(c)) return(c)
z - NULL
for (i in seq_along(c)) {
z - c(z,c[[i]])
}
if (unique) return(unique(z))
return(z)
}
or some such. Still, I suggest you stick with built in functions whenever
possible.
I tend to encourage people to write functions.
In addition, writing functions yourself is a good way to exercise your R
skills. On the other hand, built in functions often solve the problem
faster and are more generic. In my experience it takes quite a bit of R
knowledge before one is good enough to beat a general, builtin function.
Often a lengthy self written function can be replaced by one call to a
built in function. In addition, it saves a lot of time when you use the
already present functions. Several times I wanted something done in R,
only to find out that it was already done. This meant getting the job
done in 1 hour instead of two days of programming.
In general I tend to agree with Michael and encourage people to stick
with the builtin functions.
my 2 cts ;),
regards,
Paul
I suppose you may end up reinventing the wheel,
but once you get used to writing functions it
is often faster to write a specialized one than
to find one that meets your needs. When you
discover a new idiom for your task (e.g., calling
unlist() instead of the for loop), you just edit
one function (OurUnlist) instead of editing all
your scripts that used the old idiom).
Once you get used to writing functions (and using
them), you are ready to document them and package
them up for others to use.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Michael Weylandt
PS -- Can you email me (off list) and let me know what this is for? We've
been asked this question a couple of times over the last few days and I'm
just wondering why it's a calculation of interest to so many.
On Thu, Aug 11, 2011 at 1:00 PM, Srinivas Iyyer
srini_iyyer_...@yahoo.comwrote:
Thank you. that was very easy.
-srini
--- On *Thu, 8/11/11, R. Michael Weylandt
michael.weyla...@gmail.com*wrote:
From: R. Michael Weylandt michael.weyla...@gmail.com
Subject: Re: [R] help with loops
To: Srinivas Iyyer srini_iyyer_...@yahoo.com
Cc: r-help@r-project.org
Date: Thursday, August 11, 2011, 12:49 PM
unlist()
Michael Weylandt
On Thu, Aug 11, 2011 at 12:46 PM, Srinivas Iyyer
srini_iyyer_...@yahoo.com http://mc/compose?to=srini_iyyer_...@yahoo.com
wrote:
hi I need help with list object.
I have a list object
a - c('apple','orange','grape')
b - c('car','truck','jeep')
c - list(a,b)
names(c) - c('fruit','vehicle')
c
$fruit
[1] apple orange grape
$vehicle
[1] car truck jeep
I want to write all the elements of this list in one object 'z'.
z
[1] apple orange grape car truck jeep
How can I write the elements of c to z
I tried using a for loop. Could any one help me please. thanks
z - ''
for (i in 1:length(c)){
+ k - c[[i]]
+ z - c(z,k)
+ return(z)}
Error: no function to return from, jumping to top level
Thank you.
srini
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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--
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Global Climate Division
Royal Netherlands Meteorological Institute (KNMI)
Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39
P.O. Box 201 | 3730 AE | De Bilt
tel: +31 30 2206 494
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[R] help with loops
hi I need help with list object.
I have a list object
a - c('apple','orange','grape')
b - c('car','truck','jeep')
c - list(a,b)
names(c) - c('fruit','vehicle')
c
$fruit
[1] apple orange grape
$vehicle
[1] car truck jeep
I want to write all the elements of this list in one object 'z'.
z
[1] apple orange grape car truck jeep
How can I write the elements of c to z
I tried using a for loop. Could any one help me please. thanks
z - ''
for (i in 1:length(c)){
+ k - c[[i]]
+ z - c(z,k)
+ return(z)}
Error: no function to return from, jumping to top level
Thank you.
srini
__
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] help with loops
unlist()
Michael Weylandt
On Thu, Aug 11, 2011 at 12:46 PM, Srinivas Iyyer
srini_iyyer_...@yahoo.comwrote:
hi I need help with list object.
