如何生成列表的所有排列?例如:
permutations([])
[]
permutations([1])
[1]
permutations([1, 2])
[1, 2]
[2, 1]
permutations([1, 2, 3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
如何生成列表的所有排列?例如:
permutations([])
[]
permutations([1])
[1]
permutations([1, 2])
[1, 2]
[2, 1]
permutations([1, 2, 3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
当前回答
def pzip(c, seq):
result = []
for item in seq:
for i in range(len(item)+1):
result.append(item[i:]+c+item[:i])
return result
def perm(line):
seq = [c for c in line]
if len(seq) <=1 :
return seq
else:
return pzip(seq[0], perm(seq[1:]))
其他回答
对于性能,一个由Knuth启发的numpy解决方案(第22页):
from numpy import empty, uint8
from math import factorial
def perms(n):
f = 1
p = empty((2*n-1, factorial(n)), uint8)
for i in range(n):
p[i, :f] = i
p[i+1:2*i+1, :f] = p[:i, :f] # constitution de blocs
for j in range(i):
p[:i+1, f*(j+1):f*(j+2)] = p[j+1:j+i+2, :f] # copie de blocs
f = f*(i+1)
return p[:n, :]
复制大量内存可节省时间-它比列表(itertools.permutations(range(n))快20倍:
In [1]: %timeit -n10 list(permutations(range(10)))
10 loops, best of 3: 815 ms per loop
In [2]: %timeit -n100 perms(10)
100 loops, best of 3: 40 ms per loop
另一种方法(无libs)
def permutation(input):
if len(input) == 1:
return input if isinstance(input, list) else [input]
result = []
for i in range(len(input)):
first = input[i]
rest = input[:i] + input[i + 1:]
rest_permutation = permutation(rest)
for p in rest_permutation:
result.append(first + p)
return result
输入可以是字符串或列表
print(permutation('abcd'))
print(permutation(['a', 'b', 'c', 'd']))
功能性风格
def addperm(x,l):
return [ l[0:i] + [x] + l[i:] for i in range(len(l)+1) ]
def perm(l):
if len(l) == 0:
return [[]]
return [x for y in perm(l[1:]) for x in addperm(l[0],y) ]
print perm([ i for i in range(3)])
结果:
[[0, 1, 2], [1, 0, 2], [1, 2, 0], [0, 2, 1], [2, 0, 1], [2, 1, 0]]
这里有一个算法,它在不创建新的中间列表的情况下处理列表,类似于Ber在https://stackoverflow.com/a/108651/184528.
def permute(xs, low=0):
if low + 1 >= len(xs):
yield xs
else:
for p in permute(xs, low + 1):
yield p
for i in range(low + 1, len(xs)):
xs[low], xs[i] = xs[i], xs[low]
for p in permute(xs, low + 1):
yield p
xs[low], xs[i] = xs[i], xs[low]
for p in permute([1, 2, 3, 4]):
print p
您可以在这里亲自尝试代码:http://repl.it/J9v
def permutate(l):
for i, x in enumerate(l):
for y in l[i + 1:]:
yield x, y
if __name__ == '__main__':
print(list(permutate(list('abcd'))))
print(list(permutate([1, 2, 3, 4])))
#[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
#[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]