Re: [R] filling the matrix row by row in the order from lower to larger elements
Sorry, I didn't have time to check the speed, indeed.
However - isn't apply the same as a loop, just hidden?
D.
On Fri, Apr 6, 2012 at 6:59 PM, ilai ke...@math.montana.edu wrote:
On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
This works great:
Really ? surprising given it is the EXACT same for-loop as in your
original problem with counter i replaced by k and reorder to
matrix[!100]- 0 instead of matrix(0)[i]- 100
You didn't even attempt to implement Carl's suggestion to use apply
family for looping (which I still think is completely unnecessary).
The only logical conclusion is N=nrow(input) was not large enough to
pose a problem in the first place. In the future please use some brain
power before waisting ours.
Cheers
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
result-input
N-nrow(input)
for (k in 1:N){
foo - which (input == k,arr.ind=T)
result[k,foo[2]] -100
}
result[result !=100]-0
Dimitri
On Fri, Apr 6, 2012 at 5:14 PM, Carl Witthoft c...@witthoft.com wrote:
I think the OP wants to fill values in an arbitrarily large matrix. Now,
first of all, I'd like to know what his real problem is, since this seems
like a very tedious and unproductive matrix to produce. But in the
meantime, since he also left out important information, let's assume the
input matrix is N rows by M columns, and that he wants therefore to end up
with N instances of 100, not counting the original value of 100 that is
one of his ranking values (a bad BAD move IMHO).
Then either loop or lapply over an equation like (I've expanded things more
than necessary for clarity
result-inmatrix
for (k in 1:N){
foo - which (inmatrix == k,arr.ind=T)
result[k,foo[2]] -100
}
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all. Something like this maybe?
(input-matrix(c(5,1,3,7,2,6,4,8),nc=2))
output - matrix(0,max(input),2)
output[input[,1],1] - 100
output[input[,2],2] - 100
output
Cheers
On Fri, Apr 6, 2012 at 1:49 PM, Dimitri Liakhovitski
dimitri.liakhovitski at gmail.com wrote:
Hello, everybody!
I have a matrix input (see example below) - with all unique entries
that are actually unique ranks (i.e., start with 1, no ties).
I want to assign a value of 100 to the first row of the column that
contains the minimum (i.e., value of 1).
Then, I want to assign a value of 100 to the second row of the column
that contains the value of 2, etc.
The results I am looking for are in desired.results.
My code (below) does what I need. But it's using a loop through all
the rows of my matrix and searches for a matrix element every time.
My actual matrix is very large. Is there a way to do it more efficiently?
Thank you very much for the tips!
Dimitri
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
(input)
desired.result-as.matrix(data.frame(a=c(100,0,100,0),b=c(0,100,0,100)))
(desired.result)
result-as.matrix(data.frame(a=c(0,0,0,0),b=c(0,0,0,0)))
for(i in 1:nrow(input)){ # i-1
mymin-i
mycoords-which(input==mymin,arr.ind=TRUE)
result[i,mycoords[2]]-100
input[mycoords]-max(input)
}
(result)
--
Sent from my Cray XK6
Quidvis recte factum, quamvis humile, praeclarum.
__
R-help@r-project.org mailing list
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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.
--
Dimitri Liakhovitski
marketfusionanalytics.com
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Dimitri Liakhovitski
marketfusionanalytics.com
__
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Re: [R] filling the matrix row by row in the order from lower to larger elements
Thank you very much, Rui.
This definitely produces the result needed.
Again, I have not checked the speed yet.
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
f - function(x){
nr - nrow(x)
result - matrix(0, nrow=nr, ncol=ncol(x))
colnames(result) - colnames(x)
inp.ord - order(x)[1:nr] - 1 # Keep only one per row, must be zero based
inx - cbind(1:nr, inp.ord %/% nr + 1) # Index matrix
result[inx] - 100
result
}
f(input)
Dimitri
On Fri, Apr 6, 2012 at 7:02 PM, Rui Barradas rui1...@sapo.pt wrote:
Hello,
Oops!