I have a list object
a - c('apple','orange','grape')
b - c('car','truck','jeep')
c - list(a,b)
names(c) - c('fruit','vehicle')
c
$fruit
[1] apple orange grape
$vehicle
[1] car truck jeep
I want to write all the elements of this list in one object 'z'.
z
[1] apple orange grape car truck jeep
How can I write the elements of c to z
I tried using a for loop. Could any one help me please. thanks
z - ''
for (i in 1:length(c)){
+ k - c[[i]]
+ z - c(z,k)
+ return(z)}
Error: no function to return from, jumping to top level
Thank you.
srini
__
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]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with loops
Thank you. that was very easy.
-srini
--- On Thu, 8/11/11, R. Michael Weylandt michael.weyla...@gmail.com wrote:
From: R. Michael Weylandt michael.weyla...@gmail.com
Subject: Re: [R] help with loops
To: Srinivas Iyyer srini_iyyer_...@yahoo.com
Cc: r-help@r-project.org
Date: Thursday, August 11, 2011, 12:49 PM
unlist()
Michael Weylandt
On Thu, Aug 11, 2011 at 12:46 PM, Srinivas Iyyer srini_iyyer_...@yahoo.com
wrote:
hi I need help with list object.
I have a list object
a - c('apple','orange','grape')
b - c('car','truck','jeep')
c - list(a,b)
names(c) - c('fruit','vehicle')
c
$fruit
[1] apple orange grape
$vehicle
[1] car truck jeep
I want to write all the elements of this list in one object 'z'.
z
[1] apple orange grape car truck jeep
How can I write the elements of c to z
I tried using a for loop. Could any one help me please. thanks
z - ''
for (i in 1:length(c)){
+ k - c[[i]]
+ z - c(z,k)
+ return(z)}
Error: no function to return from, jumping to top level
Thank you.
srini
__
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]]
__
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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] help with loops
No problem,
By the way, you can't (or at least shouldn't) use return() outside of a
function -- that was the source of your old error message.
If you, for whatever reason, couldn't use unlist() you would write:
OurUnlist - function(c, unique = F) {
if (!is.list(c)) return(c)
z - NULL
for (i in seq_along(c)) {
z - c(z,c[[i]])
}
if (unique) return(unique(z))
return(z)
}
or some such. Still, I suggest you stick with built in functions whenever
possible.
Michael Weylandt
PS -- Can you email me (off list) and let me know what this is for? We've
been asked this question a couple of times over the last few days and I'm
just wondering why it's a calculation of interest to so many.
On Thu, Aug 11, 2011 at 1:00 PM, Srinivas Iyyer
srini_iyyer_...@yahoo.comwrote:
Thank you. that was very easy.
-srini
--- On *Thu, 8/11/11, R. Michael Weylandt michael.weyla...@gmail.com*wrote:
From: R. Michael Weylandt michael.weyla...@gmail.com
Subject: Re: [R] help with loops
To: Srinivas Iyyer srini_iyyer_...@yahoo.com
Cc: r-help@r-project.org
Date: Thursday, August 11, 2011, 12:49 PM
unlist()
Michael Weylandt
On Thu, Aug 11, 2011 at 12:46 PM, Srinivas Iyyer
srini_iyyer_...@yahoo.com http://mc/compose?to=srini_iyyer_...@yahoo.com
wrote:
hi I need help with list object.
I have a list object
a - c('apple','orange','grape')
b - c('car','truck','jeep')
c - list(a,b)
names(c) - c('fruit','vehicle')
c
$fruit
[1] apple orange grape
$vehicle
[1] car truck jeep
I want to write all the elements of this list in one object 'z'.
z
[1] apple orange grape car truck jeep
How can I write the elements of c to z
I tried using a for loop. Could any one help me please. thanks
z - ''
for (i in 1:length(c)){
+ k - c[[i]]
+ z - c(z,k)
+ return(z)}
Error: no function to return from, jumping to top level
Thank you.
srini
__
R-help@r-project.org http://mc/compose?to=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]]
__
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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] help with loops
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of R. Michael
Weylandt
Sent: Thursday, August 11, 2011 10:09 AM
To: Srinivas Iyyer
Cc: r-help@r-project.org
Subject: Re: [R] help with loops
No problem,
By the way, you can't (or at least shouldn't) use return() outside of a
function -- that was the source of your old error message.
