猜一个字python循环for不从零开始_Python-多处理-巨⼤的for循环下午好，
我对Python还是很陌⽣，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，但我需要这样做8次...(是的，我知道OS迭代的次数完全是疯狂的。
我的机器相当不错，所以我想使⽤多处理python功能来加快速度。
您是否知道任何简单的解决⽅案可以从我的机器的处理能⼒中获利？
输⼊：
配对对：
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，您可以假设len(list1 [i])和len(list2 [i])都是整数，并且都等于198。(在实际问题中，我们实际上有21个不同的整数，但是都以相同的顺序排列-这意味着它们不会超出198。
for循环如下：
for first in ListPairs:
print(str(first))
for cond in ListPairs:
for third in ListPairs:
for fourth in ListPairs:
for fifth in ListPairs:
for sixth in ListPairs:
for venth in ListPairs:
sumA = first[0] + cond[0] + third[0] + fourth[0] + fifth[0] + sixth[0] + venth[0]
sumB = first[1] + cond[1] + third[1] + fourth[1] + fifth[1] + sixth[1] + venth[1]
for i in range(len(list1)):
if sumA == len(list1[i]) and sumB == len(list2[i]):
List7 = []
小腹疼痛男性List7 = [first, cond, third, fourth, fifth, sixth, venth]
ListsOut[i].append(List7)
for eighth in ListPairs:
sumA = first[0] + cond[0] + third[0] + fourth[0] + fifth[0] + sixth[0] + venth[0] + eighth[0]
sumB = first[1] + cond[1] + third[1] + fourth[1] + fifth[1] + sixth[1] + venth[1] + eighth[1]
for i in range(len(list1)):
if sumA == len(list1[i]) and sumB == len(list2[i]):
List8 = []
List8 = [first, cond, third, fourth, fifth, sixth, venth, eighth]
ListsOut[i].append(List8)
⾮常感谢！
解决⽅案
您发布的解决⽅案可能永远不会完成，因为它将需要经历超过10 ^ 21个元素的组合。与其使⽤多重处理，不如使⽤更快的算法。
胡萝卜炒瘦肉
使⽤您在问题中使⽤的list1，list2和lists_out，我们正在寻找组合16和36之间的整数的⽅法，以使它们求和成为list1和list2中序列的长度。组合应为[16，36]范围内的7或8个整数。
import itertools
def so43965562(list1, list2, lists_out, lower=16, upper=36):
asrt 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)元旦宣传海报
以下函数写s为和n之间的整数之l和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)))
第一夫人歌词[(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.
内容英语>>> print(list(combinations_summing_to(16, 36, 2, 34)))
[(16, 18), (17, 17)]
"""
if n == 0:
return (suffix,) if s == 0 el ()
elif n == 1:
return ((s,) + suffix,) if l <= s <= u el ()
el:
贝茜
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)
您可以按以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)
会考题print(len(lists_out[0]))
# Outputs 1795769, takes about 2.5 conds to run.
请注意，输出⼤⼩呈指数增长，当n = 7 * 16 = 112时从零开始，因此当您编写问题时，当n = 198时，计算所有组合仍将花费很长时间。
