How do I check if elements are integer by using type() in Python?

Question

Just like the title but how do I check if a certain element is an integer and square those in list while deleting elements that aren't integer?

For example, the list is [0, 2, 'Python', 'C++', 3]

And this is what I tried:

def main():
    lst = [0, 2, 'Python', 'C++', 3]

    print("Given list: ", lst)
    newLst = [item**2 for item in lst if type(lst[item]) == int]

    print("The list which only contains square of integers: ", newLst)

main()

And error happens;

TypeError: list indices must be integers or slices, not str

What am I missing?

Answer 1

The problem in your code is that you are using the element item in order to slice a list - but indices must be integers. Therefore, you should change

newLst = [item**2 for item in lst if type(lst[item]) == int]

to

newLst = [item**2 for item in lst if type(item) == int]

---

Alternatively, you can use isinstance() in order to test if the element is of type int :

lst = [0, 2, 'Python', 'C++', 3]
lst_num = [ x**2 for x in lst if isinstance(x, int)]

and the output will be:

print(lst_num)
[0, 4, 9]

Answer 2

item is the list element, not its index. You should use type(item) , not type(lst[item])

newLst = [item**2 for item in lst if type(item) == int]

It's also preferable to use isinstance() because it works with subclasses.

newLst = [item**2 for item in lst if isinstance(item, int)]

Answer 3

you can use isinstance(item, int)

Answer 4

The canonical way to do a typecheck in Python is isinstance(val,int) instead of type(val) == int as was answered in this previous post What's the canonical way to check for type in Python?