If you, for whatever reason, couldn't use unlist() you would write:
OurUnlist - function(c, unique = F) {
if (!is.list(c)) return(c)
z - NULL
for (i in seq_along(c)) {
z - c(z,c[[i]])
}
if (unique) return(unique(z))
return(z)
}
or some such. Still, I suggest you stick with built in functions whenever
possible.
I tend to encourage people to write functions.
I suppose you may end up reinventing the wheel,
but once you get used to writing functions it
is often faster to write a specialized one than
to find one that meets your needs. When you
discover a new idiom for your task (e.g., calling
unlist() instead of the for loop), you just edit
one function (OurUnlist) instead of editing all
your scripts that used the old idiom).
Once you get used to writing functions (and using
them), you are ready to document them and package
them up for others to use.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Michael Weylandt
PS -- Can you email me (off list) and let me know what this is for? We've
been asked this question a couple of times over the last few days and I'm
just wondering why it's a calculation of interest to so many.
On Thu, Aug 11, 2011 at 1:00 PM, Srinivas Iyyer
srini_iyyer_...@yahoo.comwrote:
Thank you. that was very easy.
-srini
--- On *Thu, 8/11/11, R. Michael Weylandt
michael.weyla...@gmail.com*wrote:
From: R. Michael Weylandt michael.weyla...@gmail.com
Subject: Re: [R] help with loops
To: Srinivas Iyyer srini_iyyer_...@yahoo.com
Cc: r-help@r-project.org
Date: Thursday, August 11, 2011, 12:49 PM
unlist()
Michael Weylandt
On Thu, Aug 11, 2011 at 12:46 PM, Srinivas Iyyer
srini_iyyer_...@yahoo.com http://mc/compose?to=srini_iyyer_...@yahoo.com
wrote:
hi I need help with list object.
I have a list object
a - c('apple','orange','grape')
b - c('car','truck','jeep')
c - list(a,b)
names(c) - c('fruit','vehicle')
c
$fruit
[1] apple orange grape
$vehicle
[1] car truck jeep
I want to write all the elements of this list in one object 'z'.
z
[1] apple orange grape car truck jeep
How can I write the elements of c to z
I tried using a for loop. Could any one help me please. thanks
z - ''
for (i in 1:length(c)){
+ k - c[[i]]
+ z - c(z,k)
+ return(z)}
Error: no function to return from, jumping to top level
Thank you.
srini
__
R-help@r-project.org http://mc/compose?to=R-help@r-project.org mailing
list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on loops
Thanks Jim!
--- On Tue, 12/7/10, jim holtman jholt...@gmail.com wrote:
From: jim holtman jholt...@gmail.com
Subject: Re: [R] Help on loops
To: Anup Nandialath anup_nandial...@yahoo.com
Cc: r-help@r-project.org
Date: Tuesday, December 7, 2010, 7:47 PM
use split and lapply to make it easier. Substitute in your own
calculations since this is just an example:
x
id t X1 X2
1 1 1 4 3
2 1 2 9 2
3 1 3 7 3
4 1 4 6 6
5 2 1 6 4
6 2 2 5 3
7 2 3 1 1
8 3 1 9 6
9 3 2 5 5
# first split the data by 'id'
x.s - split(x, x$id)
# then process each group
result - lapply(x.s, function(.grp){
+ output - .grp[1, 3:4] * 4 # data from the first row
+ i - 2
+ # repeat for the rest of the row
+ while (i = nrow(.grp)){
+ output - output + .grp[i, 3:4] * 2
+ i - i + 1
+ }
+ output # return value
+ })
result
$`1`
X1 X2
1 60 34
$`2`
X1 X2
5 36 24
$`3`
X1 X2
8 46 34
On Tue, Dec 7, 2010 at 5:43 PM, Anup Nandialath
anup_nandial...@yahoo.com wrote:
Dear R-helpers,
I have a basic question on using loops.
