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

[ilai]

On Sun, Apr 8, 2012 at 8:26 AM, Dimitri Liakhovitski
<dimitri.liakhovitski at 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 <keren at math.montana.edu> wrote:
>> On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
>> <dimitri.liakhovitski at 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 <carl at 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 at 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 at 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