python递增代码怎么写_在Python中递增字符串

可以简化环路很多：

def next_code(code):

code = list(code)

for i, let in enumerate(code):

if let != 'z':

code[i] = chr(ord(let) + 1)

break

code[i] = 'a'

return ''.join(code)

如果目标是刚刚产生的所有值一个接一个，从'aaa'，itertools.product可用于制作发电机：

from future_builtins import map # Only on Python 2

from itertools import product

def allcodes():

# You want the left side to vary faster, so reverse before joining

return map(''.join, map(reversed, product(string.ascii_lowercase, repeat=3)))

for code in allcodes():

print(code)

或者您可以根据需要调用该函数以获取序列中的下一个代码，而不必将其用作迭代器：

nextcode = allcodes().__next__ # .next on Py2

如果发电机应为无穷大(所以从zzz换到aaa)，只是改变allcodes要么：

# Avoid cycle if storing all 26**3 codes in memory is a bad idea

def allcodes():

while True:

yield from map(''.join, map(reversed, product(string.ascii_lowercase, repeat=3)))

# On Py2, change yield from line to:

# for code in map(''.join, map(reversed, product(string.ascii_lowercase, repeat=3))): yield code

或更高的存储成本，而且更简单：

from itertools import cycle

def allcodes():

return cycle(map(''.join, map(reversed, product(string.ascii_lowercase, repeat=3))))