python实现排列和组合
- 一、模块
- 二、python实现
一、模块
python模块实现排列和组合
- 组合
from itertools import combinations
s = "123456"
print(list(combinations(s, 4)))
# [('1', '2', '3', '4'), ('1', '2', '3', '5'), ('1', '2', '3', '6'),
# ('1', '2', '4', '5'), ('1', '2', '4', '6'), ('1', '2', '5', '6'),
# ('1', '3', '4', '5'), ('1', '3', '4', '6'), ('1', '3', '5', '6'),
# ('1', '4', '5', '6'), ('2', '3', '4', '5'), ('2', '3', '4', '6'),
# ('2', '3', '5', '6'), ('2', '4', '5', '6'), ('3', '4', '5', '6')]
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combinations(iterable,r)
iterable:可迭代对象,需要组合的对象
r:需要组合的位数
2. 排列
from itertools import permutations
s = "123456"
print(list(permutations(s, 2)))
# [('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('1', '6'), ('2', '1'),
# ('2', '3'), ('2', '4'), ('2', '5'), ('2', '6'), ('3', '1'), ('3', '2'),
# ('3', '4'), ('3', '5'), ('3', '6'), ('4', '1'), ('4', '2'), ('4', '3'),
# ('4', '5'), ('4', '6'), ('5', '1'), ('5', '2'), ('5', '3'), ('5', '4'),
# ('5', '6'), ('6', '1'), ('6', '2'), ('6', '3'), ('6', '4'), ('6', '5')]
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permutations(iterable,r)
iterable:可迭代对象,需要排列的对象
r:需要排列的位数
二、python实现
来源于官网,效率不如模块
- 组合
def combinations(iterable, r):
pool = tuple(iterable)
n = len(iterable)
if r > n:
return
indices = list(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)
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- 排列
def permutations(iterable, r=None):
pool = tuple(iterable)
n = len(iterable)
r = n if r is None else r
if r > n:
return
indices = list(range(n))
cycles = list(range(n, n - r, -1))
yield tuple(pool[i] for i in indices[:r])
while n:
for i in reversed(range(r)):
cycles[i] -= 1
if cycles[i] == 0:
indices[i:] = indices[i + 1:] + indices[i:i + 1]
cycles[i] = n - i
else:
j = cycles[i]
indices[i], indices[-j] = indices[-j], indices[i]
yield tuple(pool[i] for i in indices[:r])
break
else:
return
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