Python：如何在大于M的公共列中找到非零单元格的N行以上

Question

I have nxm matrix and I want to find programatically N or more rows that contains non zero cells in more than M common columns.

For example. Here is the matrix:

matrix([[ 0.,  0.,  1.,  1.,  1.,  0.,  1.,  0.],
        [ 1.,  0.,  1.,  0.,  1.,  1.,  0.,  1.],
        [ 1.,  0.,  0.,  0.,  0.,  1.,  0.,  0.],
        [ 0.,  0.,  0.,  0.,  1.,  0.,  1.,  0.],
        [ 0.,  1.,  0.,  1.,  1.,  0.,  1.,  0.],
        [ 0.,  1.,  0.,  1.,  0.,  0.,  0.,  0.]])

And I looking for 2 or more rows which contains non zero cells in 2 or more common colums. There are several possible results, one of them is:

row1: [ 1.,  0.,  1.,  0.,  1.,  1.,  0.,  1.],
row2: [ 1.,  0.,  0.,  0.,  0.,  1.,  0.,  0.],
      col1                     col5

Is it possible to find all rows combinations, that solve this task?

Answer 1

from itertools import combinations                   
matrix =[[ 0.,  0.,  1.,  1.,  1.,  0.,  1.,  0.],   
         [ 1.,  0.,  1.,  0.,  1.,  1.,  0.,  1.],   
         [ 1.,  0.,  0.,  0.,  0.,  1.,  0.,  0.],   
         [ 0.,  0.,  0.,  0.,  1.,  0.,  1.,  0.],   
         [ 0.,  1.,  0.,  1.,  1.,  0.,  1.,  0.],   
         [ 0.,  1.,  0.,  1.,  1.,  0.,  0.,  0.],   
         [ 0.,  1.,  0.,  1.,  1.,  0.,  0.,  1.]]   

m = 2                                                
n = 6                                                
req_rows = []                                        
ncm = [x for x in combinations(matrix,m)]            
for x in ncm:                                        
    if sum([1 for l in zip(*x) if not 0 in l])>=m:   
        print x

output

([0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 0.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 1.0])
([0.0, 0.0, 1.0, 1.0, 1.0, 0.0, 1.0, 0.0], [1.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0])
([0.0, 0.0, 1.0, 1.0, 1.0, 0.0, 1.0, 0.0], [0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 1.0, 0.0])
([0.0, 0.0, 1.0, 1.0, 1.0, 0.0, 1.0, 0.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0])
([0.0, 0.0, 1.0, 1.0, 1.0, 0.0, 1.0, 0.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 0.0])
([0.0, 0.0, 1.0, 1.0, 1.0, 0.0, 1.0, 0.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 1.0])
([1.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0], [1.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0])
([1.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 1.0])
([0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 1.0, 0.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0])
([0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 0.0])
([0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 1.0])
([0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 0.0], [0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 1.0])

Answer 2

from pprint import pprint
from itertools import combinations

def solve(lst, m):

    col, n = {}, len(lst)
    for i, x in enumerate(lst):
        col[i] = [j for j, y in enumerate(x) if y]

    for s in xrange(n, m-1, -1):
        for c in combinations(xrange(n), s):
            values  = set(col[c[0]]).intersection(*(col[k] for k in c[1:]))
            if len(values) >= m:
                yield [lst[k] for k in c]

for x in solve(matrix, 2):
    pprint(x)

Output:

[[0, 0, 1, 1, 1, 0, 1, 0],
 [0, 0, 0, 0, 1, 0, 1, 0],
 [0, 1, 0, 1, 1, 0, 1, 0]]
[[0, 0, 1, 1, 1, 0, 1, 0], [1, 0, 1, 0, 1, 1, 0, 1]]
[[0, 0, 1, 1, 1, 0, 1, 0], [0, 0, 0, 0, 1, 0, 1, 0]]
[[0, 0, 1, 1, 1, 0, 1, 0], [0, 1, 0, 1, 1, 0, 1, 0]]
[[1, 0, 1, 0, 1, 1, 0, 1], [1, 0, 0, 0, 0, 1, 0, 0]]
[[0, 0, 0, 0, 1, 0, 1, 0], [0, 1, 0, 1, 1, 0, 1, 0]]
[[0, 1, 0, 1, 1, 0, 1, 0], [0, 1, 0, 1, 0, 0, 0, 0]]