What happened to the function 'f'?
Forgot to copy and pasted only the rest, now complete.
f - function(x){
nr - nrow(x)
result - matrix(0, nrow=nr, ncol=ncol(x))
colnames(result) - colnames(x)
inp.ord - order(x)[1:nr] - 1 # Keep only one per row, must be zero
based
inx - cbind(1:nr, inp.ord %/% nr + 1) # Index matrix
result[inx] - 100
result
}
# Original example
input - as.matrix(data.frame(a=c(5,1,3,7), b=c(2,6,4,8)))
(input)
desired.result - as.matrix(data.frame(a=c(100,0,100,0), b=c(0,100,0,100)))
(desired.result)
all.equal(f(input), desired.result)
# Two other examples
set.seed(123)
(x - matrix(sample(10, 10), ncol=2))
f(x)
(y - matrix(sample(40, 40), ncol=5))
f(y)
Note that there's no loops (or apply, which is also a loop.)
Rui Barradas
--
View this message in context:
http://r.789695.n4.nabble.com/filling-the-matrix-row-by-row-in-the-order-from-lower-to-larger-elements-tp4538171p4538486.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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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.
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] filling the matrix row by row in the order from lower to larger elements
Preoccupation with speed of execution is typically (certainly not
always) misplaced. Provided you have used sensible basic
vectorization, loops in whatever form work adequately. First get
working code. Then, if necessary, parallelization, byte compilation,
or complex vectorization strategies can be employed. And, of course,
there's always C code for those who know it.
-- Bert
On Sun, Apr 8, 2012 at 7:31 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Thank you very much, Rui.
This definitely produces the result needed.
Again, I have not checked the speed yet.
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
f - function(x){
nr - nrow(x)
result - matrix(0, nrow=nr, ncol=ncol(x))
colnames(result) - colnames(x)
inp.ord - order(x)[1:nr] - 1 # Keep only one per row, must be zero
based
inx - cbind(1:nr, inp.ord %/% nr + 1) # Index matrix
result[inx] - 100
result
}
f(input)
Dimitri
On Fri, Apr 6, 2012 at 7:02 PM, Rui Barradas rui1...@sapo.pt wrote:
Hello,
Oops!
What happened to the function 'f'?
Forgot to copy and pasted only the rest, now complete.
f - function(x){
nr - nrow(x)
result - matrix(0, nrow=nr, ncol=ncol(x))
colnames(result) - colnames(x)
inp.ord - order(x)[1:nr] - 1 # Keep only one per row, must be zero
based
inx - cbind(1:nr, inp.ord %/% nr + 1) # Index matrix
result[inx] - 100
result
}
# Original example
input - as.matrix(data.frame(a=c(5,1,3,7), b=c(2,6,4,8)))
(input)
desired.result - as.matrix(data.frame(a=c(100,0,100,0), b=c(0,100,0,100)))
(desired.result)
all.equal(f(input), desired.result)
# Two other examples
set.seed(123)
(x - matrix(sample(10, 10), ncol=2))
f(x)
(y - matrix(sample(40, 40), ncol=5))
f(y)
Note that there's no loops (or apply, which is also a loop.)
Rui Barradas
--
View this message in context:
http://r.789695.n4.nabble.com/filling-the-matrix-row-by-row-in-the-order-from-lower-to-larger-elements-tp4538171p4538486.html
Sent from the R help mailing list archive at Nabble.com.
__
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.
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
R-help@r-project.org mailing list
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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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
__
R-help@r-project.org mailing list
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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] filling the matrix row by row in the order from lower to larger elements
On Sun, Apr 8, 2012 at 8:26 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Sorry, I didn't have time to check the speed, indeed.
However - isn't apply the same as a loop, just hidden?
D.
