python中具有不同元素的组合

Question

I'm struggling when calculating different types of combinations.

Let's explain with an example, I have this array or it could be a dataframe and I want different combinations of some columns from it.

As I will then multiply this matrix by the combination to sum the numbers.

test = np.array ([[10,11,12,21,22,31,32,33],
                   [10,11,12,21,22,31,32,33],
                   [10,11,12,21,22,31,32,33],
                    [10,11,12,21,22,31,32,33],
                    [10,11,12,21,22,31,32,33],
                    [10,11,12,21,22,31,32,33],
                    [10,11,12,21,22,31,32,33]])

The possible combinations for the first three columns are [1,0,0],[0,1,0],[0,0,1], so I need 10 or 11 or 12 Following columns, 21 or 22, therefore, combinations [1,0], [0,1] And the last three columns, 31,3 2, 33, then it will be [1,0,0],[0,1,0],[0,0,1]

So, I get the possible combination by using this function i found in another question.

n=3
for i in range(2**n):
    s = bin(i)[2:]
    s = "0" * (n-len(s)) + s
    print (list(s))

Which gives me:

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

All possible combinations, including the zeros. Although I managed to delete those.

It calculates more than the combinations I need, and I find myself eliminating too many combinations I do not need.

When I only need these cases:

[1,0,0,   1,0,   1,0,0]
[0,1,0,   1,0,   1,0,0]
[0,0,1,   1,0,   1,0,0]
[1,0,0,   0,1,   1,0,0]
[0,1,0,   0,1,   1,0,0]
[0,0,1,   0,0,   1,0,0]
 etc....

I need to delete many rows which are not relevant on the 8 cases, and delete rows where I find more than three 1's, and select where the 1's are positioned correctly,etc... not efficient at all. I'm a bit lost.

Answer 1

I really don't understand the logic behind the example, but does this solve your problem?

from itertools import product,permutations

a = set(permutations([0,0,1]))
b = set(permutations([0,1]))

comb = []
for t1,t2,t3 in product(a,b,a):
    comb.append([*t1,*t2,*t3])

print(comb)

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