如何检查列表中是否存在 integer 或字符串

Question

I want to know how I can check if an integer is in a list. Here is some code I made:

# "1" is a string. 10 is an integer.
my_list = ["1", 10]
if int in my_list:
    print("Integer in the list!")

What is wrong with this code? And how do I make it work?

Answer 1

As mentioned by Cargcigenicate, both int == "1" and int == 10 are False , so the overall check is False . What you want to do is check whether or not an element in the list is an instance of int , not whether or not they are equal to int .

This can be accomplished a bit more concisely by using the builtin any along with a generator expression :

if any(isinstance(x, int) for x in my_list):
    ...

Note that in general using isinstance should be prefered over type when reasoning about the type of an object since the former takes inheritance into account.

Answer 2

The problem with this is, int == 10 is false. int is a class, and 10 is an instance of the class. That doesn't mean however that they equate to each other.

You'd need to do some preprocessing first. There's many ways to do this, and it ultimately depends on what your end goal is, but as an example:

my_list = ["1", 10]

# Create a new list that holds the type of each element in my_list
types = [type(x) for x in my_list]

if int in types:  # And check against the types, not the instances themselves
    print("Integer in the list!")