Yes ?apply is a loop but not the same as ?for, see Intro to R. As
Bert Gunter pointed out the issue here was not speed but that your
problem could have been vectorized to avoid loops all together (which
would have the added benefit of speed). This was given to you in my
first response which assumed a small number of columns in the matrix,
but Rui gave an elegant expansion to use on any ncol(matrix). Using
apply was suggested at some point by others, my comment was simply
that you failed to meet even that adjustment.
Cheers
On Fri, Apr 6, 2012 at 6:59 PM, ilai ke...@math.montana.edu wrote:
On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
This works great:
Really ? surprising given it is the EXACT same for-loop as in your
original problem with counter i replaced by k and reorder to
matrix[!100]- 0 instead of matrix(0)[i]- 100
You didn't even attempt to implement Carl's suggestion to use apply
family for looping (which I still think is completely unnecessary).
The only logical conclusion is N=nrow(input) was not large enough to
pose a problem in the first place. In the future please use some brain
power before waisting ours.
Cheers
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
result-input
N-nrow(input)
for (k in 1:N){
foo - which (input == k,arr.ind=T)
result[k,foo[2]] -100
}
result[result !=100]-0
Dimitri
On Fri, Apr 6, 2012 at 5:14 PM, Carl Witthoft c...@witthoft.com wrote:
I think the OP wants to fill values in an arbitrarily large matrix. Now,
first of all, I'd like to know what his real problem is, since this seems
like a very tedious and unproductive matrix to produce. But in the
meantime, since he also left out important information, let's assume the
input matrix is N rows by M columns, and that he wants therefore to end up
with N instances of 100, not counting the original value of 100 that is
one of his ranking values (a bad BAD move IMHO).
Then either loop or lapply over an equation like (I've expanded things more
than necessary for clarity
result-inmatrix
for (k in 1:N){
foo - which (inmatrix == k,arr.ind=T)
result[k,foo[2]] -100
}
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all. Something like this maybe?
(input-matrix(c(5,1,3,7,2,6,4,8),nc=2))
output - matrix(0,max(input),2)
output[input[,1],1] - 100
output[input[,2],2] - 100
output
Cheers
On Fri, Apr 6, 2012 at 1:49 PM, Dimitri Liakhovitski
dimitri.liakhovitski at gmail.com wrote:
Hello, everybody!
I have a matrix input (see example below) - with all unique entries
that are actually unique ranks (i.e., start with 1, no ties).
I want to assign a value of 100 to the first row of the column that
contains the minimum (i.e., value of 1).
Then, I want to assign a value of 100 to the second row of the column
that contains the value of 2, etc.
The results I am looking for are in desired.results.
My code (below) does what I need. But it's using a loop through all
the rows of my matrix and searches for a matrix element every time.
My actual matrix is very large. Is there a way to do it more efficiently?
Thank you very much for the tips!
Dimitri
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
(input)
desired.result-as.matrix(data.frame(a=c(100,0,100,0),b=c(0,100,0,100)))
(desired.result)
result-as.matrix(data.frame(a=c(0,0,0,0),b=c(0,0,0,0)))
for(i in 1:nrow(input)){ # i-1
mymin-i
mycoords-which(input==mymin,arr.ind=TRUE)
result[i,mycoords[2]]-100
input[mycoords]-max(input)
}
(result)
--
Sent from my Cray XK6
Quidvis recte factum, quamvis humile, praeclarum.
__
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.
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
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.
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
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] filling the matrix row by row in the order from lower to larger elements
Agreed, Rui provided a very elegant vectorized solution - and I am
very thankful to him. Unfortunately (for myself), I am not as
proficient in vectorization - otherwise, I would not have asked the
question.
Why I did not follow on the original hint to use apply? For this reason (quote):
However - isn't apply the same as a loop, just hidden?
with slight caveats, yes.
Bert
Dimitri
On Sun, Apr 8, 2012 at 3:16 PM, ilai ke...@math.montana.edu wrote:
On Sun, Apr 8, 2012 at 8:26 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Sorry, I didn't have time to check the speed, indeed.
However - isn't apply the same as a loop, just hidden?
D.