I have a panel data set with different variables measured for n firms over
t time periods. A snapshot of the data is given below
id t X1 X2
1 1 4 3
1 2 9 2
1 3 7 3
1 4 6 6
2 1 6 4
2 2 5 3
2 3 1 1
3 1 9 6
3 2 5 5
thus total sample n=9
My problem is as follows. I need to do some computations on the data where
the first observation for each firm (ie. id=1 and t=1; id=2 and t=1; id=3 and
t=1) requires a specific operation on x1 and x2 and the subsequent operations
are based on the computed value of the first operation.
For example the pseudocode is as follows
##define output matrix
output - rep(0,n)
## define coefficient vector
b - c(1,1)
for (i in 1:number of simulations)
{
for (j in 1:id)
{
for(k in 1:t)
{
if(Data$t[,2]==1)
{
meanvec - Data[k,3:4]%*%b
output[k] = calc (meanvec) # output from calc is a scalar
}
else
output[k]= calc(a*output[k-1]+Data[k,3:4]%*%b)
}
}
}
Thus the end result should be a vector output with nine observations based
on the computations.
I hope the problem is clear and I greatly appreciate any help in solving this
problem . Thanks in advance for your help.
Kind Regards
Anup
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] Help on loops
Dear R-helpers,
I have a basic question on using loops.
I have a panel data set with different variables measured for n firms over
t time periods. A snapshot of the data is given below
id t X1 X2
1 1 4 3
1 2 9 2
1 3 7 3
1 4 6 6
2 1 6 4
2 2 5 3
2 3 1 1
3 1 9 6
3 2 5 5
thus total sample n=9
My problem is as follows. I need to do some computations on the data where the
first observation for each firm (ie. id=1 and t=1; id=2 and t=1; id=3 and t=1)
requires a specific operation on x1 and x2 and the subsequent operations are
based on the computed value of the first operation.
For example the pseudocode is as follows
##define output matrix
output - rep(0,n)
## define coefficient vector
b - c(1,1)
for (i in 1:number of simulations)
{
for (j in 1:id)
{
for(k in 1:t)
{
if(Data$t[,2]==1)
{
meanvec - Data[k,3:4]%*%b
output[k] = calc (meanvec) # output from calc is a scalar
}
else
output[k]= calc(a*output[k-1]+Data[k,3:4]%*%b)
}
}
}
Thus the end result should be a vector output with nine observations based on
the computations.
I hope the problem is clear and I greatly appreciate any help in solving this
problem . Thanks in advance for your help.
Kind Regards
Anup
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Re: [R] Help on loops
use split and lapply to make it easier. Substitute in your own
calculations since this is just an example:
x
id t X1 X2
1 1 1 4 3
2 1 2 9 2
3 1 3 7 3
4 1 4 6 6
5 2 1 6 4
6 2 2 5 3
7 2 3 1 1
8 3 1 9 6
9 3 2 5 5
# first split the data by 'id'
x.s - split(x, x$id)
# then process each group
result - lapply(x.s, function(.grp){
+ output - .grp[1, 3:4] * 4 # data from the first row
+ i - 2
+ # repeat for the rest of the row
+ while (i = nrow(.grp)){
+ output - output + .grp[i, 3:4] * 2
+ i - i + 1
+ }
+ output # return value
+ })
result
$`1`
X1 X2
1 60 34
$`2`
X1 X2
5 36 24
$`3`
X1 X2
8 46 34
On Tue, Dec 7, 2010 at 5:43 PM, Anup Nandialath
anup_nandial...@yahoo.com wrote:
Dear R-helpers,
I have a basic question on using loops.
I have a panel data set with different variables measured for n firms over
t time periods. A snapshot of the data is given below
id t X1 X2
1 1 4 3
1 2 9 2
1 3 7 3
1 4 6 6
2 1 6 4
2 2 5 3
2 3 1 1
3 1 9 6
3 2 5 5
thus total sample n=9
My problem is as follows. I need to do some computations on the data where
the first observation for each firm (ie. id=1 and t=1; id=2 and t=1; id=3 and
t=1) requires a specific operation on x1 and x2 and the subsequent operations
are based on the computed value of the first operation.
