我很难处理群组分区问题。有人流了吗 请给我一些亮点？
让我简化一下我的问题。我想分十个数字（即0, 1, ..., 9）
分为三组，每组有（4, 3, 3）个数字。条件是：
组内序列无关紧要。例如，[（0, 1, 2, 3），（4, 5,
6），（7, 8, 9）]将被视为与[（3, 0, 1, 2）相同，
（5, 6, 4），（7, 8, 9）]。
我希望（1, 2, 3）始终位于同一组中，（7, 8）也是如此。
如何列出符合上述条件的所有可能的分组方案？非常感谢！
我正在使用Python 2.7。
答案 0 :(得分：3)
所以你想要分成3个大小为4,3,3的块，一个块中有（1,2,3），一个块中有（7,8）。
这意味着1,2,3和7,8不能在同一个区块中。
首先忘记键盘并分析问题
恕我直言，你应该将3个案件分开：总计：5 * 4 = 20个不同的分区
<案例2总计：5 * 4/2 = 10个不同的分区（/ 2因为你想要组合而不是排列）
<案例3总计：5个不同的分区
所以你知道你将有35个不同的分区
Python代码：
def gen():
B1 = [1,2,3]
B2 = [7,8]
C = [x for x in range(10) if x not in B1 + B2 ]
def gen1():
for x in C:
c = C[:]
b1 = B1[:]
b1.append(x)
c.remove(x)
for y in c:
c1 = c[:]
b2 = B2[:]
b2.append(y)
c1.remove(y)
yield(b1, b2, c1)
def gen2():
for i in range(len(C)-1):
for j in range(i+1, len(C)):
b2 = B2 + [C[i], C[j]]
c = [C[k] for k in range(len(C)) if k not in (i,j)]
yield (B1, b2, c)
def gen3():
for x in C:
b2 = B2[:]
c = C[:]
c.remove(x)
b2.append(x)
yield(B1, b2, c)
for g in (gen1, gen2, gen3):
for t in g():
yield t
你说得对：
>>> list(gen())
[([1, 2, 3, 0], [7, 8, 4], [5, 6, 9]), ([1, 2, 3, 0], [7, 8, 5], [4, 6, 9]),
([1, 2, 3, 0], [7, 8, 6], [4, 5, 9]), ([1, 2, 3, 0], [7, 8, 9], [4, 5, 6]),
([1, 2, 3, 4], [7, 8, 0], [5, 6, 9]), ([1, 2, 3, 4], [7, 8, 5], [0, 6, 9]),
([1, 2, 3, 4], [7, 8, 6], [0, 5, 9]), ([1, 2, 3, 4], [7, 8, 9], [0, 5, 6]),
([1, 2, 3, 5], [7, 8, 0], [4, 6, 9]), ([1, 2, 3, 5], [7, 8, 4], [0, 6, 9]),
([1, 2, 3, 5], [7, 8, 6], [0, 4, 9]), ([1, 2, 3, 5], [7, 8, 9], [0, 4, 6]),
([1, 2, 3, 6], [7, 8, 0], [4, 5, 9]), ([1, 2, 3, 6], [7, 8, 4], [0, 5, 9]),
([1, 2, 3, 6], [7, 8, 5], [0, 4, 9]), ([1, 2, 3, 6], [7, 8, 9], [0, 4, 5]),
([1, 2, 3, 9], [7, 8, 0], [4, 5, 6]), ([1, 2, 3, 9], [7, 8, 4], [0, 5, 6]),
([1, 2, 3, 9], [7, 8, 5], [0, 4, 6]), ([1, 2, 3, 9], [7, 8, 6], [0, 4, 5]),
([1, 2, 3], [7, 8, 0, 4], [5, 6, 9]), ([1, 2, 3], [7, 8, 0, 5], [4, 6, 9]),
([1, 2, 3], [7, 8, 0, 6], [4, 5, 9]), ([1, 2, 3], [7, 8, 0, 9], [4, 5, 6]),
([1, 2, 3], [7, 8, 4, 5], [0, 6, 9]), ([1, 2, 3], [7, 8, 4, 6], [0, 5, 9]),
([1, 2, 3], [7, 8, 4, 9], [0, 5, 6]), ([1, 2, 3], [7, 8, 5, 6], [0, 4, 9]),
([1, 2, 3], [7, 8, 5, 9], [0, 4, 6]), ([1, 2, 3], [7, 8, 6, 9], [0, 4, 5]),
([1, 2, 3], [7, 8, 0], [4, 5, 6, 9]), ([1, 2, 3], [7, 8, 4], [0, 5, 6, 9]),
([1, 2, 3], [7, 8, 5], [0, 4, 6, 9]), ([1, 2, 3], [7, 8, 6], [0, 4, 5, 9]),
([1, 2, 3], [7, 8, 9], [0, 4, 5, 6])]
（手动格式化以方便阅读......）
答案 1 :(得分：2)
For comb in combination k=3 in (0,4,5,6,9), remaining a, b:
(g1+a, g2+b, comb) (g1+b, g2+a, comb)
(g2+a+b, g3, g1)
For comb in combination k=4 in (0,4,5,6,9), remaining a:
(comb, g1, g2+a)
from itertools import combinations, permutations
def partition_generator():
wildcards = (0,4,5,6,9)
g1, g2 = (1,2,3), (7,8)
for comb in combinations(wildcards, 3):
unused = remaining(wildcards, comb)
for r in permutations(unused):
yield part(g1, g2, comb, r)
yield part(g2, g1, comb, unused)
for comb in combinations(wildcards, 4):
yield part(comb, g1, g2, remaining(wildcards, comb))
def remaining(a, b):
return [ x for x in a if x not in b ]
def part(x,y,z,remaining):
q = list(remaining)
while len(x) < 4:
x = x + (q.pop(0),)
if len(y) < 3:
y = y + (q.pop(0),)
if len(z) < 3:
z = z + (q.pop(0),)
return (x,y,z)
>>> for partition in partition_generator():
... print(partition)
...
