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

*To*: mathgroup at smc.vnet.net*Subject*: [mg99515] Re: [mg99492] Given a matrix, find position of first non-zero element*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Thu, 7 May 2009 06:33:39 -0400 (EDT)*References*: <200905060929.FAA02194@smc.vnet.net>

Hi, Range@Length@A /. Map[#[[1]] -> # &, Position[A, _?(# != 0 &)]] or Split[Position[A, _?(# != 0 &)], #1[[1]] == #2[[1]] &][[All, 1]] or Transpose[{Range[Length[A]], LengthWhile[#, # == 0 &] + 1 & /@ A}] or Map[Scan[If[#[[1]] != 0, Return@#[[2]]] &, #] &, MapIndexed[List, A, {2}]] or (purely patterns, but quite a hack) Module[{n = 1}, A1 /. {0 -> 1, _Integer -> 0} /. {{x___, 0, ___} :> {n++, x + 1}, {___Integer} :> (n++ /. _ :> Sequence[])}] or... Regards, Leonid On Wed, May 6, 2009 at 2:29 AM, Nasser Abbasi <nma at 12000.org> 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 > > > >

**References**:**Given a matrix, find position of first non-zero element in each row***From:*"Nasser Abbasi" <nma@12000.org>