[R] searching for specific row in matrix

[jim holtman]

This should work for you:

> create_bin_string <- function(len)
+ {
+  sample(0:1, len, replace=T)
+ }
>
> ROWS = 10
> COLS =  5
> set.seed(2)
> pop = matrix(create_bin_string(ROWS*COLS), ROWS, COLS, byrow=T)
>
>
>
> target=c(1, 1, 0, 1, 1)
>
> # my population
> print(pop)
      [,1] [,2] [,3] [,4] [,5]
 [1,]    0    1    1    0    1
 [2,]    1    0    1    0    1
 [3,]    1    0    1    0    0
 [4,]    1    1    0    0    0
 [5,]    1    0    1    0    0
 [6,]    0    0    0    1    0
 [7,]    0    0    1    1    1
 [8,]    1    1    0    1    0
 [9,]    1    0    0    0    1
[10,]    1    1    0    1    1
>
> # determine which data matches
> matches <- t(pop) == target  # 't' due to matching in column order
>
> # colSums equal to COLS will indicate matches
> which(colSums(matches) == COLS)
[1] 10
>

On Wed, Jun 11, 2008 at 7:58 AM, Esmail Bonakdarian <esmail.js at gmail.com> wrote:
> Hi,
>
> I have matrix of bits and a target vector. Is there an
> efficient way to search the rows of the matrix for the target?
> I am interested in the first row index where target is found.
>
> Example:
>
>> source("lookup.R")
>      [,1] [,2] [,3] [,4] [,5]
>  [1,]    1    0    1    1    0
>  [2,]    1    1    0    1    0
>  [3,]    0    0    1    0    0
>  [4,]    1    0    0    1    1
>  [5,]    1    0    1    1    1
>  [6,]    1    1    0    0    1
>  [7,]    1    0    0    1    1
>  [8,]    0    0    1    1    1
>  [9,]    0    1    1    0    1
> [10,]    0    0    0    1    0
>
> target:  1 1 0 1 1
>
> Should return -1 (or some other indicator) since the
> target was not found in any of the rows.
>
>
>
>> source("lookup.R")
>      [,1] [,2] [,3] [,4] [,5]
>  [1,]    0    0    1    1    0
>  [2,]    1    0    0    0    0
>  [3,]    1    0    0    0    0
>  [4,]    1    1    0    0    0
>  [5,]    1    1    1    0    0
>  [6,]    0    0    1    1    0
>  [7,]    0    1    1    1    0
>  [8,]    0    0    1    1    0
>  [9,]    1    1    0    1    1
> [10,]    1    0    1    0    0
>
> target:  1 1 0 1 1
>
> Should return 9 since the target was  found in row 9
>
>
> If the target is found, it is no longer necessary to keep
> searching the rest of the matrix (which may be quite large)
>
> The data/size etc may change of course, but target will
> always have the same number of "columns" as the matrix.
>
> I tried variations of "which", and a for loop
> comparing pop[i,] to target without much success, nor
> did google yield any results. I am hoping someone here
> can provide a suggestion.
>
> Thanks,
>
> EB
>
>
> ---------------------------------------------
>
> # Here is the code that generates the above data
>
> create_bin_string <- function(len)
> {
>  sample(0:1, len, replace=T)
> }
>
> ROWS = 10
> COLS =  5
> pop = matrix(create_bin_string(ROWS*COLS), ROWS, COLS, byrow=T)
>
>
>
> target=c(1, 1, 0, 1, 1)
>
> # my population
> print(pop)
>
> # I am looking for the index of this in pop
> # if present (else -1?)
> cat("\ntarget: ", target, "\n")
>
>
>
> ##
> ## this is NOT working
> ## plus it would continue to search
> ## after it found the target
> ##
> for(i in ROWS)
>   if (pop[i,] == target)
>      cat("\nfound in row: ", i, "\n\n")
>
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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.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

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