将字符串转换为长度为1的列表 [英] Convert a string to a list of length one

问题描述

我创建了一个需要列表才能正常工作的方法.但是，您可以发送列表或简单字符串.我想将该字符串转换为包含整个字符串作为元素的列表.例如，如果我有:

I created a method that requires a list in order to work properly. However, you can send in a list OR a simple string. I want to turn that string into a list that contains that entire string as an element. For example, if I have:

"I am a string"

我想将其转换为:

["I am a string"]

我可以这样做:

"I am a string".split("!@#$%^&*")

因为我永远不会有符号的组合，所以它将始终将其转换为列表而不会删​​除任何字符.但是，这似乎并不是一种很好的方法.还有另一种方法吗?

Because I will never have that combination of symbols, it will always convert it to a list without removing any characters. However, this doesn't seem like that great of a way to do it. Is there another way?

推荐答案

>>> "abc"
'abc'
>>> ["abc"]
['abc']
>>> abc = "abc"
>>> abc
'abc'
>>> [abc]
['abc']
>>> "I am a string".split("!@#$%^&*") == ["I am a string"]
True

将值放在方括号中可以使列表包含一个项目，就像多个值可以使列表中包含多个项目一样.唯一不遵循此模式的容器是元组，因为圆括号也用于分组.在这种情况下，只需在单个项目后添加一个逗号:

Putting the value in square brackets makes a list with one item, just as multiple values makes a list with multiple items. The only container which does not follow this pattern is the tuple, as the round brackets are also used for grouping. In that case, just add a comma after the single item:

>>> abc
'abc'
>>> (abc)
'abc'
>>> (abc,)
('abc',)

如果您希望函数在幕后处理列表和字符串的方式有所不同，请对函数进行编码，例如:

If you want your function to handle list and strings differently under the cover, code your function like:

def f(maybe_list):
    if not isinstance(maybe_list, list):
        maybe_list = [maybe_list]
    # carry on, you have a list.

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