CODEWITHSUNDEEP
pythonoptimization
FluidMechanics Potential Flows
FluidMechanics Potential Flows

Reputation: 588

Maximum and minimum of two values

I'm writing a code that will be called several times and I was wondering if there was something faster (computationally wise) than this:

maximum = max(a, b)  # a and b are integers
minimum = min(a, b)

The code works but I guess once I called max we are wasting in a way computational power by calling min. I tried to look at the problem on the internet but I'm having trouble putting words on it.

Upvotes: 2

Views: 3560

Answers (4)

FluidMechanics Potential Flows
FluidMechanics Potential Flows

Reputation: 588

I've ran this code with your suggestions (all very clever ones, thank you!!) to try and see what's faster. The differences are actually quite impressive:

import time

def whosfaster(a, b):
    start_time = time.time()
    maximum = max(a, b)
    minimum = min(a, b)
    method0 = time.time() - start_time


    start_time = time.time()
    if a > b:
        maximum, minimum = a, b
    else:
        maximum, minimum = b, a
    method1 = time.time() - start_time

    start_time = time.time()
    maximum, minimum = max(a, b), min(a, b)
    method2 = time.time() - start_time

    start_time = time.time()
    maximum, minimum = (a, b) if a > b else (b, a)
    method3 = time.time() - start_time

    start_time = time.time()
    if max(a, b) == a:
        maximum = a
        minimum = b
    else:
        maximum = b
        minimum = a
    method4 = time.time() - start_time

    return maximum, minimum, method0, method1, method2, method3, method4

total = [0, 0, 0, 0, 0]
for a in range(1, 5000):
    for b in range(1, 5000):
        maximum, minimum, method0, method1, method2, method3, method4 = whosfaster(a, b)
        total[0] += method0
        total[1] += method1
        total[2] += method2
        total[3] += method3
        total[4] += method4

print(total)

The results I obtained are:

    Run number 1 : 
[7.562410593032837, 2.4336302280426025, 7.17744255065918, 3.160186529159546, 5.583981037139893]
    Run number 2 but my laptop was on battery saver mode :
[12.757804155349731, 4.507119178771973, 13.357162237167358, 5.251198768615723, 8.934610605239868]

If someone wants to run it on their end to compare, you're welcome to do so, my laptop isn't the fastest!!

EDIT : I didn't really know how to implement Mark Ransom's answer into practice, so I didn't try it. Feel welcome to edit my post and change my code and to rerun it on your hand if you know how to do so.

Upvotes: 1

Adon Bilivit
Adon Bilivit

Reputation: 26976

This would rather depend on what data types a and b are. If they are simple types (e.g., numeric) then:

if a > b:
  maximum, minimum = a, b
else:
  maximum, minimum = b, a

...might be faster than calling the min and max functions.

Or, if you like one-liners:

maximum, minimum = (a, b) if a > b else (b, a)

Upvotes: 5

Mark Ransom
Mark Ransom

Reputation: 308138

Mathematically min(a,b) is the same as a+b-max(a,b). Whether this is an improvement is debatable.

Upvotes: 1

HaseebGarfinkel
HaseebGarfinkel

Reputation: 50

If you are worried about computational power here it should not be a problem. No matter what you would need to define two different variables, maximum and minimum, so the way you did it is fine. If you really do not want to use min, then something like this could work but it is already more complex than what you did:

if max(a, b) == a:
    maximum = a
    minimum = b
else:
    maximum = b
    minimum = a

Upvotes: 2

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