下午好,
我对Python很陌生,我必须解决一个需要尝试数十亿个假设的问题...更具体地说,我需要迭代440个元素的列表,但我需要做8次。 ..(是的,我知道,迭代次数完全是疯了)。
我的机器非常好,所以我想使用多处理python功能来加速这一点。
您是否知道任何可以从我的机器处理功能中获利的简单解决方案?
输入:
ListPairs:
for ind1 in range(16,37):
for ind2 in range(16,37):
ListPairsAux = []
ListPairsAux.append(ind1)
ListPairsAux.append(ind2)
ListPairs.append(ListPairsAux)
为了简化问题,你可以假设len(list1 [i])和len(list2 [i])都是整数,并且都等于198.(在实际问题中我们实际上会有21个不同的整数,但都是以相同的顺序 - 意味着它们不会超过198。
for循环如下:
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)
非常感谢你!
答案 0 :(得分:1)
您发布的解决方案可能永远不会完成,因为它需要经过超过10 ^ 21个元素组合。您应该使用更快的算法,而不是使用多处理。
使用list1,list2和lists_out 你在问题中使用,我们正在寻找结合两者之间的整数的方法 在图16和36中,它们总和为list1和list2中序列的长度。 组合应该是[16,36]范围内的7或8个整数。
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)
以下函数将s
写为n
和l
之间u
个整数的总和。
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)
您可以按如下方式运行解决方案:
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.
请注意,输出大小呈指数级增长,从...开始 当n = 7 * 16 = 112时没有任何东西,所以它仍然需要很长时间才能计算出来 当你在问题中写下n = 198时的所有组合。
答案 1 :(得分:0)
如果你想要组合(或排列), 检查python itertools
https://docs.python.org/2/library/itertools.html#itertools.combinations
否则请修改算法。
for combination in itertools.combinations_with_replacement(ListPairs, 8):
# combination is a tuple
for i in combination:
#check your condition