Re: Using implicit information about row indices
- To: mathgroup at smc.vnet.net
- Subject: [mg68314] Re: [mg68254] Using implicit information about row indices
- From: "Carl K. Woll" <carlw at wolfram.com>
- Date: Tue, 1 Aug 2006 06:59:40 -0400 (EDT)
- References: <200607310745.DAA26729@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Diamond, Mark wrote: > I am wanting to fill a square matrix M (say, r by r) using information from > a list, L. > The list L is also of length r, but is not a square matrix. It contains > lists of ordered pairs {col, num} > It typically looks something like this; I have arranged it into rows so that > it is more obvious > { > {{1, a}, {2, b}}, > > {{1, c}, {2, d}, {3, e}, {4, f}}, > > {{2, g}, {3, h}, {4, i}}, > > {{3, k}, {4, l}, {5,m}}, > > {{4, n}, {5, p}} > > } > > Now what I am trying to do is, for example, set M[[1,1]]=a, M[[1,2]]=b, > M[[2,1]]=c, M[[2.2]]=d, M[[2,3]]=e and so forth, and leave (or set) all the > other entries in M to zero. The row information for M is implicit in the > structure of the list L; additionally, the column indices as they appear > within sublist of L are guraranteed to be sequential. It seems with all this > info, there should be an easy way to fill M correctly, but I have struggled > without success to do anything other than an iterative process with Do[]. I > would appreciate any guidance you might have. > > Thank you. > > Mark Diamond > > > To extract the position information one can use MapIndexed: L = {{{1, a}, {2, b}}, {{1, c}, {2, d}, {3, e}, {4, f}}, {{2, g}, {3, h}, {4, i}}, {{3, k}, {4, l}, {5, m}}, {{4, n}, {5, p}}}; Here I will use an inert function in MapIndexed so you get an idea of what is returned: MapIndexed[f,L,{2}] {{f({1,a},{1,1}),f({2,b},{1,2})},{ f({1,c},{2,1}),f({2,d},{2,2}),f({3,e},{2,3}),f({4,f},{2,4})},{f({2, g},{3,1}),f({3,h},{3,2}),f({4,i},{3,3})},{f({3,k},{4,1}), f({4,l},{4,2}),f({5,m},{4,3})},{f({4,n},{5,1}),f({5,p},{5,2})}} The first argument of f is the element of L being mapped over, and the second argument is the position of that element. In particular, the second argument's first element is the row number. So, we can convert L to a list of pos->val rules as follows: rules = MapIndexed[{#2[[1]], #1[[1]]} -> #1[[2]] & , L, {2}] // Flatten {{1, 1} -> a, {1, 2} -> b, {2, 1} -> c, {2, 2} -> d, {2, 3} -> e, {2, 4} -> f, {3, 2} -> g, {3, 3} -> h, {3, 4} -> i, {4, 3} -> k, {4, 4} -> l, {4, 5} -> m, {5, 4} -> n, {5, 5} -> p} A list of pos->val rules is the input expected by SparseArray: matrix = SparseArray[rules]; Here is the normal display of the sparse array: matrix//Normal {{a,b,0,0,0},{c,d,e,f,0},{0,g,h,i,0},{0,0,k,l,m},{0,0,0,n,p}} Carl Woll Wolfram Research