Re: Trying to use ReplacePart

*To*: mathgroup at smc.vnet.net*Subject*: [mg119970] Re: Trying to use ReplacePart*From*: Heike Gramberg <heike.gramberg at gmail.com>*Date*: Sat, 2 Jul 2011 05:02:10 -0400 (EDT)*References*: <201107010040.UAA24254@smc.vnet.net>

I don't know if this is more natural, but it does seem to produce the same matrix as your code without the need to initialise matrix first and with a (in my opinion) more transparent way to select the elements from matrix1 nNum = 3; nQnum = 2; cNum = 2^(nNum - nQnum); cQnum = 2^(nNum - 1); matrix1 = RandomInteger[{1, 10}, {2^nNum, 2^nNum}] list2 = Flatten[Table[(2 j - 1) cNum + i, {j, cQnum/cNum}, {i, cNum}], 1]; matrix = matrix1[[list2, list2]] Heike. On 1 Jul 2011, at 01:40, Iv=E1n Lazaro wrote: > Hi! > > I needed some days ago to replace elements in a matrix. My first > thought was to use ReplacePart, but I found myself trying to tell it > to make all replacements at once, instead of returning multiple copies > of the original matrix with only one element changed. As I was in a > hurry, I solved the problem in a different way. However I'm still > wondering how can I use a more "natural" code to solve this problem. > > > nNum = 3; > nQnum = 2; > > cNum = 2^(nNum - nQnum); > cQnum = 2^(nNum - 1); > > list = Table[i, {i, 1, cQnum}]; > list1 = Partition[list, cNum]; > matrix = ConstantArray[0, {cQnum, cQnum}]; > matrix1 = RandomInteger[{1, 10}, {2^nNum, 2^nNum}] > > > (matrix[[#[[1]], #[[2]]]] = > matrix1[[ > cNum*Position[list1, #[[1]]][[1, 1]] + #[[1]], cNum*Position[list1, > #[[2]]][[1, 1]] + #[[2]]]]) & /@ Tuples[{list, list}]; > > > or a faster solution, > > Table[matrix[[i, j]] = > matrix1[[ > cNum*Position[list1, i][[1, 1]] + i, cNum*Position[list1, j][[1, 1]] + > j]], {i, 1, cQnum}, {j, 1,cQnum}]; >