Despite having looked over various posts concerning optimizing nested For loop expressions into a functional programming equivalent I still can not figure out how to actually to do this. Such constructs are inefficient in Mathematica and hence, I seek to find a more efficient approach. Sadly, having started with programming with FORTRAN (prior to IV), some ideas die hard in an aging brain.
I have a very large array that I must evaluate. This array m, is 31x2754, but for purposes of example I have down-sampled so that it is only a 30 x 32 matrix. It produces a square matrix whose order is the size of the columns (here 32, when downsampled, but 2754x2754 when fully sampled). Although a doubly nested set of For loops provides the answer I seek and is relatively efficient when the matrix is only 32x32, it's efficiency degrades rapidly as n approaches 2754. I am at a loss to how to convert the following procedural code snippet into a functional programming equivalent that is more efficient.
The test input (ignoring how I get there, which itself is rather complicated) is:
m = {{0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1,
0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0,
0}, {1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0,
0, 1, 0, 0, 1, 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, 1, 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, 1, 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, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 1, 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, 1, 0, 0, 0, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {0, 0, 1, 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, 1, 0, 1, 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, 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, 1, 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, 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, 1, 0, 1, 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, 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, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 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, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1,
0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0}, {0, 1, 1, 0, 1, 1, 1, 0, 0, 0,
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0,
0}, {0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1,
1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0}, {0, 1, 0, 0, 0, 1, 1, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1,
0}, {0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1,
0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1,
0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 1, 1,
1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1}}
Dimensions[m]
{30, 32}
Although not essential, to be sure that the output is properly created I create a constant array and fill the matrix c, which will be the eventual output matrix with a value that WILL NOT be present in the matrix upon completion, so that by quick observation I can be sure that the matrix has been properly populated.
This is done as follows and establishes the order of the square matrix, where the length of columnlabels for this snippet is 32 (the actual column labels are irrelevant here):
c = ConstantArray[3, {Length[columnlabels], Length[columnlabels]}];
{characterrows, charactercolumns} = Dimensions[c]
The loop construct I wish to replace with a functional approach effectively compares two column of m, each of which has only two distinct values (either 0 or 1), and for which their product partial order potentially generates 4 distinct possibilities [vertices] ({0,0},{1,0},{0,1},{1,1}} and for which I want to assign to the each pair of columns a 1 if less than 4 of these possibilities are actually present in the comparison and a 0 if all 4 possibilities are present when comparing the two columns). This construct is as follows:
t0 = AbsoluteTime[];
For[i = 1, i <= Length[columnlabels], i++,
For[j = 1, j <= Length[columnlabels], j++,
f = m[[All, i]];
g = m[[All, j]];
c[[i, j]] = If[Length[Tally[Table[{g[[n]], f[[n]]}, {n, 1, Length[g]}]]] < 4, 1, 0];
]
]
t1 = AbsoluteTime[];
timeelapsed = UnitConvert[Quantity[t1 - t0, "Seconds"], "Minutes"]
What is the most efficient functional programming expression needed to replace this nested set of For loops?