Faster for loop with only if in python

I’m dealing with a big dataset and want to basically this:

test = np.random.rand(int(1e7))-0.5
def test0(test):
    return [0 if c<0 else c for c in test]

which is doing this:

def test1(test):
    for i,dat in enumerate(test):
        if dat<0: 
            test[i] = 0
        else:
            test[i] = dat
    return test

Is there a way to modify test0 to skip the else request so i works like this:

def test1(test):
    for i,dat in enumerate(test):
        if dat<0: test[i] = 0
    return test

Thanks in advance!

>Solution :

just do which seems to be fastest option:

test[test < 0] = 0

test:

import numpy as np
import timeit
from copy import copy
from functools import partial


def create_data():
    return np.random.rand(int(1e7))-0.5


def func1(data):
    data[data < 0] = 0


def func2(data):
    np.putmask(data, data < 0, 0)


def func3(data):
    np.maximum(data, 0)


if __name__ == '__main__':
    n_loops = 1000
    test = create_data()

    t1 = timeit.Timer(partial(func1, copy(test)))
    t2 = timeit.Timer(partial(func2, copy(test)))
    t3 = timeit.Timer(partial(func3, copy(test)))

    print(f"func1 timeit {t1.timeit(n_loops)} num test loops {n_loops}")
    print(f"func2 timeit {t2.timeit(n_loops)} num test loops {n_loops}")
    print(f"func3 timeit {t3.timeit(n_loops)} num test loops {n_loops}")

test results:

execution1:

func1 timeit 6.547473503 num test loops 1000
func2 timeit 12.467706045000002 num test loops 1000
func3 timeit 22.388469115 num test loops 1000

execution2:

func1 timeit 6.785610083000002 num test loops 1000
func2 timeit 12.850003099999999 num test loops 1000
func3 timeit 24.912014532 num test loops 1000