python - Python - 从大量组合中构建满足某些标准的子列表

Question

读了很长时间，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，我想确定一组 20 个字符串，它们都符合相对于集合中其他字符串的特定标准。虽然 itertools.combinations 会给我所有可能的组合，但并非所有条件都值得检查。

例如，如果 [list[0], list[1], etc] 因为 list[0] 和 list[1] 不能在一起而失败，那么其他 18 个字符串是什么都没关系，该集合每次都会失败，那是一大堆浪费的检查。

目前我有 20 个嵌套的 for 循环，但似乎必须有更好/更快的方法来做到这一点。：

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)

第 20 个 for 循环中有打印语句来提醒我是否存在一组 20 个。for 语句至少允许我在单个加法失败时尽早跳过集合，但不记得更大的组合何时失败：

例如[list[0], list[1]]失败，所以跳到[list[0], [list[2]]哪个通过。接下来是[list[0], list[2], list[1]]哪个会失败，因为 0 和 1 再次在一起，所以它将移动到[list[0], list[2], list[3]]可能通过或不通过的哪个。我担心的是最终它也会测试：

• [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]]

所有这些组合都将具有与以前的组合相同的结果。基本上，我交换了 itertools.combinations 的恶魔测试所有我知道失败的集合组合，因为早期值失败了 for 循环的恶魔，当我不关心它们的顺序时，它们将值的顺序视为一个因素。这两种方法都显着增加了我的代码完成所需的时间。

任何关于如何摆脱恶魔的想法将不胜感激。

Answer 1

使用您当前的方法，但也要跟踪索引，以便在您的内部循环中您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)

Answer 2

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))

n这将从l中选择基数的子集test(subset) == False。

它试图避免不必要的工作。但是，鉴于从 93 个元素中选择 20 个元素有 1e20 种方法，您可能需要重新考虑您的整体方法。

Answer 3

您应该基于您的解决方案，itertools.combinations因为这将解决订购问题；短路滤波比较容易解决。

递归解决方案

让我们快速回顾一下如何实施combinations作品；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, ())

应用过滤器现在很容易：

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, ())

让我们检查一下：

>>> list(combinations_filterfalse(lambda t: sum(t) % 2 == 1, range(5), 2))
[(0, 2), (0, 4), (2, 4)]

迭代解决方案

文档itertools.combinations中列出的实际实现使用迭代循环：

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))
[()]

比较

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

Answer 4

您可以利用问题的两个方面：

1. 顺序无关紧要
2. if test_function(L)is Truethentest_function的任何子列表Lwill also beTrue

您还可以通过处理索引 0-92 而不是list[0]-来简化一些事情list[92]- 只有在test_function我们可能关心列表的内容是什么的情况下。

下面的代码首先找到可行的对，然后是四组、八组和十六组。最后，它找到了 16 和 4 的所有可行组合来获得 20 的列表。但是有超过 100,000 组 8 组，所以它仍然太慢，我放弃了。可能您可以按照相同的方式做一些事情，但可以加快速度itertools，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]

编辑：我找到了一个更好的解决方案。首先，识别范围 (0-92) 内符合 的所有值对test_function，保持对的顺序。大概第一对的第一个值必须是解决方案的第一个值，最后一对的第二个值必须是解决方案的最后一个值（但检查......这个假设是否正确test_function？如果这不是一个安全的假设，那么您将需要重复find_paths 开始和结束的所有可能值）。然后找到一条从第一个值到最后一个值的路径，该路径的长度为 20 个值，并且也符合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])