Filter list's elements by type of each element

Question

I have list with different types of data (string, int, etc.). I need to create a new list with, for example, only int elements, and another list with only string elements. H

Answer 1

Sort the list by type, and then use groupby to group it:

>>> import itertools
>>> l = ['a', 1, 2, 'b', 'e', 9.2, 'l']
>>> l.sort(key=lambda x: str(type(x)))
>>> lists = [list(v) for k,v in itertools.groupby(l, lambda x: str(type(x)))]
>>> lists
[[9.2], [1, 2], ['a', 'b', 'e', 'l']]

Answer 2

You can accomplish this with list comprehension:

integers = [elm for elm in data if isinstance(elm, int)]

Where data is the data. What the above does is create a new list, populate it with elements of data (elm) that meet the condition after the if, which is checking if element is instance of an int. You can also use filter:

integers = list(filter(lambda elm: isinstance(elm, int), data))

The above will filter out elements based on the passed lambda, which filters out all non-integers. You can then apply it to the strings too, using isinstance(elm, str) to check if instance of string.