filling the matrix row by row in the order from lower to larger elements

[Dimitri Liakhovitski-2]

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

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[ilai-2]

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
<[hidden email]> 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
>
> ______________________________________________
> [hidden email] 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.

______________________________________________
[hidden email] 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.

[Rui Barradas]

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

[Carl Witthoft]

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."

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[Carl Witthoft]

Ok, how's this:


Rgames> foo
      [,1] [,2] [,3] [,4]
[1,]    3    6    1   16
[2,]   10   14   12    5
[3,]   11    7   15    9
[4,]    8    4   13    2

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,]    3    3    3    3
  [2,]   10   10   10   10
  [3,]   11   11  100   11
  [4,]    8    8    8    8
  [5,]    6    6    6    6
  [6,]   14   14   14   14
  [7,]    7    7    7    7
  [8,]    4    4    4  100
  [9,]  100    1    1    1
[10,]   12   12   12   12
[11,]   15   15   15   15
[12,]   13   13   13   13
[13,]   16   16   16   16
[14,]    5  100    5    5
[15,]    9    9    9    9
[16,]    2    2    2    2


Rgames> rab<-matrix(apply(bar,1,max),4,4)
Rgames> rab
      [,1] [,2] [,3] [,4]
[1,]    3    6  100   16
[2,]   10   14   12  100
[3,]  100    7   15    9
[4,]    8  100   13    2


--

Sent from my Cray XK6
"Quidvis recte factum, quamvis humile, praeclarum."

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[Dimitri Liakhovitski-2]

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 <[hidden email]> 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."
>
>
> ______________________________________________
> [hidden email] 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

______________________________________________
[hidden email] 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.

[ilai-2]

On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
<[hidden email]> 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 <[hidden email]> 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."
>>
>>
>> ______________________________________________
>> [hidden email] 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
>
> ______________________________________________
> [hidden email] 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.

______________________________________________
[hidden email] 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.

[Rui Barradas]

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

[Dimitri Liakhovitski-2]

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 <[hidden email]> wrote:

> On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
> <[hidden email]> 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 <[hidden email]> 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."
>>>
>>>
>>> ______________________________________________
>>> [hidden email] 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
>>
>> ______________________________________________
>> [hidden email] 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

______________________________________________
[hidden email] mailing list
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and provide commented, minimal, self-contained, reproducible code.

[Dimitri Liakhovitski-2]

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 <[hidden email]> 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.
>
> ______________________________________________
> [hidden email] 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

______________________________________________
[hidden email] 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.

[Bert Gunter]

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
<[hidden email]> 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 <[hidden email]> 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.
>>
>> ______________________________________________
>> [hidden email] 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
>
> ______________________________________________
> [hidden email] 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.

--

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: [phone]
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

______________________________________________
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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.

[ilai-2]

On Sun, Apr 8, 2012 at 8:26 AM, Dimitri Liakhovitski
<[hidden email]> 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 <[hidden email]> wrote:
>> On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
>> <[hidden email]> 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 <[hidden email]> 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."
>>>>
>>>>
>>>> ______________________________________________
>>>> [hidden email] 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
>>>
>>> ______________________________________________
>>> [hidden email] 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

______________________________________________
[hidden email] 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-2]

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 <[hidden email]> wrote:

> On Sun, Apr 8, 2012 at 8:26 AM, Dimitri Liakhovitski
> <[hidden email]> 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 <[hidden email]> wrote:
>>> On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
>>> <[hidden email]> 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 <[hidden email]> 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."
>>>>>
>>>>>
>>>>> ______________________________________________
>>>>> [hidden email] 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
>>>>
>>>> ______________________________________________
>>>> [hidden email] 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

--
Dimitri Liakhovitski
marketfusionanalytics.com

______________________________________________
[hidden email] 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.