假设我有一个像(1,2,3,4,5)这样的元组
我想把它分组成1,1,0，即(1组2个)，(1组3个)和(0组4个)
结果是(1,2), (3,4,5),(2,3), (1,4,5),(3,4), (1,2,5),(4,5), (1,2,3),(1,3), (2,4,5)等等
我如何实现这个?这是可能的itertools吗?
您可以使用组合，但是您仍然需要实现您自己的组分区逻辑:
from itertools import combinations
def group(T,g234):
if sum(g234) == 0: yield [];return # end resursion
size = next(g for g,n in enumerate(g234,2) if n>0) # first count>0
g234 = list(g234) # remaining groups
g234[size-2] -= 1
for combo in combinations(range(len(T)),size): # combine tuple indexes
part = tuple(T[i] for i in combo) # subgroup of combination
others = [v for i,v in enumerate(T) if i not in combo] # remaining
yield from ([part]+rest for rest in group(others,g234)) # assemble parts
输出:
print(*group((1,2,3,4,5),[1,1,0]),sep="n")
[(1, 2), (3, 4, 5)]
[(1, 3), (2, 4, 5)]
[(1, 4), (2, 3, 5)]
[(1, 5), (2, 3, 4)]
[(2, 3), (1, 4, 5)]
[(2, 4), (1, 3, 5)]
[(2, 5), (1, 3, 4)]
[(3, 4), (1, 2, 5)]
[(3, 5), (1, 2, 4)]
[(4, 5), (1, 2, 3)]
print(*group((1,2,3,4,5,6),(0,2,0)),sep="n")
[(1, 2, 3), (4, 5, 6)]
[(1, 2, 4), (3, 5, 6)]
[(1, 2, 5), (3, 4, 6)]
[(1, 2, 6), (3, 4, 5)]
[(1, 3, 4), (2, 5, 6)]
[(1, 3, 5), (2, 4, 6)]
...
print(*group((1,2,3,4,5,6),(3,0,0)),sep="n")
[(1, 2), (3, 4), (5, 6)]
[(1, 2), (3, 5), (4, 6)]
[(1, 2), (3, 6), (4, 5)]
[(1, 2), (4, 5), (3, 6)]
[(1, 2), (4, 6), (3, 5)]
[(1, 2), (5, 6), (3, 4)]
...
print(*group((1,2,3,4,5,6,7,8,9),(1,1,1)),sep="n")
[(1, 2), (3, 4, 5), (6, 7, 8, 9)]
[(1, 2), (3, 4, 6), (5, 7, 8, 9)]
[(1, 2), (3, 4, 7), (5, 6, 8, 9)]
[(1, 2), (3, 4, 8), (5, 6, 7, 9)]
...
[编辑]迭代版本。使用deque代替递归的调用堆栈:
from itertools import combinations
from collections import deque
def group(T,g234):
stack = deque([(T,g234,[],None)]) #tuple, groups, parts, combo_iterator
while stack:
T,g234,parts,cIter = stack.pop()
#print(len(T),g234,len(parts),bool(cIter))
if cIter is None:
if not sum(g234): yield parts;continue # partition complete
size = next(g for g,n in enumerate(g234,2) if n>0) # first count>0
g234 = list(g234) # remaining groups
g234[size-2] -= 1
cIter = combinations(range(len(T)),size) # combine tuple indexes
combo = next(cIter,None)
if combo is None: continue
stack.append((T,g234,parts,cIter))
part = tuple(T[i] for i in combo) # subgroup of combination
others = [v for i,v in enumerate(T) if i not in combo] # remaining
stack.append((others,g234,parts+[part],None)) # assemble parts
输出:
for parts in group(range(1000),(500, 0,0)):
print(parts[:3],"...",parts[-3:])
[(0, 1), (2, 3), (4, 5)] ... [(994, 995), (996, 997), (998, 999)]
[(0, 1), (2, 3), (4, 5)] ... [(994, 995), (996, 998), (997, 999)]
[(0, 1), (2, 3), (4, 5)] ... [(994, 995), (996, 999), (997, 998)]
[(0, 1), (2, 3), (4, 5)] ... [(994, 995), (997, 998), (996, 999)]
[(0, 1), (2, 3), (4, 5)] ... [(994, 995), (997, 999), (996, 998)]
[(0, 1), (2, 3), (4, 5)] ... [(994, 995), (998, 999), (996, 997)]
...
假设原始列表不包含重复数字，则
l = {1,2,3,4,5}
[(i,j) for i in itertools.combinations(l,2) for j in itertools.combinations(l.difference(i),3)]
[((1, 2), (3, 4, 5)),
((1, 3), (2, 4, 5)),
((1, 4), (2, 3, 5)),
((1, 5), (2, 3, 4)),
((2, 3), (1, 4, 5)),
((2, 4), (1, 3, 5)),
((2, 5), (1, 3, 4)),
((3, 4), (1, 2, 5)),
((3, 5), (1, 2, 4)),
((4, 5), (1, 2, 3))]