((1, 2, 3, 6), (7, 8, 9), (0, 4, 5))
((1, 2, 3, 9), (7, 8, 6), (0, 4, 5))
((7, 8, 6, 9), (1, 2, 3), (0, 4, 5))
((1, 2, 3, 5), (7, 8, 9), (0, 4, 6))
((1, 2, 3, 9), (7, 8, 5), (0, 4, 6))
((7, 8, 5, 9), (1, 2, 3), (0, 4, 6))
((1, 2, 3, 5), (7, 8, 6), (0, 4, 9))
((1, 2, 3, 6), (7, 8, 5), (0, 4, 9))
((7, 8, 5, 6), (1, 2, 3), (0, 4, 9))
((1, 2, 3, 4), (7, 8, 9), (0, 5, 6))
((1, 2, 3, 9), (7, 8, 4), (0, 5, 6))
((7, 8, 4, 9), (1, 2, 3), (0, 5, 6))
((1, 2, 3, 4), (7, 8, 6), (0, 5, 9))
((1, 2, 3, 6), (7, 8, 4), (0, 5, 9))
((7, 8, 4, 6), (1, 2, 3), (0, 5, 9))
((1, 2, 3, 4), (7, 8, 5), (0, 6, 9))
((1, 2, 3, 5), (7, 8, 4), (0, 6, 9))
((7, 8, 4, 5), (1, 2, 3), (0, 6, 9))
((1, 2, 3, 0), (7, 8, 9), (4, 5, 6))
((1, 2, 3, 9), (7, 8, 0), (4, 5, 6))
((7, 8, 0, 9), (1, 2, 3), (4, 5, 6))
((1, 2, 3, 0), (7, 8, 6), (4, 5, 9))
((1, 2, 3, 6), (7, 8, 0), (4, 5, 9))
((7, 8, 0, 6), (1, 2, 3), (4, 5, 9))
((1, 2, 3, 0), (7, 8, 5), (4, 6, 9))
((1, 2, 3, 5), (7, 8, 0), (4, 6, 9))
((7, 8, 0, 5), (1, 2, 3), (4, 6, 9))
((1, 2, 3, 0), (7, 8, 4), (5, 6, 9))
((1, 2, 3, 4), (7, 8, 0), (5, 6, 9))
((7, 8, 0, 4), (1, 2, 3), (5, 6, 9))
((0, 4, 5, 6), (1, 2, 3), (7, 8, 9))
((0, 4, 5, 9), (1, 2, 3), (7, 8, 6))
((0, 4, 6, 9), (1, 2, 3), (7, 8, 5))
((0, 5, 6, 9), (1, 2, 3), (7, 8, 4))
((4, 5, 6, 9), (1, 2, 3), (7, 8, 0))
答案 2 :(得分：0)
如果我理解你的问题是正确的，那么你应该按照你的要求行事：
from copy import deepcopy
# slice into appropriate chunks
def slice_lst(size_tup, lst):
for index, length in enumerate(size_tup):
running_total = sum(size_tup[:index])
yield lst[running_total:running_total+length]
# perform integer partition
def integer_partition(num):
return {(x,) + y for x in range(1, num) for y in integer_partition(num-x)} | {(num,)}
# create all partitions given the starting list
def subsets(lst):
for partition in integer_partition(len(lst)):
yield list(slice_lst(partition, deepcopy(lst)))
# check that 1,2 and 3 are always contained within the same subset
def triplet_case(lst):
bool_arr = [1 in lst, 2 in lst, 3 in lst]
return all(bool_arr) or not any (bool_arr)
# check that 7 and 8 are always contained within the same subset
def duplet_case(lst):
bool_arr = [7 in lst, 8 in lst]
return all(bool_arr) or not any(bool_arr)
for subset in subsets([1,2,3,4,5,6,7,8,9]):
if all([triplet_case(s) for s in subset]) and all([duplet_case(s) for s in subset]):
print subset
如果您有任何问题，请随时提出后续问题！