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 > > >

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