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