For example the pseudocode is as follows
##define output matrix
output - rep(0,n)
## define coefficient vector
b - c(1,1)
for (i in 1:number of simulations)
{
for (j in 1:id)
{
for(k in 1:t)
{
if(Data$t[,2]==1)
{
meanvec - Data[k,3:4]%*%b
output[k] = calc (meanvec) # output from calc is a scalar
}
else
output[k]= calc(a*output[k-1]+Data[k,3:4]%*%b)
}
}
}
Thus the end result should be a vector output with nine observations based
on the computations.
I hope the problem is clear and I greatly appreciate any help in solving this
problem . Thanks in advance for your help.
Kind Regards
Anup
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What is the problem that you are trying to solve?
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[R] Help with Loops
Hi
I have tried many attempts but cant get the loop right, as I am not a strong
programmer. What I am basically trying to do is compare 2 spreadsheets. The
problem is that one of them only contain a portion of the overall data
(TESTSAMP), where the other has a full datasetFULLSAMP. From the complete set I
would like to remove the rows of data which are not in the TESTSAMP. Column 1
contains the sample numbers which can be used to identify samples. Does anyone
have any suggestions?
I have tried various things like double loops and so on, but I am sure there is
an easier way or function to do this.
i tried this method, but Im not sure how to only keep looping until a match is
found. I dont understand how repeat loops work in R.
for (i in 1:length(FULLSAMP[,1])) {
if (FULLSAMP[i,1] != TESTSAMP[i,1]) {
FULLSAMP - FULLSAMP[-i,]
}
Thanks in advance
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Re: [R] Help with Loops
You don't need a loop for this, I think. Since you don't provide an
example it's hard to know how your data are set up, but look at this:
FULLSAMP - data.frame(A = 1:10, B=letters[1:10])
TESTSAMP - data.frame(A = c(2,4,5,8), C=1:4)
FULLSAMP
A B
1 1 a
2 2 b
3 3 c
4 4 d
5 5 e
6 6 f
7 7 g
8 8 h
9 9 i
10 10 j
TESTSAMP
A C
1 2 1
2 4 2
3 5 3
4 8 4
FULLSAMP[FULLSAMP$A %in% TESTSAMP$A,]
A B
2 2 b
4 4 d
5 5 e
8 8 h
Sarah
On Thu, May 13, 2010 at 10:49 AM, Amit Patel amitrh...@yahoo.co.uk wrote:
Hi
I have tried many attempts but cant get the loop right, as I am not a strong
programmer. What I am basically trying to do is compare 2 spreadsheets. The
problem is that one of them only contain a portion of the overall data
(TESTSAMP), where the other has a full datasetFULLSAMP. From the complete set
I would like to remove the rows of data which are not in the TESTSAMP. Column
1 contains the sample numbers which can be used to identify samples. Does
anyone have any suggestions?
I have tried various things like double loops and so on, but I am sure there
is an easier way or function to do this.
i tried this method, but Im not sure how to only keep looping until a match
is found. I dont understand how repeat loops work in R.
for (i in 1:length(FULLSAMP[,1])) {
if (FULLSAMP[i,1] != TESTSAMP[i,1]) {
FULLSAMP - FULLSAMP[-i,]
}
Thanks in advance
--
Sarah Goslee
http://www.functionaldiversity.org
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Re: [R] Help with Loops
Hi,
On Thu, May 13, 2010 at 10:49 AM, Amit Patel amitrh...@yahoo.co.uk wrote:
Hi
I have tried many attempts but cant get the loop right, as I am not a strong
programmer. What I am basically trying to do is compare 2 spreadsheets. The
problem is that one of them only contain a portion of the overall data
(TESTSAMP), where the other has a full datasetFULLSAMP. From the complete set
I would like to remove the rows of data which are not in the TESTSAMP. Column
1 contains the sample numbers which can be used to identify samples. Does
anyone have any suggestions?
I have tried various things like double loops and so on, but I am sure there
is an easier way or function to do this.
i tried this method, but Im not sure how to only keep looping until a match
is found. I dont understand how repeat loops work in R.
for (i in 1:length(FULLSAMP[,1])) {
if (FULLSAMP[i,1] != TESTSAMP[i,1]) {
FULLSAMP - FULLSAMP[-i,]
}
You want to not use for loops as much as possible.
