如何使用 isinstance() 根据对象的类型将列表分成两个列表?

How to separate a list into two lists according to the objects' types using isinstance()?

列表如下:["apple", "orange", 5, "banana", 8, 9]

如何使用 isinstance() 将字符串 (str) 放入一个列表并将整数 (int) 放入另一个列表?

这边-

a = ["apple", "orange", 5, "banana", 8, 9]

b1 = [el for el in a if isinstance(el, str)]
b2 = [el for el in a if isinstance(el, int)]

使用列表理解:

lst = ["apple", "orange", 5, "banana", 8, 9]
strings = [s for s in lst if isinstance(s, str)]
integers = [n for n in lst if isinstance(n, int)]

或者,为了避免使用两个 for 循环,您也可以只遍历列表并根据需要附加到相应的列表:

strings = list()
integers = list()

for l in lst:
    if isinstance(l, str):
        strings.append(l)
    elif isinstance(l, int):
        integers.append(l)

这是使用 itertools.groupbytype 的通用解决方案。

我在这里选择 return 字典,因为它很容易通过名称获取元素,但您也可以 return 列表的列表。

from itertools import groupby

l = ["apple", "orange", 5, "banana", 8, 9]

grouper = lambda x: type(x).__name__

{k:list(g) for k,g in groupby(sorted(l, key=grouper), grouper)}

输出:

{'int': [5, 8, 9], 'str': ['apple', 'orange', 'banana']}

如列表:

ints, strings = [list(g) for k,g in groupby(sorted(l, key=grouper), grouper)]

输出:

>>> ints
[5, 8, 9]
>>> strings
['apple', 'orange', 'banana']