7

假设我有一个包含 8 个对象的列表，编号为 1-8。

物品被放入三个盒子中，一个盒子3个，另一个盒子3个，最后一个盒子2个。根据数学，有 8C3*5C3=560 种方法可以做到这一点。我想循环浏览 560 个项目。Python有什么办法吗？

结果应如下所示：

list=['12','345',678'], ['12','346','578'], ..., etc.

请注意，['12','345','678']和['12','354',876']被认为与此目的相同。

我想为这个列表制作一个for循环。Python有什么办法吗？

这是我得到的解决方案，但看起来很难看。

import itertools
for c1,c2 in itertools.combinations(range(8),2):
            l2=list(range(8))
            l2.pop(c2)
            l2.pop(c1)
            for c3,c4,c5 in itertools.combinations(l2,3):
                l3=l2[:]
                l3.remove(c5)
                l3.remove(c4)
                l3.remove(c3)
                c6,c7,c8=l3
                print(c1,c2,c3,c4,c5,c6,c7,c8)
4

3 回答 3

2 
def F(seq, parts, indexes=None, res=[], cur=0):
    if indexes is None: # indexes to use for combinations
        indexes = range(len(seq))

    if cur >= len(parts): # base case
        yield [[seq[i] for i in g] for g in res]
        return    

    for x in combinations(indexes, r=parts[cur]):
        set_x = set(x)
        new_indexes = [i for i in indexes if i not in set_x]
        for comb in F(seq, parts, new_indexes, res=res + [x], cur=cur + 1):
            yield comb

it = F('12345678', parts=(2,3,3))
for i in range(10):
    print [''.join(g) for g in next(it)]

['12', '345', '678']
['12', '346', '578']
['12', '347', '568']
['12', '348', '567']
['12', '356', '478']
['12', '357', '468']
['12', '358', '467']
['12', '367', '458']
['12', '368', '457']
['12', '378', '456']

另一个例子：

for c in F('1234', parts=(2,2)):
    print [''.join(g) for g in c]

['12', '34']
['13', '24']
['14', '23']
['23', '14']
['24', '13']
['34', '12']
于 2013-05-02T05:58:41.753 回答
0

您可以置换所有 8 个值（如之前的答案所示）。为此，请使用此先前的答案（也在以下代码中）。

然后将每个组合分配为一个元组，这样它们就可以被散列和唯一，因为你必须对它们进行排序，所以它们也可以唯一地进行比较。

def all_perms(elements):
    if len(elements) <=1:
        yield elements
    else:
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                #nb elements[0:1] works in both string and list contexts
                yield perm[:i] + elements[0:1] + perm[i:]


v = [1,2,3,4,5,6,7,8]

a = {}
for i in all_perms(v):
    k = (tuple(sorted([i[0],i[1]])) , tuple(sorted([i[2],i[3],i[4]])) , tuple(sorted([i[5],i[6],i[7]])))

    if k not in a:
        a[k] = [str(i[0])+str(i[1]), str(i[2])+str(i[3])+str(i[4]), str(i[5])+str(i[6]) + str(i[7])]

x = 0
for i in a.values():
    print x, i
    x+=1

对于 8 个值的示例，这提供了 560 种组合。

于 2013-05-02T06:08:32.203 回答
-1

l 将是八个对象的列表，在此示例中为字符串：

l = ["O1","02","03","04","04","06","07","08"]
for group in [l[:3],l[3:6],l[6:]]: #get 3 slices of the list into 3's and a 2
    print(group)

产生：

>>> 
['O1', '02', '03']
['04', '04', '06']
['07','08']
于 2013-05-02T05:24:08.197 回答