What is the most optimal solution (time wise) for multiple if statements

Question

The examples below are from my chess game but the question is more general for runtime optimizing if-statements in Python. What type of code will run faster, and is there a huge difference? Bare in mind that in this case I will calculated this inside a minimax algorithm (thousands of times). Case 3 if of course nicer, but is it faster to run?

Another question would be if there is a difference in the order of which I put these values in? In this case the pawn option will come up the most times, is it then better to have that one first in the if-statements/dict?

# Case 1:
  
if piece_type == 'K':
    value = 20000
elif piece_type == 'Q':
    value = 900
elif piece_type == 'B':
    value = 330
elif piece_type == 'N':
    value = 320
elif piece_type == 'R':
    value = 500
elif piece_type == 'p':
    value = 100

return value

# Case 2:

if piece_type == 'K':
    return 20000
elif piece_type == 'Q':
    return 900
elif piece_type == 'B':
    return 330
elif piece_type == 'N':
    return 320
elif piece_type == 'R':
    return 500
elif piece_type == 'p':
    return 100

# Case 3:

piece_values = {'K': 20000, 'Q': 900, 'B': 330, 'N': 320, 'R': 500, 'p': 100}
return piece_values.get('K')

`piece_values[k]` is probably faster than using the `.get` method. — kaya3, Oct 13 '20 at 05:01
Case 3 is O(1) and will be the fastest. https://stackoverflow.com/questions/1963507/time-complexity-of-accessing-a-python-dict — Alexander, Oct 13 '20 at 05:03
The return in case 3 suggests that `piece_values` is regenerated inside of a function. Is it created only once? — tdelaney, Oct 13 '20 at 05:13
Thanks @Alexander, and is there any difference for the order of the dict? E.g. put pawns first? — eligolf, Oct 13 '20 at 05:30

score 1 · Accepted Answer · answered Oct 13 '20 at 05:42

import timeit
piece_type = 'p'
def case1():
    if piece_type == 'K':
        value = 20000
    elif piece_type == 'Q':
        value = 900
    elif piece_type == 'B':
        value = 330
    elif piece_type == 'N':
        value = 320
    elif piece_type == 'R':
        value = 500
    elif piece_type == 'p':
        value = 100
    return value


def case2():
    if piece_type == 'K':
        return 20000
    elif piece_type == 'Q':
        return 900
    elif piece_type == 'B':
        return 330
    elif piece_type == 'N':
        return 320
    elif piece_type == 'R':
        return 500
    elif piece_type == 'p':
        return 100

piece_values = {'K': 20000, 'Q': 900, 'B': 330, 'N': 320, 'R': 500, 'p': 100}
def case3():
    return piece_values.get(piece_type)

N = 1000
r1 = 0
r2 = 0
r3 = 0
for i in range(N):
    r1 += timeit.timeit("case1()", setup="from __main__ import case1", number=100000)
    r2 += timeit.timeit("case2()", setup="from __main__ import case2", number=100000)
    r3 += timeit.timeit("case3()", setup="from __main__ import case3", number=100000)
print(r1/N)
print(r2/N)
print(r3/N)

Results

For piece_type = 'p'

0.043659273699999786
0.042489134299999745
0.025282950800000023

For peice_type = 'K'

0.017085672599999883
0.015899844999999937
0.02493290909999991

As if we see the results, accessing elements of dictionary doesn't matter where the element is located.
Accessing elements down the if-else ladder cost more time compared to the elements at the top of ladder.

So if the conditionals are less then if-else is better, if there are large number of conditionals then dictionary is better as dictionary is always O(1).

We also see that that ordering of the conditionals matters. There's a lot more pawns than Kings, so to optimize the if-conditionals they should come first, follwed by Knights, Rooks and Bishops (in any order), and then finally the Queen and King. — BurnNote, Oct 13 '20 at 05:50
Thank you so much! Also showing me how to time similar things in the future :D — eligolf, Oct 13 '20 at 05:53

What is the most optimal solution (time wise) for multiple if statements

1 Answers1