Imagine your samples are identified as letters, so FULLSAMP[,1] will
be letters A..Z, and TESTSAMP[,1] will be some random 15 letters. Now
the job is to match the rows in TESTAMP to the rows in FULLSAMP, and
remove any extra rows in FULLSAMP that don' appear in testamp.
## Making some data
R fullsamp - data.frame(id=LETTERS, something=sample(1:100,
length(letters)), stringsAsFactors=FALSE)
R testsamp - data.frame(id=sample(LETTERS, 15),
something=sample(1:100, 15), stringsAsFactors=FALSE)
## Let's find where the testamp rows appear in fullsamp
R xref - match(testsamp[,1], fullsamp[,1])
## Now reduce fullsamp to have only the data corresponding to testsamp
## (and in the same order
R fullsamp.sub - fullsamp[xref,]
Notice that fullsamp.sub now has only rows with IDs appearing in
testsamp and they are also in the same order as testsamp.
Now go ahead and read the help you'll find in ?match
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
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Re: [R] Help with Loops
On May 13, 2010, at 10:49 AM, Amit Patel wrote:
Hi
I have tried many attempts but cant get the loop right, as I am not
a strong programmer. What I am basically trying to do is compare 2
spreadsheets. The problem is that one of them only contain a portion
of the overall data (TESTSAMP), where the other has a full
datasetFULLSAMP. From the complete set I would like to remove the
rows of data which are not in the TESTSAMP. Column 1 contains the
sample numbers which can be used to identify samples. Does anyone
have any suggestions?
I have tried various things like double loops and so on, but I am
sure there is an easier way or function to do this.
i tried this method, but Im not sure how to only keep looping until
a match is found. I dont understand how repeat loops work in R.
for (i in 1:length(FULLSAMP[,1])) {
if (FULLSAMP[i,1] != TESTSAMP[i,1]) {
FULLSAMP - FULLSAMP[-i,]
}
Abandon the loop. Use merge.
... or the %in% function.
--
David Winsemius, MD
West Hartford, CT
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[R] help with loops
Hi guys, my doubt is quite simple,
I´ll try to explain:
test = matrix(0, nrow = 783, ncol = 12)
for (x in 1:9){
for (y in 1:12){
### In the original script for each y its generated a vector (87x1)
### 87 times 9(x) = 783 (equals the number of rows I want to fill in the
test matrix
}
}
What I would like to do is to fill the test matrix with the ys in the
collumns and the 87 elements of each x in the rows.
for example, x = 1
y = 1
test[1:87, 1] = vector generated when y is 1
x = 2
y = 1
test [88:174,1] = another vector generated when y is 2
Could anybody help me? The script is quite long so i couldn´t just paste it
here I think what i wrote above is undestandable.
Thank you very much
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[R] Help with loops
Hi
I am trying to create a loop which averages replicates in my data.
The original data has many rows. and consists of 40 column zz[,2:41] plus row
headings in zz[,1]
I am trying to average each set of values (i.e. zz[1,2:3] averaged and placed
in average_value[1,2] and so on.
below is my script but it seems to be stuck in an endless loop
Any suggestions??
for (i in 1:length(average_value[,1])) {
average_value[i] - i^100; print(average_value[i])
#calculates Meanss
#Sample A
average_value[i,2] - rowMeans(zz[i,2:3])
average_value[i,3] - rowMeans(zz[i,4:5])
average_value[i,4] - rowMeans(zz[i,6:7])
average_value[i,5] - rowMeans(zz[i,8:9])
average_value[i,6] - rowMeans(zz[i,10:11])
#Sample B
average_value[i,7] - rowMeans(zz[i,12:13])
average_value[i,8] - rowMeans(zz[i,14:15])
average_value[i,9] - rowMeans(zz[i,16:17])
average_value[i,10] - rowMeans(zz[i,18:19])
average_value[i,11] - rowMeans(zz[i,20:21])
#Sample C
average_value[i,12] - rowMeans(zz[i,22:23])
average_value[i,13] - rowMeans(zz[i,24:25])
average_value[i,14] - rowMeans(zz[i,26:27])
average_value[i,15] - rowMeans(zz[i,28:29])
average_value[i,16] - rowMeans(zz[i,30:31])
#Sample D
average_value[i,17] - rowMeans(zz[i,32:33])
average_value[i,18] - rowMeans(zz[i,34:35])
average_value[i,19] - rowMeans(zz[i,36:37])
average_value[i,20] - rowMeans(zz[i,38:39])
average_value[i,21] - rowMeans(zz[i,40:41])
}
thanks
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Re: [R] Help with loops
I'm not quite sure what you want to do, but this might help:
d=data.frame(replicate(40, rnorm(20)))
d$sample=rep(c('a','b','c','d'),each=5)
lib(doBy)
summaryBy(.~sample,da=d)
David Freedman
Amit Patel-7 wrote:
Hi
I am trying to create a loop which averages replicates in my data.
