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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg99497] Re: Given a matrix, find position of first non-zero element in each
  • From: Oliver Ruebenkoenig <ruebenko at googlemail.com>
  • Date: Thu, 7 May 2009 06:30:18 -0400 (EDT)
  • References: <gtrl9k$242$1@smc.vnet.net>
  • Reply-to: oliver.ruebenkoenig at web.de

On Wed, 6 May 2009, 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
>
>
>
>

Here is one,

First /@ SplitBy[Drop[ArrayRules[SparseArray[A]][[All, 1]], -1],
    First]

perhaps one with a better complexity could created taking into account 
that only the first non 0 is needed in each row.

Oliver


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