Question

我正在编写一个函数来增加一个3个字母（a-z）的字符串。例如：
输入： aaa
输出：baa

输入： zba
输出：aca

所以订单如下

aaa
baa
...
zaa
aba
bba
cba
...
zba
aca
bca
cca
...
zca
ada
...
zzz
aaa

我编写了以下函数next_code()并且它有效，但我想知道是否有更优雅的方法来实现它而不是循环遍历字符串中的单个字母：

# 0 = a; 25 = z
def digit_to_char(digit):
    return chr(ord('a') + digit)

# a = 0; z = 25
def char_to_digit(char):
    return ord(char)-ord('a')

def next_code(code):
    # if used up all codes, loop from start
    if code == 'zzz':
        return next_code('aaa')
    else:
        code = list(code)
        # loop over letters and see which one we can increment
        for (i, letter) in enumerate(code):
            if letter == 'z':
                # go on to the next letter
                code[i] = 'a'
                continue
            else:
                # increment letter
                code[i] = digit_to_char(char_to_digit(letter) + 1)
                return ("".join(code))
                break



print (next_code('aab'))

Answer 1

只使用itertools产品

>>> import itertools
>>> from string import ascii_lowercase
>>> strings = itertools.product(*[ascii_lowercase]*3)
>>> "".join(next(strings,"No More Combos..."))
'aaa'
>>> "".join(next(strings,"No More Combos..."))
'aab'
>>> "".join(next(strings,"No More Combos..."))
'aac'
...

我可能会这样做

如果你想回到'aaa＆＃39;在结束之后你可以使用itertools.cycle

strings = itertools.cycle(itertools.product(*[ascii_lowercase]*3))

Answer 2

你可以大量简化循环：

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)

如果目标是逐个生成所有值，从'aaa'开始，itertools.product可用于生成生成器：

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)

或者你可以根据需要调用它来获取序列中的下一个代码而不将其用作迭代器：

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

如果生成器应该是无限的（因此它从zzz包裹到aaa），只需将allcodes更改为：

# 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

或更高的内存成本，但更简单：

from itertools import cycle

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

增加Python中的字符串

2 个答案: