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Re: Given a matrix, find position of first non-zero element in each row

  • To: mathgroup at smc.vnet.net
  • Subject: [mg99553] Re: [mg99492] Given a matrix, find position of first non-zero element in each row
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 7 May 2009 06:40:35 -0400 (EDT)
  • References: <200905060929.FAA02194@smc.vnet.net>

One way:

A = {{0, 0, 5}, {50, 0, 100}, {0, 75, 100}, {75, 100, 0}, {0, 75,  
100}, {0,
    75, 100}};

ls = MapIndexed[Flatten[{#2, Position[#1, x_ /; x != 0, 1, 1]}] & , A]
{{1, 3}, {2, 1}, {3, 2}, {4, 1}, {5, 2}, {6, 2}}

checking:

Extract[A, ls]
{5, 50, 75, 75, 75, 75}

Andrzej Kozlowski


On 6 May 2009, at 18:29, Nasser Abbasi wrote:

> This is a little problem I saw in another forum, and I am trying to  
> also
> solve it in Mathematica.
>
> Given a Matrix, I need to find the position of the first occurance  
> of a
> value which is not zero in each row.
>
> The position found will be the position in the orginal matrix  
> ofcourse.
>
> So, given this matrix,
>
> A = {
> {0, 0, 5},
> {50, 0, 100},
> {0, 75, 100},
> {75, 100, 0},
> {0, 75, 100},
> {0, 75, 100}
> };
>
> The result should be
>
> {{1, 3}, {2, 1}, {3, 2}, {4, 1}, {5, 2}, {6, 2}}
>
> This is how I solved this problem and after a bit of  struggle. I  
> wanted to
> see if I could avoid using a Table, and solve it just using Patterns  
> and
> Position and Select, but could not so far.
>
>
> Table[Flatten[{i, Flatten[Position[A[[i,All]], _?(#1 != 0 & ), 1,  
> 1]]}], {i,
> 1, 6}]
>
> Out[174]= {{1, 3}, {2, 1}, {3, 2}, {4, 1}, {5, 2}, {6, 2}}
>
> I am not happy with the above solution. I am sure there is a better  
> one (the
> above also do not work well when one row has all zeros).
>
> Do you see a better and more elegant way to do this?
>
> thanks,
> --Nasser
>
>
>



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