Call int() function on every list element?

Question:

I have a list with numeric strings, like so:

numbers = ['1', '5', '10', '8'];

I would like to convert every list element to integer, so it would look like this:

numbers = [1, 5, 10, 8];

I could do it using a loop, like so:

new_numbers = [];
for n in numbers:
    new_numbers.append(int(n));
numbers = new_numbers;

Does it have to be so ugly? I’m sure there is a more pythonic way to do this in a one line of code. Please help me out.

Asked By: Silver Light

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Source

Answers:

This is what list comprehensions are for:

numbers = [ int(x) for x in numbers ]
Answered By: adamk

In Python 2.x another approach is to use map:

numbers = map(int, numbers)

Note: in Python 3.x map returns a map object which you can convert to a list if you want:

numbers = list(map(int, numbers))
Answered By: Mark Byers

Another way,

for i, v in enumerate(numbers): numbers[i] = int(v)
Answered By: Nick Dandoulakis

If you are intending on passing those integers to a function or method, consider this example:

sum(int(x) for x in numbers)

This construction is intentionally remarkably similar to list comprehensions mentioned by adamk. Without the square brackets, it’s called a generator expression, and is a very memory-efficient way of passing a list of arguments to a method. A good discussion is available here: Generator Expressions vs. List Comprehension

Answered By: Tim McNamara

just a point,

numbers = [int(x) for x in numbers]

the list comprehension is more natural, while

numbers = map(int, numbers)

is faster.

Probably this will not matter in most cases

Useful read: LP vs map

Answered By: renatopp

Another way to make it in Python 3:

numbers = [*map(int, numbers)]

However ideally you may be happy just with map, as it is returning an iterator:

numbers = map(int, numbers)

Answered By: zhukovgreen

Thought I’d consolidate the answers and show some timeit results.

Python 2 sucks pretty bad at this, but map is a bit faster than comprehension.

Python 2.7.13 (v2.7.13:a06454b1afa1, Dec 17 2016, 20:42:59) [MSC v.1500 32 bit (Intel)] on win32
Type "copyright", "credits" or "license()" for more information.
>>> import timeit
>>> setup = """import random
random.seed(10)
l = [str(random.randint(0, 99)) for i in range(100)]"""
>>> timeit.timeit('[int(v) for v in l]', setup)
116.25092001434314
>>> timeit.timeit('map(int, l)', setup)
106.66044823117454

Python 3 is over 4x faster by itself, but converting the map generator object to a list is still faster than comprehension, and creating the list by unpacking the map generator (thanks Artem!) is slightly faster still.

Python 3.6.1 (v3.6.1:69c0db5, Mar 21 2017, 17:54:52) [MSC v.1900 32 bit (Intel)] on win32
Type "copyright", "credits" or "license()" for more information.
>>> import timeit
>>> setup = """import random
random.seed(10)
l = [str(random.randint(0, 99)) for i in range(100)]"""
>>> timeit.timeit('[int(v) for v in l]', setup)
25.133059591551955
>>> timeit.timeit('list(map(int, l))', setup)
19.705547827217515
>>> timeit.timeit('[*map(int, l)]', setup)
19.45838406513076

Note: In Python 3, 4 elements seems to be the crossover point (3 in Python 2) where comprehension is slightly faster, though unpacking the generator is still faster than either for lists with more than 1 element.

Answered By: Jim

It may also be worth noting that NumPy will do this on the fly when creating an array:

import numpy as np

numbers = ['1', '5', '10', '8']
numbers = np.array(numbers,
                   dtype=int)
numbers

array([ 1,  5, 10,  8])
Answered By: Arne