划分列表元素迭代

发布于 2025-01-24 08:43:22 字数 1015 浏览 0 评论 0原文

我想拆分列表名称元素。更重要的是，我只想用Oscar Muller将字符串分开

names = ['Oscar Muller Some other Name', 'Oscar Muller', 'Peter Pan']
expected_names = ['Oscar Muller', 'Some other Name', 'Oscar Muller', 'Peter Pan']

d = "Oscar Muller "
for line in names:
    s = [e+d for e in line.split(d) if e]

。

[list（filter（none，re.split（r'oscar \ smuller \ s'，i），i））））in name]

也没有做任何事情。

d1 = re.compile(r"Oscar\sMuller\s")
d = d1.search(names)
for line in names:
    if d:
        s = [e+d for e in line.split(d) if e]

但是它引起了输入.split（）的问题。错误：typeError：必须是str或none，而不是re.pattern。因此，我将其更改为处理每个列表元素。

d1 = re.compile(r"Oscar\sMuller\s")
d = list(filter(d1.match, names))
for line in names:
    if d:
        s = [e+d for e in line.split(d) if e]

但是它也无法正常工作，返回typeError：必须是str或none，不列表

问题：我在做什么错？

原文

I want to split a list names element. More precicely i only want to split the strings with Oscar Muller

names = ['Oscar Muller Some other Name', 'Oscar Muller', 'Peter Pan']
expected_names = ['Oscar Muller', 'Some other Name', 'Oscar Muller', 'Peter Pan']

d = "Oscar Muller "
for line in names:
    s = [e+d for e in line.split(d) if e]

That didnt do anything.

[list(filter(None, re.split(r'Oscar\sMuller\s', i))) for i in names]

didnt do anything either.

d1 = re.compile(r"Oscar\sMuller\s")
d = d1.search(names)
for line in names:
    if d:
        s = [e+d for e in line.split(d) if e]

but it caused issues with input .split(). Error: TypeError: must be str or None, not re.Pattern. So i changed it to process each list element.

d1 = re.compile(r"Oscar\sMuller\s")
d = list(filter(d1.match, names))
for line in names:
    if d:
        s = [e+d for e in line.split(d) if e]

But it didnt work either, returning TypeError: must be str or None, not list

Question: What am i doing wrong?

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花开半夏魅人心 2025-01-31 08:43:22

您还可以使用列表理解使其成为一行：

import re
[j for i in [re.split(r"(?<=Oscar Muller)", k) for k in names] for j in i if j]

You can also use list comprehension to make it one line:

import re
[j for i in [re.split(r"(?<=Oscar Muller)", k) for k in names] for j in i if j]

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薄荷梦 2025-01-31 08:43:22

从本质上讲，您需要做的是为原始列表中的每个项目生成1或2个项目的订书机，然后将列表变成单个迭代。

您可以做几种方法。您可以使用生成器函数，也可以巧妙地使用itertools

import re

def my_generator(names):
    for name in names:
        sublist = re.split(r"(?<=Oscar Muller) ", name)
        yield from sublist

names = ['Oscar Muller Some other Name', 'Oscar Muller', 'Peter Pan']
expected_names = list(my_generator(names))

，也可以使用itertools：

import itertools
import re

names = ['Oscar Muller Some other Name', 'Oscar Muller', 'Peter Pan']
expected_names = list(itertools.chain.from_iterable(re.split(r"(?<=Oscar Muller) ", s) for s in names))

Essentially, what you need to do is generate 1 or 2 item sublists for each item in the original list, and then flatten the list into a single iterable.

A couple ways you could do this. You could use a generator function, or some clever use of itertools

import re

def my_generator(names):
    for name in names:
        sublist = re.split(r"(?<=Oscar Muller) ", name)
        yield from sublist

names = ['Oscar Muller Some other Name', 'Oscar Muller', 'Peter Pan']
expected_names = list(my_generator(names))

Or you could one-liner it with itertools:

import itertools
import re

names = ['Oscar Muller Some other Name', 'Oscar Muller', 'Peter Pan']
expected_names = list(itertools.chain.from_iterable(re.split(r"(?<=Oscar Muller) ", s) for s in names))

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