Hi,
I need to compare two matrices with each other. If you can get one of
them out of the other one by resorting the rows and/or the columns, then
both of them are equal, otherwise they're not. A matrix could look like
this:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 0 1 1 1 0 1 1 0
[2,] 1 1 0 0 0 1 0 1
[3,] 1 0 1 0 0 0 1 1
[4,] 1 1 0 0 1 0 0 0
[5,] 1 0 1 1 1 0 0 0
[6,] 0 1 0 1 1 0 0 0
[7,] 0 0 0 0 0 1 1 1
Note that each matrix consists of ones and zeros, in each row and in
each column there are at least three ones and one zero and each pair of
rows/columns may have at most two positions where both are ones (e.g.
for the 1. and 2. rows those positions are 2 and 6).
I was advised to sort both matrices in the same way and then to compare
them element by element. But I don't manage to get them sorted... My
approach is as following:
1. Sort the rows after the row sums (greater sums first).
2. Sort the columns after the first column (columns with ones in the
first row go left, columns with zeros go right).
3. Save the left part (all columns with ones in the first row) and the
right part in separate matrices.
4. Repeat steps 2 and 3 with both of the created matrices (now taking
the second row for sorting), repeat until all fragments consist of a
single column.
5. Compose the columns to a sorted matrix.
This algorithm has several problems:
1. How to make a loop that is branching out in two subloops on each
iteration?
2. How to organize the intermediate results and compose them without
losing the order? Maybe save them in lists and sublists?
3. A fundamental problem: If there are rows with equal sums the result
may depend on which of them is sorted after first. Maybe this algorithm
won't work at all because of this problem?
Thanks in advance for your input,
Michael
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