Yes ?apply is a loop but not the same as ?for, see Intro to R. As
Bert Gunter pointed out the issue here was not speed but that your
problem could have been vectorized to avoid loops all together (which
would have the added benefit of speed). This was given to you in my
first response which assumed a small number of columns in the matrix,
but Rui gave an elegant expansion to use on any ncol(matrix). Using
apply was suggested at some point by others, my comment was simply
that you failed to meet even that adjustment.
Cheers
On Fri, Apr 6, 2012 at 6:59 PM, ilai ke...@math.montana.edu wrote:
On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
This works great:
Really ? surprising given it is the EXACT same for-loop as in your
original problem with counter i replaced by k and reorder to
matrix[!100]- 0 instead of matrix(0)[i]- 100
You didn't even attempt to implement Carl's suggestion to use apply
family for looping (which I still think is completely unnecessary).
The only logical conclusion is N=nrow(input) was not large enough to
pose a problem in the first place. In the future please use some brain
power before waisting ours.
Cheers
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
result-input
N-nrow(input)
for (k in 1:N){
foo - which (input == k,arr.ind=T)
result[k,foo[2]] -100
}
result[result !=100]-0
Dimitri
On Fri, Apr 6, 2012 at 5:14 PM, Carl Witthoft c...@witthoft.com wrote:
I think the OP wants to fill values in an arbitrarily large matrix. Now,
first of all, I'd like to know what his real problem is, since this seems
like a very tedious and unproductive matrix to produce. But in the
meantime, since he also left out important information, let's assume the
input matrix is N rows by M columns, and that he wants therefore to end up
with N instances of 100, not counting the original value of 100 that is
one of his ranking values (a bad BAD move IMHO).
Then either loop or lapply over an equation like (I've expanded things
more
than necessary for clarity
result-inmatrix
for (k in 1:N){
foo - which (inmatrix == k,arr.ind=T)
result[k,foo[2]] -100
}
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all. Something like this maybe?
(input-matrix(c(5,1,3,7,2,6,4,8),nc=2))
output - matrix(0,max(input),2)
output[input[,1],1] - 100
output[input[,2],2] - 100
output
Cheers
On Fri, Apr 6, 2012 at 1:49 PM, Dimitri Liakhovitski
dimitri.liakhovitski at gmail.com wrote:
Hello, everybody!
I have a matrix input (see example below) - with all unique entries
that are actually unique ranks (i.e., start with 1, no ties).
I want to assign a value of 100 to the first row of the column that
contains the minimum (i.e., value of 1).
Then, I want to assign a value of 100 to the second row of the column
that contains the value of 2, etc.
The results I am looking for are in desired.results.
My code (below) does what I need. But it's using a loop through all
the rows of my matrix and searches for a matrix element every time.
My actual matrix is very large. Is there a way to do it more efficiently?
Thank you very much for the tips!
Dimitri
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
(input)
desired.result-as.matrix(data.frame(a=c(100,0,100,0),b=c(0,100,0,100)))
(desired.result)
result-as.matrix(data.frame(a=c(0,0,0,0),b=c(0,0,0,0)))
for(i in 1:nrow(input)){ # i-1
mymin-i
mycoords-which(input==mymin,arr.ind=TRUE)
result[i,mycoords[2]]-100
input[mycoords]-max(input)
}
(result)
--
Sent from my Cray XK6
Quidvis recte factum, quamvis humile, praeclarum.
__
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.
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
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.
--
Dimitri
Re: [R] filling the matrix row by row in the order from lower to larger elements
Hello,
Oops!
What happened to the function 'f'?
