Do if statements take significant amounts of time in python?

Question

I am creating a graphing program which has to iterate values through a calculation 10000-1000000 times, and then append part of that output to a list. In order to change which list it is appended to, there are ~3 if statements inside that loop. While it would logically be faster to use the if statement first, is there a significant amount of time saved?

As an example:

output = []
append_to = "pol"
for i in range(10000):
    if append_to == "pol":
        output.append(np.cos(i))
    else:
        output.append(np.sin(i))

Would this be significantly slower than:

output = []
append_to = "pol"
if append_to == "pol":
    for i in range(10000):
        output.append(np.cos(i))
else:
    for i in range(10000):
        output.append(np.sin(i))

Regardless of which one is faster, keep in mind they are not entirely equivalent (assuming the value of append_to may change, otherwise the whole question has no point) — DeepSpace
– DeepSpace, Commented Nov 7, 2017 at 9:31
Does append_to stay the same during the entire loop? In that case, you could define f = np.cos if append_to == "pol" else np.sin and then do output = list(map(f, range(10000)) — tobias_k
– tobias_k, Commented Nov 7, 2017 at 9:33
Yes, append_to stays the same over the entire loop, why does this invalidate the question? I am asking about optimization. Thank you for your suggestion, i will have to look into the map function. — Evan Quiney
– Evan Quiney, Commented Nov 8, 2017 at 9:16

Jose A. García · Accepted Answer · 2017-11-07 10:34:18Z

2

Why dont just try?

import numpy as np
import timeit

def one():
    output = []
    append_to = "pol"
    for i in range(10000):
        if append_to == "pol":
            output.append(np.cos(i))
        else:
            output.append(np.sin(i))

def two():
    output = []
    append_to = "pol"
    if append_to == "pol":
        for i in range(10000):
            output.append(np.cos(i))
    else:
        for i in range(10000):
            output.append(np.sin(i))

print(timeit.timeit('f()', 'from __main__ import one as f', number=1000))
print(timeit.timeit('f()', 'from __main__ import two as f', number=1000))

Output:
9.042721510999854
8.626055914000062

So yes, it is faster, as expected. And just for you to know the lookup also takes a bit of time, so if you do ap = output.append and then call ap instead of output.append you get a marginal improvement.

edited Nov 7, 2017 at 10:34

answered Nov 7, 2017 at 9:27

Jose A. García

8885 silver badges13 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Polymer · Accepted Answer · 2017-11-07 09:28:13Z

2

Give it a try!

import math, time

time_start = time.time()

output = []
append_to = "pol"
for i in range(10000000):
    if append_to == "pol":
        output.append(math.cos(i))
    else:
        output.append(math.sin(i))

print("End: " + str(time.time() - time_start))

For that run, I got 4.278s. For this run:

import math, time

time_start = time.time()

output = []
append_to = "pol"
if append_to == "pol":
    for i in range(10000000):
        output.append(math.cos(i))
else:
    for i in range(10000000):
        output.append(math.sin(i))

print("End: " + str(time.time() - time_start))

I got 3.751s.

So there you go!

answered Nov 7, 2017 at 9:28

Polymer

1,1181 gold badge9 silver badges17 bronze badges

Collectives™ on Stack Overflow

Do if statements take significant amounts of time in python?

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related