What is the best way to reset (re-initialize) a list of integers (to 0) in Python

Question

Given a list of integers ( list_nb ) of fixed size ( n ), what is the best way to reset (re-initialize) all the values to 0.

I already tried two methods, 1) using a loop to reset all elements one by one, 2) a new list with the same size (where all elements are at 0).

The second method clearly gives better results in terms of execution time.

import time

n = 200_000_000
list_nb = [0]*n

# some lines of code that change the list_nb ....

# method 1 to reset
start_time = time.time()
for i in range(len(list_nb)):
    list_nb[i] = 0
end_time = time.time()
print("Method 1 time = ",end_time-start_time," s")

# method 2 to reset
start_time = time.time()
list_nb = [0]*n
end_time = time.time()
print("Method 2 time = ",end_time-start_time," s")

Result:

• Method 1 time = 16.05620765686035 s
• Method 2 time = 1.279524564743042 s

I don't know if there is any other way to do it.

Thanks.

Answer 1

Method 2 is clearly superior on speed, and it's easy to understand for anyone familiar with Python sequence multiplication. A warning: It's not resetting the contents of the list ; it's making a whole new list and rebinding the name to that new list . So if list_nb was an argument to a function, the caller's copy of the list would not be changed. If that's what you want, great, but if you want the contents to be changed in place, change it to:

list_nb[:] = [0]*n  # Or itertools.repeat(0, n), though CPython doesn't speed it up

which uses slice assignment to replace the contents of the existing list in-place. It takes about twice as long (it builds a new list , copies from it, then throws it away) as list_nb = [0]*n , but if you need in-place mutation, it's worth it.

Answer 2

I think the best way to do it is overwriting the list as you did in method #2. It's easier to overwrite the existing list than replacing every element in the list with 0. I believe there is no faster way to do it, but I might be wrong. Nice job !

Answer 3

Using numpy is pretty quick.

import numpy as np
import time

n= 200000000

start_time = time.time()
list_nb = np.zeros(n)
end_time = time.time()
print(end_time-start_time)

This takes me 0.003000497817993164

Answer 4

If you are ready to use numpy:

start_time = time.time()
a = numpy.zeros(n, dtype=object) 
end_time = time.time()
print("Method 0 time = ",end_time-start_time," s")

gives

('Method 0 time = ', 0.8265918731689453, ' s')