The original data has many rows. and consists of 40 column zz[,2:41] plus
row headings in zz[,1]
I am trying to average each set of values (i.e. zz[1,2:3] averaged and
placed in average_value[1,2] and so on.
below is my script but it seems to be stuck in an endless loop
Any suggestions??
for (i in 1:length(average_value[,1])) {
average_value[i] - i^100; print(average_value[i])
#calculates Meanss
#Sample A
average_value[i,2] - rowMeans(zz[i,2:3])
average_value[i,3] - rowMeans(zz[i,4:5])
average_value[i,4] - rowMeans(zz[i,6:7])
average_value[i,5] - rowMeans(zz[i,8:9])
average_value[i,6] - rowMeans(zz[i,10:11])
#Sample B
average_value[i,7] - rowMeans(zz[i,12:13])
average_value[i,8] - rowMeans(zz[i,14:15])
average_value[i,9] - rowMeans(zz[i,16:17])
average_value[i,10] - rowMeans(zz[i,18:19])
average_value[i,11] - rowMeans(zz[i,20:21])
#Sample C
average_value[i,12] - rowMeans(zz[i,22:23])
average_value[i,13] - rowMeans(zz[i,24:25])
average_value[i,14] - rowMeans(zz[i,26:27])
average_value[i,15] - rowMeans(zz[i,28:29])
average_value[i,16] - rowMeans(zz[i,30:31])
#Sample D
average_value[i,17] - rowMeans(zz[i,32:33])
average_value[i,18] - rowMeans(zz[i,34:35])
average_value[i,19] - rowMeans(zz[i,36:37])
average_value[i,20] - rowMeans(zz[i,38:39])
average_value[i,21] - rowMeans(zz[i,40:41])
}
thanks
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Re: [R] Help with loops
Dear Amit,
The following should get you started. What I'm doing is creating an
identifiers (g) with the names of the columns you want to group for and
then use a combination of apply() and tapply() to get the mean for each row
in the levels of g. In your case, you have more columns than I have in my
example, but with slightly modifications you can adapt the code below to
your needs.
See ?apply and ?rep for more information.
HTH,
Jorge
# Some data
set.seed(123)
X - matrix(rnorm(100), ncol=10)
colnames(X) - paste('x',1:10,sep=)
rownames(X) - paste('sample_',1:10,sep=)
# Defining the groups using seq()
g - rep(1:(ncol(X)/2), each = 2 )
# Calculating the means
res - t( apply(X, 1, tapply, g, mean) )
res
# res[1,1] is the mean for X[1, 1:2]
mean(X[1,1:2])
# [1] 0.2408457
On Fri, May 15, 2009 at 8:17 AM, Amit Patel amitrh...@yahoo.co.uk wrote:
Hi
I am trying to create a loop which averages replicates in my data.