Forgot to copy and pasted only the rest, now complete.
f - function(x){
nr - nrow(x)
result - matrix(0, nrow=nr, ncol=ncol(x))
colnames(result) - colnames(x)
inp.ord - order(x)[1:nr] - 1 # Keep only one per row, must be zero
based
inx - cbind(1:nr, inp.ord %/% nr + 1) # Index matrix
result[inx] - 100
result
}
# Original example
input - as.matrix(data.frame(a=c(5,1,3,7), b=c(2,6,4,8)))
(input)
desired.result - as.matrix(data.frame(a=c(100,0,100,0), b=c(0,100,0,100)))
(desired.result)
all.equal(f(input), desired.result)
# Two other examples
set.seed(123)
(x - matrix(sample(10, 10), ncol=2))
f(x)
(y - matrix(sample(40, 40), ncol=5))
f(y)
Note that there's no loops (or apply, which is also a loop.)
Rui Barradas
--
View this message in context:
http://r.789695.n4.nabble.com/filling-the-matrix-row-by-row-in-the-order-from-lower-to-larger-elements-tp4538171p4538486.html
Sent from the R help mailing list archive at Nabble.com.
__
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] filling the matrix row by row in the order from lower to larger elements
Hello, I maybe missing something but this seems like an indexing problem which doesn't require a loop at all. Yes, but with 'order'. # Original example input - as.matrix(data.frame(a=c(5,1,3,7), b=c(2,6,4,8))) (input) desired.result - as.matrix(data.frame(a=c(100,0,100,0), b=c(0,100,0,100))) (desired.result) all.equal(f(input), desired.result) # Two other examples set.seed(123) (x - matrix(sample(10, 10), ncol=2)) f(x) (y - matrix(sample(40, 40), ncol=5)) f(y) Hope this helps, Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/filling-the-matrix-row-by-row-in-the-order-from-lower-to-larger-elements-tp4538171p4538334.html Sent from the R help mailing list archive at Nabble.com. __ 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] filling the matrix row by row in the order from lower to larger elements
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all. Something like this maybe?
(input-matrix(c(5,1,3,7,2,6,4,8),nc=2))
output - matrix(0,max(input),2)
output[input[,1],1] - 100
output[input[,2],2] - 100
output
Cheers
On Fri, Apr 6, 2012 at 1:49 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Hello, everybody!
I have a matrix input (see example below) - with all unique entries
that are actually unique ranks (i.e., start with 1, no ties).
I want to assign a value of 100 to the first row of the column that
contains the minimum (i.e., value of 1).
Then, I want to assign a value of 100 to the second row of the column
that contains the value of 2, etc.
The results I am looking for are in desired.results.
My code (below) does what I need. But it's using a loop through all
the rows of my matrix and searches for a matrix element every time.
My actual matrix is very large. Is there a way to do it more efficiently?
Thank you very much for the tips!
Dimitri
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
(input)
desired.result-as.matrix(data.frame(a=c(100,0,100,0),b=c(0,100,0,100)))
(desired.result)
result-as.matrix(data.frame(a=c(0,0,0,0),b=c(0,0,0,0)))
for(i in 1:nrow(input)){ # i-1
mymin-i
mycoords-which(input==mymin,arr.ind=TRUE)
result[i,mycoords[2]]-100
input[mycoords]-max(input)
}
(result)
--
Dimitri Liakhovitski
__
R-help@r-project.org mailing list
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] filling the matrix row by row in the order from lower to larger elements
I think the OP wants to fill values in an arbitrarily large matrix.
Now, first of all, I'd like to know what his real problem is, since this
seems like a very tedious and unproductive matrix to produce. But in
the meantime, since he also left out important information, let's
assume the input matrix is N rows by M columns, and that he wants
therefore to end up with N instances of 100, not counting the original
value of 100 that is one of his ranking values (a bad BAD move IMHO).
Then either loop or lapply over an equation like (I've expanded things
more than necessary for clarity
result-inmatrix
for (k in 1:N){
foo - which (inmatrix == k,arr.ind=T)
result[k,foo[2]] -100
}
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all. Something like this maybe?
(input-matrix(c(5,1,3,7,2,6,4,8),nc=2))
output - matrix(0,max(input),2)
output[input[,1],1] - 100
output[input[,2],2] - 100
output
Cheers
On Fri, Apr 6, 2012 at 1:49 PM, Dimitri Liakhovitski
dimitri.liakhovitski at gmail.com wrote:
Hello, everybody!
