将字符串(不带任何分隔符)转换为列表 [英] Convert string (without any separator) to list

问题描述

我有一个电话号码(字符串)，例如我想将其变成"+ 123-456-7890"，该列表看起来像:[+，1，2，3，-，....，0].

I have a phone number(string), e.g. "[phone]", that I want to turn into a list that looks like: [+, 1, 2, 3, -, ...., 0].

为什么?因此，我可以遍历该列表并删除所有符号，因此只剩下一个数字列表，然后可以将其转换回字符串.

Why? So I can go iterate through the list and remove all the symbols, so I'm left with a list of only digits, which I can then convert back to a string.

解决此问题的最佳方法是什么?我遇到的所有解决方案均不适用，因为在数字之间没有任何特殊字符(因此我无法在其中分割字符串).

What's the best way to solve this problem? None of the solutions I've come across are applicable, because I don't have any special characters in-between the digits (so I can't split the string there.)

有什么想法吗?我真的很感激！

Any ideas? I really appreciate it!

编辑-这是我尝试过的:

Edit - this is what I've tried:

x = row.translate(None, string.digits)
list = x.split()

也:

filter(lambda x: x isdigit())

推荐答案

您的意思是您想要类似的东西:

You mean that you want something like:

''.join(n for n in phone_str if n.isdigit())

这利用了字符串可迭代的事实.当您遍历它们时，它们每次产生1个字符.

This uses the fact that strings are iterable. They yield 1 character at a time when you iterate over them.

关于您的努力

该字符实际上从字符串中删除了所有数字，只剩下非数字.

This one actually removes all of the digits from the string leaving you with only non-digits.

x = row.translate(None, string.digits)

这是在空格处而不是在每个字符之后分割字符串:

This one splits the string on runs of whitespace, not after each character:

list = x.split()

这篇关于将字符串(不带任何分隔符)转换为列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！