The original data has many rows. and consists of 40 column zz[,2:41] plus
row headings in zz[,1]
I am trying to average each set of values (i.e. zz[1,2:3] averaged and
placed in average_value[1,2] and so on.
below is my script but it seems to be stuck in an endless loop
Any suggestions??
for (i in 1:length(average_value[,1])) {
average_value[i] - i^100; print(average_value[i])
#calculates Meanss
#Sample A
average_value[i,2] - rowMeans(zz[i,2:3])
average_value[i,3] - rowMeans(zz[i,4:5])
average_value[i,4] - rowMeans(zz[i,6:7])
average_value[i,5] - rowMeans(zz[i,8:9])
average_value[i,6] - rowMeans(zz[i,10:11])
#Sample B
average_value[i,7] - rowMeans(zz[i,12:13])
average_value[i,8] - rowMeans(zz[i,14:15])
average_value[i,9] - rowMeans(zz[i,16:17])
average_value[i,10] - rowMeans(zz[i,18:19])
average_value[i,11] - rowMeans(zz[i,20:21])
#Sample C
average_value[i,12] - rowMeans(zz[i,22:23])
average_value[i,13] - rowMeans(zz[i,24:25])
average_value[i,14] - rowMeans(zz[i,26:27])
average_value[i,15] - rowMeans(zz[i,28:29])
average_value[i,16] - rowMeans(zz[i,30:31])
#Sample D
average_value[i,17] - rowMeans(zz[i,32:33])
average_value[i,18] - rowMeans(zz[i,34:35])
average_value[i,19] - rowMeans(zz[i,36:37])
average_value[i,20] - rowMeans(zz[i,38:39])
average_value[i,21] - rowMeans(zz[i,40:41])
}
thanks
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Re: [R] Help with loops and how R stores data
On 02-Feb-08 21:16:41, R-novice wrote:
I am trying to make an array c(3,8) that contains the averages of what
is in another array c(9,8). I want to average rows 1:3, 4:6, 7:9 and
have a loop replace the generic 1:24 values in my array with the
average of the three rows.
The problem I am having is that R only replaces the last value from the
loops, and I am not experienced enough with R to know how it stores
data when looping and how to overcome this problem. By having it print
the avg, I was expecting to see 24 numbers, but it only gives me one.
Here is my code.
cts.sds-array(1:24,c(3,8))
for(i in 3){
for(j in 8){
avg-sum(exprdata[3*i-2:3*i,j]/3)
cts.sds[i,j]-avg
print(avg)
}
}
print(cts.sds)
Any help with this pesky matter will be greatly appreciated.
I think you have made two types of mistake here.
1. You wrote for(i in 3) and for(i in 3). This means that
'i' will take only one value, namely 3; and 'j' will take only
one value, namely 8. If you wanted to loop over i = 1,2,3 and
j=1,2,3,4,5,6,7,8 then you should write
for(i in (1:3)){ for(j in (1:8){ ... }}
2. This is more subtle, and you're far from the only person
to have tripped over this one.
Because of the order of precedence of the operators, in the
epression
3*i-2:3*i
for each value of 'i' it will first evaluate 2:3, namely {2,3},
and then evaluate the remainder. So, for say i=2, you will get
two numbers 3*i-2*i = 2 and 3*i-3*i = 0, whereas what you
presumably intended was (3*i-2):(3*i) = 4:6 = {4,5,6}.
The way to avoid this trap is to use parantheses to force the
order of evaluation you want:
avg-sum(exprdata[(3*i-2):(3*i,j)]/3)
Example:
i-2
3*i-2:3*i
# [1] 2 0
(3*i-2):(3*i)
# [1] 4 5 6
The reason is that the : operator takes precedence over the
binary operators *, + and -.
Enter
?Syntax
to get a listing of the various operators in order of
precedence.
Hoping this helps,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 04-Feb-08 Time: 08:18:05
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[R] Help with loops and how R stores data
I am trying to make an array c(3,8) that contains the averages of what is in
another array c(9,8). I want to average rows 1:3, 4:6, 7:9 and have a loop
replace the generic 1:24 values in my array with the average of the three
rows.
The problem I am having is that R only replaces the last value from the
loops, and I am not experienced enough with R to know how it stores data
when looping and how to overcome this problem. By having it print the avg, I
was expecting to see 24 numbers, but it only gives me one.
Here is my code.
cts.sds-array(1:24,c(3,8))
for(i in 3){
for(j in 8){
avg-sum(exprdata[3*i-2:3*i,j]/3)
cts.sds[i,j]-avg
print(avg)
}
}
print(cts.sds)
Any help with this pesky matter will be greatly appreciated.
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