示例

有一个看起来像这样的python字符串：

[1]输入：

'they <are,are not> sleeping'

它必须变成一个列表，看起来像：

[1]输出：

['they are sleeping', 'they are not sleeping']

另一个例子

[2]输入：

'hello <stupid,smart> <people,persons,animals>'

[2]输出：

['hello stupid people', 'hello stupid persons', 'hello stupid animals', 'hello smart people', 'hello smart persons', 'hello smart animals']

标签中存在替代选项，并且需要考虑所有可能情况来生成新字符串。

Answer 1

尝试一下：

import itertools

def all_choices(text):
    """
    >>> list(all_choices('they <are,are not> sleeping'))
    ['they are sleeping', 'they are not sleeping']
    >>> list(all_choices('hello <stupid,smart> <people,persons,animals>'))
    ['hello stupid people', 'hello stupid persons', 'hello stupid animals', 'hello smart people', 'hello smart persons', 'hello smart animals']
    """
    tokens = (block2 for block1 in text.split('<')
                     for block2 in block1.split('>'))

    decisions = []
    literals = []

    try:
        while True:
            literal = next(tokens)
            literals.append(literal)
            options = next(tokens).split(',')
            decisions.append(options)
    except StopIteration:
        pass

    decisions.append(('',))

    for choices in itertools.product(*decisions):
        yield ''.join(x for pair in zip(literals, choices)
                        for x in pair)

Answer 2

我借用了丹尼斯（Dennis）的doctest，但这是一个使用re.split()的表述：）

import itertools
import re


def all_choices(text):
    """
    >>> list(all_choices('they <are,are not> sleeping'))
    ['they are sleeping', 'they are not sleeping']
    >>> list(all_choices('hello <stupid,smart> <people,persons,animals>'))
    ['hello stupid people', 'hello stupid persons', 'hello stupid animals', 'hello smart people', 'hello smart persons', 'hello smart animals']
    """

    # Split the text into choice and non-choice bits
    # Wrapping the pattern in parentheses makes re.split also return the literal parts.
    # Since there is only 1 group, the odd indices in bits will be the choice bits,
    bits = re.split("(<.+?>)", text)
    # ... but we can just as well peek into the bits themselves to find that out:
    # This splits each bit into an iterable; an 1-tuple for literal parts, a list for choices.
    bits = [
        bit.strip("<>").split(",")
        if bit.startswith("<") and bit.endswith(">")
        else (bit,)
        for bit in bits
    ]
    # Itertools.product generates all combinations...
    for choices in itertools.product(*bits):
        # ... which we can join into a string and yield.
        yield "".join(choices)

Answer 3

您可以使用正则表达式：

import re

s = 'hello <stupid,smart> <people,persons,animals>'

def all_choices(my_string):
    str_list = [my_string]
    for _ in range(len(re.findall(r'<(.*?)>', my_string))):
        tmp_list = []
        for x in str_list:
            mask = re.findall(r'<(.*?)>', x)[0]
            for i in re.findall(mask, x)[0].split(','):
                tmp_list.append(re.sub(f'<({mask})>', i, x))
        str_list = tmp_list
    return str_list

print(all_choices(s))
# ['hello stupid people', 'hello stupid persons', 'hello stupid animals', 'hello smart people', 'hello smart persons', 'hello smart animals']

python字符串怎么分解

示例

另一个例子

3 个答案: