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

*To*: mathgroup at smc.vnet.net*Subject*: [mg99540] Re: Given a matrix, find position of first non-zero element in each*From*: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>*Date*: Thu, 7 May 2009 06:38:13 -0400 (EDT)*References*: <gtrl9k$242$1@smc.vnet.net>

Hi Nasser, What do you want the output to be when there is no non-zero element in a row? The following just prints the row number, but it may not be what you want. A = {{0, 0, 5}, {50, 0, 100}, {0, 75, 100}, {75, 100, 0}, {0, 75, 100}, {0, 75, 100}, {0, 0, 0}}; MapIndexed[Flatten[{#2, Position[#1, x_ /; x != 0, 1, 1]}] &, A] Out[60]= {{1, 3}, {2, 1}, {3, 2}, {4, 1}, {5, 2}, {6, 2}, {7}} Cheers -- Sjoers On May 6, 11:29 am, "Nasser Abbasi" <n... 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 wante= d 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