将迭代器转换为列表的最快方法

Question

拥有iterator对象，是否有比列表理解更快、更好或更正确的方法来获取迭代器返回的对象列表？

user_list = [user for user in user_iterator]

Answer 1

list(your_iterator)

Answer 2

since python 3.5 you can use * iterable unpacking operator:

user_list = [*your_iterator]

but the pythonic way to do it is:

user_list  = list(your_iterator)

Answer 3

@Robino was suggesting to add some tests which make sense, so here is a simple benchmark between 3 possible ways (maybe the most used ones) to convert an iterator to a list:

1. by type constructor

list(my_iterator)

1. by unpacking

[*my_iterator]

3. using list comprehension

[e for e in my_iterator]

I have been using simple_bechmark library

from simple_benchmark import BenchmarkBuilder
from heapq import nsmallest

b = BenchmarkBuilder()

@b.add_function()
def convert_by_type_constructor(size):
    list(iter(range(size)))

@b.add_function()
def convert_by_list_comprehension(size):
    [e for e in iter(range(size))]

@b.add_function()
def convert_by_unpacking(size):
    [*iter(range(size))]


@b.add_arguments('Convert an iterator to a list')
def argument_provider():
    for exp in range(2, 22):
        size = 2**exp
        yield size, size

r = b.run()
r.plot()

As you can see there is very hard to make a difference between conversion by the constructor and conversion by unpacking, conversion by list comprehension is the “slowest” approach.

---

I have been testing also across different Python versions (3.6, 3.7, 3.8, 3.9) by using the following simple script:

import argparse
import timeit

parser = argparse.ArgumentParser(
    description='Test convert iterator to list')
parser.add_argument(
    '--size', help='The number of elements from iterator')

args = parser.parse_args()

size = int(args.size)
repeat_number = 10000

# do not wait too much if the size is too big
if size > 10000:
    repeat_number = 100


def test_convert_by_type_constructor():
    list(iter(range(size)))


def test_convert_by_list_comprehension():
    [e for e in iter(range(size))]


def test_convert_by_unpacking():
    [*iter(range(size))]


def get_avg_time_in_ms(func):
    avg_time = timeit.timeit(func, number=repeat_number) * 1000 / repeat_number
    return round(avg_time, 6)


funcs = [test_convert_by_type_constructor,
         test_convert_by_unpacking, test_convert_by_list_comprehension]

print(*map(get_avg_time_in_ms, funcs))

The script will be executed via a subprocess from a Jupyter Notebook (or a script), the size parameter will be passed through command-line arguments and the script results will be taken from standard output.

from subprocess import PIPE, run

import pandas

simple_data = {'constructor': [], 'unpacking': [], 'comprehension': [],
        'size': [], 'python version': []}


size_test = 100, 1000, 10_000, 100_000, 1_000_000
for version in ['3.6', '3.7', '3.8', '3.9']:
    print('test for python', version)
    for size in size_test:
        command = [f'python{version}', 'perf_test_convert_iterator.py', f'--size={size}']
        result = run(command, stdout=PIPE, stderr=PIPE, universal_newlines=True)
        constructor, unpacking,  comprehension = result.stdout.split()
        
        simple_data['constructor'].append(float(constructor))
        simple_data['unpacking'].append(float(unpacking))
        simple_data['comprehension'].append(float(comprehension))
        simple_data['python version'].append(version)
        simple_data['size'].append(size)

df_ = pandas.DataFrame(simple_data)
df_

You can get my full notebook from here .

In most of the cases, in my tests, unpacking shows to be faster, but the difference is so small that the results may change from a run to the other. Again, the comprehension approach is the slowest, in fact, the other 2 methods are up to ~ 60% faster.