Re: Transforming matrices

*To*: mathgroup at smc.vnet.net*Subject*: [mg29685] Re: [mg29665] Transforming matrices*From*: Tomas Garza <tgarza01 at prodigy.net.mx>*Date*: Tue, 3 Jul 2001 04:40:40 -0400 (EDT)*References*: <200107020620.CAA02313@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi, Michael! I thought I'd try to give an answer to your problem. The idea goes as follows: given your matrix, say, mat, take a new matrix, newMat, where you substitute all 1's and 2's by the symbol "*". Then (in the first row, for example), look for all the positions in that row where you have a sequence *0. The sequence of such positions determine the positions of the consecutive rates you are going to substitute the zeros with. Then, apply this substitution to all rows, and you're done. I propose the following code, where I include the case where some rows could start with a 1 instead of a 0. Take the matrix mat {{0, 0, 1, 2, 2, 2, 0, 0, 0, 0, 0, 0, 1, 2, 0, 1, 2, 2, 2, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 2, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 2, 2, 0, 0, 0, 1, 2, 0, 1, 2, 2, 2, 0, 0, 0, 1, 2, 2, 0}} (the same you have in your query, with an extra row starting with 1). Then, In[2]:= newMat = mat /. {1 -> "*", 2 -> "*"} Out[2]= {{0, 0, "*", "*", "*", "*", 0, 0, 0, 0, 0, 0, "*", "*", 0, "*", "*", "*", "*", 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, "*", "*", "*", "*", 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {"*", "*", "*", 0, 0, 0, "*", "*", 0, "*", "*", "*", "*", 0, 0, 0, "*", "*", "*", 0}} In[3]:= trans[x_List] := Module[{ratesPos = #[[1]] & /@ StringPosition[StringJoin @@ ToString /@ x, "*0"]}, runs = Split[x]; class = If[runs[[1, 1]] == 0, 1, 2]; If[class == 1, runs[[1]] = runs[[1]] /. (0 -> rates[[1]]); Do[runs[[2j + 1]] = runs[[2j + 1]] /. (0 -> rates[[ratesPos[[j]]]]), {j, 1, Length[ratesPos]}], Do[runs[[2j]] = runs[[2j]] /. (0 -> rates[[ratesPos[[j]]]]), {j, 1, Length[ratesPos]}] ]; Join @@ runs /. "*" -> 0] (in the above code, the variable "class" is used to determine if a row starts with a 0 or not). You then map trans onto each row of the matrix newMat: In[4]:= trans /@ newMat Out[4]= {{40, 40, 0, 0, 0, 0, 32, 32, 32, 32, 32, 32, 0, 0, 16, 0, 0, 0, 0, 9}, {40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40}, {40, 40, 0, 0, 0, 0, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32}, {0, 0, 0, 37, 37, 37, 0, 0, 30, 0, 0, 0, 0, 20, 20, 20, 0, 0, 0, 9}} which, I hope, what you wanted to obtain. Tomas Garza Mexico City ----- Original Message ----- From: "Michael Loop" <loopm at yahoo.com> To: mathgroup at smc.vnet.net Subject: [mg29685] [mg29665] Transforming matrices > I am a relatively new user of Mathematica, and I am having some > trouble transforming a matrix. I would be appreciative of any advice. > The problem is as follows. > > Given a randomly formed matrix of form: > > 0 0 1 2 2 2 0 0 0 0 0 0 1 2 0 1 2 2 2 0 > 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > 0 0 1 2 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > > Each digit represents a different state, and therefore each digits > position must be preserved. I need to convert this matrix into a > matrix of rates instead. The zeros will be converted into constant > rates dependent on their position, and the rate will continue until a > 1 is reached. Both 1 and 2 represent a rate of 0. So given a vector > of rates: > > 40 38, 37, 36, 33, 32, 31, 30, 27, 23, 22, 21, 20, 16, 14, 13, 12, 10, > 9, 6 > > I need to convert the first randomly generated matrix to look like: > > 40 40 0 0 0 0 32 32 32 32 32 32 0 0 16 0 0 0 0 9 > 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 40 > 40 40 0 0 0 0 32 32 32 32 32 32 32 32 32 32 32 32 32 32 > > Besides the fact that the rate must stay constant until a 1 is > reached, the other subtlety of my problem is that when a rate starts > over after a 1 or 2, the new rate must begin in the position of the > next zero, but the rate must come from the position of the last 1 or > 2. I can convert a matrix of this form on an individual basis using > the Position and Table commands, but cannot find an efficient way of > converting the matrix on a large scale. The matrix I actually need to > convert is 240*1000, so if anyone can think of an efficient way of > doing this I would be grateful for your input. Thank you. > > Micahel Loop > Minneapolis, MN

**References**:**Transforming matrices***From:*loopm@yahoo.com (Michael Loop)

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