Python-多处理-巨大的for循环

Question

Good afternoon,

I am quite new to Python, and I have to solve a problem which has the need to try billions of hypothesis... More specifically I need to iterate a list of 440 elements, but I need to do it 8 times... (yes, the number os iterations is completly insane I know).

My machine is quite good, so I want to use the multiprocessing python functionalities to speed this up a lot.

Do you know any simple solution which would take profit from the processing capabilities from my machine?

Inputs:

ListPairs:

for ind1 in range(16,37):
    for ind2 in range(16,37):

        ListPairsAux = []

        ListPairsAux.append(ind1)
        ListPairsAux.append(ind2)

        ListPairs.append(ListPairsAux)

For the simplicity of the problem, you can assume that both len(list1[i]) and len(list2[i]) are integers and both are equal to 198. (In the real problem we will actually have 21 different integers, but all in the same order - meaning that they won't go much further than 198.

The for loops are the ones below:

for first in ListPairs:
    print(str(first))
    for second in ListPairs:
        for third in ListPairs:
            for fourth in ListPairs:
                for fifth in ListPairs:
                    for sixth in ListPairs:
                        for seventh in ListPairs:
                            sumA = first[0] + second[0] + third[0] + fourth[0] + fifth[0] + sixth[0] + seventh[0]
                            sumB = first[1] + second[1] + third[1] + fourth[1] + fifth[1] + sixth[1] + seventh[1]
                            for i in range(len(list1)):
                                if sumA == len(list1[i]) and sumB == len(list2[i]):
                                    List7 = []
                                    List7 = [first, second, third, fourth, fifth, sixth, seventh]
                                    ListsOut[i].append(List7)

                            for eighth in ListPairs:
                                sumA = first[0] + second[0] + third[0] + fourth[0] + fifth[0] + sixth[0] + seventh[0] + eighth[0]
                                sumB = first[1] + second[1] + third[1] + fourth[1] + fifth[1] + sixth[1] + seventh[1] + eighth[1]
                                for i in range(len(list1)):
                                    if sumA == len(list1[i]) and sumB == len(list2[i]):
                                        List8 = []
                                        List8 = [first, second, third, fourth, fifth, sixth, seventh, eighth]
                                        ListsOut[i].append(List8)

Thank you so much!

Answer 1

The solution you post will probably never finish, since it would require going through more than 10^21 combinations of elements. Rather than using multiprocessing you should use a faster algorithm.

Using the list1, list2 and lists_out that you use in your question, we are looking for ways to combine integers between 16 and 36 so that they sum to the lengths of the sequences in list1 and list2. The combinations should be of 7 or 8 integers in the range [16, 36].

import itertools
def so43965562(list1, list2, lists_out, lower=16, upper=36):
    assert len(list1) == len(list2) == len(lists_out)
    for n in (7, 8):
        for i in range(len(list1)):
            # Find all combinations of n numbers in [lower, upper]
            # that sum to len(list1[i])
            combs1 = combinations_summing_to(lower, upper, n, len(list1[i]))
            # Find all combinations of n numbers in [lower, upper]
            # that sum to len(list2[i])
            combs2 = combinations_summing_to(lower, upper, n, len(list2[i]))
            for t1, t2 in itertools.product(combs1, combs2):
                result = [(v1, v2) for v1, v2 in zip(t1, t2)]
                lists_out[i].append(result)

The following function writes s as a sum of n integers between l and u .

def combinations_summing_to(l, u, n, s, suffix=()):
    """In which ways can s be written as the sum of n integers in [l, u]?

    >>> # Write 2 as a sum of 4 integers between 0 and 5.
    >>> print(list(combinations_summing_to(0, 5, 4, 2)))
    [(0, 0, 0, 2), (0, 0, 1, 1)]
    >>> # Write 5 as a sum of 3 integers between 0 and 5.
    >>> print(list(combinations_summing_to(0, 5, 3, 5)))
    [(0, 0, 5), (0, 1, 4), (0, 2, 3), (1, 1, 3), (1, 2, 2)]
    >>> # Write 12 as a sum of 3 integers between 0 and 5.
    >>> print(list(combinations_summing_to(0, 5, 3, 12)))
    [(2, 5, 5), (3, 4, 5), (4, 4, 4)]
    >>> # Write 34 as a sum of 2 integers between 16 and 36.
    >>> print(list(combinations_summing_to(16, 36, 2, 34)))
    [(16, 18), (17, 17)]
    """
    if n == 0:
        return (suffix,) if s == 0 else ()
    elif n == 1:
        return ((s,) + suffix,) if l <= s <= u else ()
    else:
        return itertools.chain.from_iterable(
            # Combinations summing to s where the last element is k
            combinations_summing_to(l, k, n - 1, s - k, (k,) + suffix)
            for k in range(u, l-1, -1)
            # Early bailout if you can't make s with all elements <= k
            if l * n <= s <= k * n)

You can run the solution as follows:

lists_out = [[]]
so43965562(list1=[[0]*(7*16+1)], list2=[[0]*(7*16+2)], lists_out=lists_out)
for result in lists_out[0]:
    print(result)
# Outputs the following two combinations:
# [(16, 16), (16, 16), (16, 16), (16, 16), (16, 16), (16, 16), (17, 18)]
# [(16, 16), (16, 16), (16, 16), (16, 16), (16, 16), (16, 17), (17, 17)]
lists_out = [[]]
n = 133
so43965562(list1=[[0]*n], list2=[[0]*n], lists_out=lists_out)
print(len(lists_out[0]))
# Outputs 1795769, takes about 2.5 seconds to run.

Note that the output size increases exponentially, starting at nothing when n = 7*16 = 112, so it will still take a long time to compute all the combinations when n = 198 as you write in your question.

Answer 2

If you want the combinations(or permutations), check python itertools

https://docs.python.org/2/library/itertools.html#itertools.combinations

Otherwise modify your algorithm.

 for combination in itertools.combinations_with_replacement(ListPairs, 8):
     # combination is a tuple
     for i in combination:
           #check your condition