不使用 itertools 的所有组合？

Question

我想知道如何使用 Python 中的 itertools 编写一个算法，为我提供数字列表有重复和没有的所有可能组合。

例如：列表[1,2]的所有可能组合：

[[1,1],[1,2],[2,2],[2,1]]

列表[1,2,3] 的所有可能组合。然后，该算法将在一个列表中为我提供 27 (3³) 个不同的列表。

这是我的代码：

def all_possibilities(list):
    result = [list]

    for i in list:
        new_list1 = list[:]

        for k in range(len(list)):
            new_list2 = list[:]
            new_list1[k] = i
            new_list2[k] = i
            if new_list1 != list:
                result.append(new_list1)
            if new_list2 != list:
                result.append(new_list2)

    for u in result:
        for y in result:
            if u == y and (u is y) == False:
                result.remove(y)

    return (len(result),result)

Answer 1

您可以编写自己的product 方法来找出多个列表的cartesian product。

def cartesian_product(*lists):
  result = [[]]
  for list in lists:
    result = [x + [y] for x in result for y in list]
  return result  
l = [1, 2, 3]
lst = [l for i in l]
x = cartesian_product(*lst)
for tuple in x:
   print(tuple)

输出

(1, 1)
(1, 2)
(2, 1)
(2, 2)   

Answer 2

这实际上是用替换进行组合。您可以调整组合的大小。我让它返回一个元组，因为元组是不可变的，并且元组操作比列表操作快。但是，如果需要，您可以轻松地使其返回列表列表。干杯

def seq_slicer(container, step=10):
    """(container)->container
    Returns slices of container in chunks of step.
    Great selecting chunks of a sequence in predetrmined slices efficiently.

    In the event where step is greater than the length of the sequence to be sliced,
    the slice taken will be the length of the sequence.

    >>> [i for i in seq_slicer(range(40), 10)]
    [range(0, 10), range(10, 20), range(20, 30), range(30, 40)]

    >>> [i for i in seq_slicer(range(41), 10)]
    [range(0, 10), range(10, 20), range(20, 30), range(30, 40), range(40, 41)]

    >>> [c for c in seq_slicer("abcdefghijklm", 3)]
    ['abc', 'def', 'ghi', 'jkl', 'm']

    >>> [c for c in seq_slicer(list("abcdefghijklm"), 3)]
    [['a', 'b', 'c'], ['d', 'e', 'f'], ['g', 'h', 'i'], ['j', 'k', 'l'], ['m']]

    Author: Xero
    License: Apache V2
    """
    i = 0
    span = len(container)
    while i < span:
        yield container[i:i+step]
        i += step

def combinations_with_replacement(seq, num):
    """(sequence type, integer)->list
    return every possible combination of num elements from seq in lexicographic order
    >>> combinations_with_replacement([1, 2], 2)
    ((1, 1), (1, 2), (2, 1), (2, 2))


    Author: Xero
    License: Apache V2
    """
    lst = []
    _seq = tuple(seq)
    slices = [c for c in seq_slicer(_seq, num-1)]
    for elem in seq:
        for combo in [(elem,) + body for body in slices]:
            lst.append(combo)
    return tuple(lst)