def permutations(iterable, r=None):
# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
# permutations(range(3)) --> 012 021 102 120 201 210
pool = tuple(iterable)
n = len(pool)
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
got = permutations(getAllTheLetters(),4)
cnt = 0
for i in got:
cnt += 1
print ''.join(i)
print cnt
上面没有给出“zzzz”或“zzz”
我需要如下所示的内容:
A B C D..
aa、ab、ac、..
啊啊啊啊……
但是 do_perm() 被硬编码为循环四次,这是我不想做的。
def getAllTheLetters(begin='a', end='z'):
beginNum = ord(begin)
endNum = ord(end)
yield ''
for number in xrange(beginNum, endNum+1):
yield chr(number)
def do_perm(l):
s = set()
for a in getAllTheLetters():
for b in getAllTheLetters():
for c in getAllTheLetters():
for d in getAllTheLetters():
to_add = "%s%s%s%s" % (a,b,c,d)
if to_add != "":
s.add(to_add)
return s
got = do_perm(1)
cnt = 0
for i in sorted(got):
cnt +=1
print i
print cnt
最佳答案
您可以简单地使用 itertools.product
,像这样
from itertools import product
def get_strings(letters, max_length):
for i in range(1, max_length + 1):
for value in product(letters, repeat=i):
yield "".join(value)
当你像这样调用它时
print(list(get_strings("ab", 2)))
你会得到
['a', 'b', 'aa', 'ab', 'ba', 'bb']
如果你想获取从a
到z
的所有值,你可以调用get_strings
,就像这样
from string import ascii_lowercase
print(list(get_strings(ascii_lowercase, 4)))
注意:这将创建大量字符串,因此您的计算机可能会停止响应。如果您只想遍历字符串,请使用 for
循环和 get_strings
如下所示,并且不要创建列表。
for current_string in get_strings(ascii_lowercase, 4):
# Process the current_string
关于python:获取所有可能的字母表达到给定的长度,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28666239/