读了很久，我第一次找不到我正在做的事情的答案。
我有一个包含93个字符串的列表，每个字符串长度为6个字符。从这93个字符串中，我想要识别一组20个，它们都满足相对于集合中其他条件的特定标准。虽然itertools.combinations将为我提供所有可能的组合，但并非所有条件都值得检查。
例如，如果[list [0]，list [1]等]失败，因为list [0]和list [1]不能在一起，那么其他18个字符串是什么并不重要，该集合将失败每一次，这都是浪费大量的检查。
目前我已经使用了20个嵌套for循环，但似乎必须有更好/更快的方法来实现它。：
for n1 in bclist:
building = [n1]
n2bclist = [bc for bc in bclist if bc not in building]
for n2 in n2bclist: #this is the start of what gets repeated 19 times
building.append(n2)
if test_function(building): #does set fail? (counter intuitive, True when fail, False when pass)
building.remove(n2)
continue
n3bclist = [bc for bc in bclist if bc not in building]
#insert the additional 19 for loops, with n3 in n3, n4 in n4, etc
building.remove(n2)
在20日的循环中有打印语句，如果一组20甚至存在，则提醒我。 for语句至少允许我在单个添加失败时提前跳过集合，但是当没有更大的组合失败时没有记忆：
例如[list[0], list[1]]失败，所以跳到[list[0], [list[2]]通过。接下来是[list[0], list[2], list[1]]，它将失败，因为0和1再次在一起，所以它将移动到[list[0], list[2], list[3]]，可能会或不会通过。我担心的是最终还会测试：
[list[0], list[3], list[2]] [list[2], list[0], list[3]] [list[2], list[3], list[0]] [list[3], list[0], list[2]] [list[3], list[2], list[0]] 所有这些组合将与之前的组合具有相同的结果。基本上我交易了itertools.combinations的恶魔，测试我知道失败的所有集合的组合，因为早期的值对于for循环的恶魔而言失败，当我不关心它们的顺序时，它将值的顺序视为一个因素。这两种方法都会显着增加我的代码完成所需的时间。
任何有关如何摆脱恶魔的想法都将不胜感激。
答案 0 :(得分：1)
使用您当前的方法，但也要跟踪索引，以便在内部循环中可以跳过您已经检查过的元素：
bcenum = list(enumerate(bclist))
for i1, n1 in bcenum:
building = [n1]
for i2, n2 in bcenum[i1+1:]: #this is the start of what gets repeated 19 times
building.append(n2)
if test_function(building): #does set fail? (counter intuitive, True when fail, False when pass)
building.remove(n2)
continue
for i3, n3 in bcenum[i2+1:]:
# more nested loops
building.remove(n2)
答案 1 :(得分：1)
def gen(l, n, test, prefix=()):
if n == 0:
yield prefix
else:
for i, el in enumerate(l):
if not test(prefix + (el,)):
for sub in gen(l[i+1:], n - 1, test, prefix + (el,)):
yield sub
def test(l):
return sum(l) % 3 == 0 # just a random example for testing
print list(gen(range(5), 3, test))
这将从n中选择基数l的子集，以便test(subset) == False。
它试图避免不必要的工作。但是，鉴于有93种方法可以选择20种元素，您可能需要重新考虑整体方法。
答案 2 :(得分：0)
您可以利用问题的两个方面：
test_function(L)为True，那么test_function的任何子列表中的L也将为True 您还可以通过处理索引0-92而不是list[0] - list[92]来简化一些事情 - 它只在test_function内，我们可能会关心列表的内容是
以下代码通过首先找到可行对，然后是四组，八组和十六组来完成。最后，它找到了16和4的所有可行组合，以获得20个列表。然而，有超过100,000套8个，所以它仍然太慢，我放弃了。可能你可以沿着同一条线做某些事情但是用itertools来加快速度，但可能还不够。
target = range(5, 25)
def test_function(L):
for i in L:
if not i in target:
return True
def possible_combos(A, B):
"""
Find all possible pairings of a list within A and a list within B
"""
R = []
for i in A:
for j in B:
if i[-1] < j[0] and not test_function(i + j):
R.append(i + j)
return R
def possible_doubles(A):
"""
Find all possible pairings of two lists within A
"""
R = []
for n, i in enumerate(A):
for j in A[n + 1:]:
if i[-1] < j[0] and not test_function(i + j):
R.append(i + j)
return R
# First, find all pairs that are okay
L = range(92)
pairs = []
for i in L:
for j in L[i + 1:]:
if not test_function([i, j]):
pairs.append([i, j])
# Then, all pairs of pairs
quads = possible_doubles(pairs)
print "fours", len(quads), quads[0]
# Then all sets of eight, and sixteen
eights = possible_doubles(quads)
print "eights", len(eights), eights[0]
sixteens = possible_doubles(eights)
print "sixteens", len(sixteens), sixteens[0]
# Finally check all possible combinations of a sixteen plus a four
possible_solutions = possible_combos(sixteens, fours)
print len(possible_solutions), possible_solutions[0]
test_function的范围（0-92）内的所有值对，保持对顺序。据推测，第一对的第一个值必须是解的第一个值，最后一对的第二个值必须是解的最后一个值（但是检查... test_function的假设是正确的吗？如果这不是一个安全的假设，那么您需要对开始和结束的所有可能值重复find_paths。然后找到从第1个到最后一个值的路径，该路径长度为20个值，并且符合test_function。
def test_function(S):
for i in S:
if not i in target:
return True
return False
def find_paths(p, f):
""" Find paths from end of p to f, check they are the right length,
and check they conform to test_function
"""
successful = []
if p[-1] in pairs_dict:
for n in pairs_dict[p[-1]]:
p2 = p + [n]
if (n == f and len(p2) == target_length and
not test_function(p2)):
successful.append(p2)
else:
successful += find_paths(p2, f)
return successful
list_length = 93 # this is the number of possible elements
target = [i * 2 for i in range(5, 25)]
