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Turning a function with multiple if else statement into a dictionary

So I currently have a function that outputs sound barrier for the input height.

def soundbarier (y):   
    if 0 < y < 1524:
        S = 340.3
    elif 1524 < y < 3048:
        S = 334.4
    elif 3048 < y < 4572:
        S = 328.4
    elif 4572 < y < 6096:
        S = 322.2
    elif 6096 < y < 7620:
        S = 316.0
    elif 7620 < y < 9144:
        S = 309.6
    elif 9144 < y < 10668:
        S = 303.6
    elif 10668 < y < 12192:
        S = 295.4
    else:
        S = 294.5
    return (float(S))

I want to shorten it using a dictionary but I can't get it to work.

def SOUND(y):
    return {
    0 < float(y) < 1524: 340.3,
    1524 < y < 3048: 334.4,
    3048 < y < 4572: 328.4,
    4572 < y < 6096: 322.2,
    6096 < y < 7620: 316.0,
    7620 < y < 9144: 309.6,
    9144 < y < 10668: 303.6,
    10668 < y < 12192: 295.4
    }.get(y, 9)

print(SOUND(1500))

I don't really want any default values. How could I make it work? I basically need the function to output S for a certain Y within a range.

like image 358
user3613025 Avatar asked Dec 13 '25 18:12

user3613025


1 Answers

Mathemagic by @Tim.

consts = [340.3, 334.4, 328.4, 322.2, 316.0, 309.6, 303.6, 295.4, 295.5]

def soundbarrier(y):
    return consts[
        -1 if (not (0 < y < 12192) or y % 1524 == 0) else y // 1524
    ]

In [1]: soundbarrier(1500)
Out[1]: 340.3

In [2]: soundbarrier(10000)
Out[2]: 303.6

In [3]: soundbarrier(100)
Out[3]: 340.3

In [4]: soundbarrier(1524)
Out[4]: 295.5
like image 73
cs95 Avatar answered Dec 16 '25 12:12

cs95



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