不使用“itertools.combinations”的组合

Question

我需要的是一次为两个元素创建组合。

如果列表包含：seq = ['A', 'B', 'C'] 输出将是com = [['A', 'B'], ['A', 'C'], ['B', 'C']]

这一切都没有itertools.combinations 方法。

我使用以下代码进行排列。但是我该如何修改它以使其与组合一起使用？

def permute(seq):

    if len(seq) <= 1:
        perms = [seq]
    else:
        perms = []
        for i in range(len(seq)):
            sub = permute(seq[:i]+seq[i+1:]) 
            for p in sub:    
                perms.append(seq[i:i+1]+p)

return perms

Answer 1

如果您不想使用itertools，请使用documented pure-Python equivalent：

def combinations(iterable, r):
    # combinations('ABCD', 2) --> AB AC AD BC BD CD
    # combinations(range(4), 3) --> 012 013 023 123
    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)

Answer 2

这很简单直接做：

def comb2(s):
    for i, v1 in enumerate(s):
        for j in range(i+1, len(s)):
            yield [v1, s[j]]

然后：

print list(comb2(['A', 'B', 'C']))

显示：

[['A', 'B'], ['A', 'C'], ['B', 'C']]

唯一让组合变得棘手的事情是满足一次只在运行时已知的元素数量。只要您事先知道这个数字，嵌套那么深的固定数量的循环就可以很明显地完成这项工作。

Answer 3

def combo2(lst,n):
    if n==0:
        return [[]]
    l=[]
    for i in range(0,len(lst)):
        m=lst[i]
        remLst=lst[i+1:]
        for p in combo2(remLst,n-1):
            l.append([m]+p)
    return l

输入：

combo2(list('ABC'),2)

输出：

[['A', 'B'], ['A', 'C'], ['B', 'C']]

Answer 4

以下是answer 中递归 C++ 代码对类似问题的粗略翻译：

def combinations(sequence, length, NULL=object()):
    """ Find all combinations of the specified length from a sequence. """
    if length <= 0:
        combos = [NULL]
    else:
        combos = []
        for i, item in enumerate(sequence, 1):
            rem_items = sequence[i:]
            rem_combos = combinations(rem_items, length-1)
            combos.extend(item if combo is NULL else [item, combo]
                            for combo in rem_combos)
    return combos

print(combinations(['A', 'B', 'C'], 2))

输出：

[['A', 'B'], ['A', 'C'], ['B', 'C']]

Answer 5

纯python中的组合

# generate combination of two list
import copy
def extend_list(list1,list2):
    temp=[];temp=copy.deepcopy(list1);temp.append(list2)
    return temp
def combination(list1, list2):
    return [extend_list(item1,item2) if type(item1) is list else [item1,item2] for item1 in list1 for item2 in list2]

# generate all combination of list of variables
def list_combination(list_of_variables):
    '''list_of_variables is a list of list e.g [[1,2],['True','False']]'''
    node_combinations=[]
    no_of_variables=len(list_of_variables)
    node_combinations=list_of_variables[0]
    for i in range(no_of_variables-1):
        node_combination = combination(node_combinations,list_of_variables[i+1])
        node_combinations=node_combination
    return node_combinations

输出

input_list = [['TRUE', 'FALSE'], [1, 2], ['TRUE', 'FALSE']]
list_combination(input_list)

[['TRUE', 1, 'TRUE'],
 ['TRUE', 1, 'FALSE'],
 ['TRUE', 2, 'TRUE'],
 ['TRUE', 2, 'FALSE'],
 ['FALSE', 1, 'TRUE'],
 ['FALSE', 1, 'FALSE'],
 ['FALSE', 2, 'TRUE'],
 ['FALSE', 2, 'FALSE']]