# ^ this is the unknown target list we're aiming for...
target_length = len(target) # ... we only know its length
L = range(list_length - 1)
pairs = []
for i in L:
for j in L[i + 1:]:
if not test_function([i, j]):
pairs.append([i, j])
firsts = [a for a, b in pairs]
nexts = [[b for a, b in pairs if a == f] for f in firsts]
pairs_dict = dict(zip(firsts, nexts))
print "Found solution(s):", find_paths([pairs[0][0]], pairs[-1][1])
答案 3 :(得分：0)
您应该将解决方案基于itertools.combinations，因为这将解决订购问题;短路滤波相对容易解决。
让我们快速回顾一下如何实施combinations的工作;最简单的方法是采用嵌套循环方法并将其转换为递归样式：
def combinations(iterable, r):
pool = tuple(iterable)
for i in range(0, len(pool)):
for j in range(i + 1, len(pool)):
...
yield (i, j, ...)
转换为递归形式：
def combinations(iterable, r):
pool = tuple(iterable)
def inner(start, k, acc):
if k == r:
yield acc
else:
for i in range(start, len(pool)):
for t in inner(i + 1, k + 1, acc + (pool[i], )):
yield t
return inner(0, 0, ())
现在应用过滤器很简单：
def combinations_filterfalse(predicate, iterable, r):
pool = tuple(iterable)
def inner(start, k, acc):
if predicate(acc):
return
elif k == r:
yield acc
else:
for i in range(start, len(pool)):
for t in inner(i + 1, k + 1, acc + (pool[i], )):
yield t
return inner(0, 0, ())
我们来看看：
>>> list(combinations_filterfalse(lambda t: sum(t) % 2 == 1, range(5), 2))
[(0, 2), (0, 4), (2, 4)]
the documentation中列出的itertools.combinations的实际实施使用了迭代循环：
def combinations(iterable, r):
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = range(r)
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)
为了优雅地适应谓词，有必要稍微重新排序循环：
def combinations_filterfalse(predicate, iterable, r):
pool = tuple(iterable)
n = len(pool)
if r > n or predicate(()):
return
elif r == 0:
yield ()
return
indices, i = range(r), 0
while True:
while indices[i] + r <= i + n:
t = tuple(pool[k] for k in indices[:i+1])
if predicate(t):
indices[i] += 1
elif len(t) == r:
yield t
indices[i] += 1
else:
indices[i+1] = indices[i] + 1
i += 1
if i == 0:
return
i -= 1
indices[i] += 1
再次检查：
>>> list(combinations_filterfalse(lambda t: sum(t) % 2 == 1, range(5), 2))
[(0, 2), (0, 4), (2, 4)]
>>> list(combinations_filterfalse(lambda t: t == (1, 4), range(5), 2))
[(0, 1), (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (2, 3), (2, 4), (3, 4)]
>>> list(combinations_filterfalse(lambda t: t[-1] == 3, range(5), 2))
[(0, 1), (0, 2), (0, 4), (1, 2), (1, 4), (2, 4)]
>>> list(combinations_filterfalse(lambda t: False, range(5), 2))
[(0, 1), (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
>>> list(combinations_filterfalse(lambda t: False, range(5), 0))
[()]
事实证明，递归解决方案不仅更简单，而且速度更快：
In [33]: timeit list(combinations_filterfalse_rec(lambda t: False, range(20), 5))
10 loops, best of 3: 24.6 ms per loop
In [34]: timeit list(combinations_filterfalse_it(lambda t: False, range(20), 5))
10 loops, best of 3: 76.6 ms per loop