如何在python中通过保留分隔符来拆分元素

嗨，我如何在保留“（”）的同时拆分此字符串：

我还尝试了：

x = s.split("?=(")

但它没有起作用

---

谢谢

如果您只需要保留分隔符，可以在拆分后重新引入：

def split_keep_separator(source, sep):
    elements = source.split(sep)
    return [e for t in zip(elements, [sep] * (len(elements))) for e in t][:-1]

为了测试它：

s = ' playsound3Dwhenpossible(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);'

x = split_keep_separator(s, "(")
print(x)

# [' playsound3Dwhenpossible',
#  '(',
#  'soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);']

---

如果要在输出数组中保留分隔符“（”，最简单的方法是使用正则表达式拆分

import re
s = ' playsound3Dwhenpossible(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);'
x = re.split(r'(\()', s)

---

根据您的输入字符串，您的预期输出可能是

[' playsound3Dwhenpossible', '(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);']

如果是这样，这对我来说似乎是最简单的

s = ' playsound3Dwhenpossible(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);'
print s.replace("(","*(").split('*')

首先替换，然后拆分

---

您可以使用原始字符串中而不是

'*'

的任何内容来代替

。
另一个基于正则表达式的解决方案：

import re

s = ' playsound3Dwhenpossible(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);'
splits = re.findall(r'(\(?[^(]+)', s)
print(splits)

输出：

[' playsound3Dwhenpossible', '(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);']

['(', '(asda', '(', '(asfasf']

[' playsound3Dwhenpossible(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);']

或者，如果您希望具有通用（或易于使用）功能：

split_keep = lambda s, sep : re.findall('({0}?[^{0}]+)'.format(re.escape(sep)), s)
splits = split_keep(s, '(')

在具有多个连续分隔符的情况下，此操作失败。以下稍微复杂的函数修复了此问题：

split_keep = lambda s, sep : re.findall('({0}[^{0}]*|{0}?[^{0}]+)'.format(re.escape(sep)), s)
print(split_keep('((asda((asfasf', '('))

输出：

[' playsound3Dwhenpossible', '(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);']

['(', '(asda', '(', '(asfasf']

[' playsound3Dwhenpossible(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);']

您可能会将拆分稍微更改为：

s = ' playsound3Dwhenpossible(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);'

x = s.split(r'\(') #raw
print(x)

输出：

[' playsound3Dwhenpossible', '(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);']

['(', '(asda', '(', '(asfasf']

[' playsound3Dwhenpossible(soundspotpoint18, %$videos_sounds_path%/sounds/lavazza_-_auguri_cherubini__15_.mp3, true, false, -52.644483, 0.947368, 90, 1, spotpoint18);']

在您的例子中，“保留分隔符”是什么意思？分隔符作为结果列表中的一个单独元素或与相邻元素之一相连？为什么要保留分隔符？如果使用正则表达式和组从字符串中提取信息以提取它们。