I have a matrix input (see example below) - with all unique entries
that are actually unique ranks (i.e., start with 1, no ties).
I want to assign a value of 100 to the first row of the column that
contains the minimum (i.e., value of 1).
Then, I want to assign a value of 100 to the second row of the column
that contains the value of 2, etc.
The results I am looking for are in desired.results.
My code (below) does what I need. But it's using a loop through all
the rows of my matrix and searches for a matrix element every time.
My actual matrix is very large. Is there a way to do it more efficiently?
Thank you very much for the tips!
Dimitri
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
(input)
desired.result-as.matrix(data.frame(a=c(100,0,100,0),b=c(0,100,0,100)))
(desired.result)
result-as.matrix(data.frame(a=c(0,0,0,0),b=c(0,0,0,0)))
for(i in 1:nrow(input)){ # i-1
mymin-i
mycoords-which(input==mymin,arr.ind=TRUE)
result[i,mycoords[2]]-100
input[mycoords]-max(input)
}
(result)
--
Sent from my Cray XK6
Quidvis recte factum, quamvis humile, praeclarum.
__
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] filling the matrix row by row in the order from lower to larger elements
Ok, how's this:
Rgames foo
[,1] [,2] [,3] [,4]
[1,]361 16
[2,] 10 14 125
[3,] 117 159
[4,]84 132
Rgames sapply(1:4,FUN=function(k){
foo[k,which(foo==k,arr.ind=T)[2]]-100;return(foo)})-bar
Rgames bar
[,1] [,2] [,3] [,4]
[1,]3333
[2,] 10 10 10 10
[3,] 11 11 100 11
[4,]8888
[5,]6666
[6,] 14 14 14 14
[7,]7777
[8,]444 100
[9,] 100111
[10,] 12 12 12 12
[11,] 15 15 15 15
[12,] 13 13 13 13
[13,] 16 16 16 16
[14,]5 10055
[15,]9999
[16,]2222
Rgames rab-matrix(apply(bar,1,max),4,4)
Rgames rab
[,1] [,2] [,3] [,4]
[1,]36 100 16
[2,] 10 14 12 100
[3,] 1007 159
[4,]8 100 132
--
Sent from my Cray XK6
Quidvis recte factum, quamvis humile, praeclarum.
__
R-help@r-project.org mailing list
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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] filling the matrix row by row in the order from lower to larger elements
Yes, that's correct - my matrix has N rows.
Thank you very much, Carl. This works great:
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
result-input
N-nrow(input)
for (k in 1:N){
foo - which (input == k,arr.ind=T)
result[k,foo[2]] -100
}
result[result !=100]-0
Dimitri
On Fri, Apr 6, 2012 at 5:14 PM, Carl Witthoft c...@witthoft.com wrote:
I think the OP wants to fill values in an arbitrarily large matrix. Now,
first of all, I'd like to know what his real problem is, since this seems
like a very tedious and unproductive matrix to produce. But in the
meantime, since he also left out important information, let's assume the
input matrix is N rows by M columns, and that he wants therefore to end up
with N instances of 100, not counting the original value of 100 that is
one of his ranking values (a bad BAD move IMHO).
Then either loop or lapply over an equation like (I've expanded things more
than necessary for clarity
result-inmatrix
for (k in 1:N){
foo - which (inmatrix == k,arr.ind=T)
result[k,foo[2]] -100
}
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all. Something like this maybe?
(input-matrix(c(5,1,3,7,2,6,4,8),nc=2))
output - matrix(0,max(input),2)
output[input[,1],1] - 100
output[input[,2],2] - 100
output
Cheers
On Fri, Apr 6, 2012 at 1:49 PM, Dimitri Liakhovitski
dimitri.liakhovitski at gmail.com wrote:
Hello, everybody!
I have a matrix input (see example below) - with all unique entries
that are actually unique ranks (i.e., start with 1, no ties).
I want to assign a value of 100 to the first row of the column that
contains the minimum (i.e., value of 1).
Then, I want to assign a value of 100 to the second row of the column
that contains the value of 2, etc.
The results I am looking for are in desired.results.
My code (below) does what I need. But it's using a loop through all
the rows of my matrix and searches for a matrix element every time.
My actual matrix is very large. Is there a way to do it more efficiently?
Thank you very much for the tips!
Dimitri
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
(input)
desired.result-as.matrix(data.frame(a=c(100,0,100,0),b=c(0,100,0,100)))
(desired.result)
result-as.matrix(data.frame(a=c(0,0,0,0),b=c(0,0,0,0)))
for(i in 1:nrow(input)){ # i-1
mymin-i
mycoords-which(input==mymin,arr.ind=TRUE)
result[i,mycoords[2]]-100
input[mycoords]-max(input)
}
(result)
--
Sent from my Cray XK6
Quidvis recte factum, quamvis humile, praeclarum.
__
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.
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
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] filling the matrix row by row in the order from lower to larger elements
On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
This works great:
Really ? surprising given it is the EXACT same for-loop as in your
original problem with counter i replaced by k and reorder to
matrix[!100]- 0 instead of matrix(0)[i]- 100
You didn't even attempt to implement Carl's suggestion to use apply
family for looping (which I still think is completely unnecessary).
The only logical conclusion is N=nrow(input) was not large enough to
pose a problem in the first place. In the future please use some brain
power before waisting ours.
Cheers
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
result-input
N-nrow(input)
for (k in 1:N){
foo - which (input == k,arr.ind=T)
result[k,foo[2]] -100
}
result[result !=100]-0
Dimitri
On Fri, Apr 6, 2012 at 5:14 PM, Carl Witthoft c...@witthoft.com wrote:
I think the OP wants to fill values in an arbitrarily large matrix. Now,
first of all, I'd like to know what his real problem is, since this seems
like a very tedious and unproductive matrix to produce. But in the
meantime, since he also left out important information, let's assume the
input matrix is N rows by M columns, and that he wants therefore to end up
with N instances of 100, not counting the original value of 100 that is
one of his ranking values (a bad BAD move IMHO).
Then either loop or lapply over an equation like (I've expanded things more
than necessary for clarity
result-inmatrix
for (k in 1:N){
foo - which (inmatrix == k,arr.ind=T)
result[k,foo[2]] -100
}
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all. Something like this maybe?
(input-matrix(c(5,1,3,7,2,6,4,8),nc=2))
output - matrix(0,max(input),2)
output[input[,1],1] - 100
output[input[,2],2] - 100
output
Cheers
On Fri, Apr 6, 2012 at 1:49 PM, Dimitri Liakhovitski
dimitri.liakhovitski at gmail.com wrote:
Hello, everybody!
I have a matrix input (see example below) - with all unique entries
that are actually unique ranks (i.e., start with 1, no ties).
I want to assign a value of 100 to the first row of the column that
contains the minimum (i.e., value of 1).
Then, I want to assign a value of 100 to the second row of the column
that contains the value of 2, etc.
The results I am looking for are in desired.results.
My code (below) does what I need. But it's using a loop through all
the rows of my matrix and searches for a matrix element every time.
My actual matrix is very large. Is there a way to do it more efficiently?
Thank you very much for the tips!
Dimitri
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
(input)
desired.result-as.matrix(data.frame(a=c(100,0,100,0),b=c(0,100,0,100)))
(desired.result)
result-as.matrix(data.frame(a=c(0,0,0,0),b=c(0,0,0,0)))
for(i in 1:nrow(input)){ # i-1
mymin-i
mycoords-which(input==mymin,arr.ind=TRUE)
result[i,mycoords[2]]-100
input[mycoords]-max(input)
}
(result)
--
Sent from my Cray XK6
Quidvis recte factum, quamvis humile, praeclarum.
__
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.
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
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-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.
