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

*To*: mathgroup at smc.vnet.net*Subject*: [mg99512] Re: [mg99492] Given a matrix, find position of first non-zero*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Thu, 7 May 2009 06:33:06 -0400 (EDT)*Reply-to*: hanlonr at cox.net

A = {{0, 0, 5}, {50, 0, 100}, {0, 75, 100}, {75, 100, 0}, {0, 75, 100}, {0, 75, 100}}; Union[Position[A, _?(# != 0 &)], SameTest -> (#1[[1]] == #2[[1]] &)] {{1, 3}, {2, 1}, {3, 2}, {4, 1}, {5, 2}, {6, 2}} Bob Hanlon ---- 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