How to compute a percentage increase/decrease multiplier in python without if statements

Question

I have a scenario in which the user enter a positive, 0, or negative value which represents a percentage increase, no change, or percentage decrease, respectively. Example: 2.5

Answer 1

if and else can actually be used in expressions on one line: x = a if b else c. You can't use elif, though, so you'll have to first refactor your code so it doesn't use them:

if U == 0:
   multiplier = 1
else:
    if U > 0:
       multiplier = U/100.0 + 1
    else:
       multiplier = (100 - (U * -1.0))/100.0

This reduces to

if U == 0:
   multiplier = 1
else:
   multiplier = U/100.0 + 1 if U > 0 else (100 - (U * -1.0))/100.0

Which reduces to

multiplier = 1 if U == 0 else (U/100.0 + 1 if U > 0 else (100 - (U * -1.0))/100.0)

Answer 2

You just need to use the distributive property of division:

multiplier = 1 + U/100.0

This is actually what your code does. You just separated three branches of a if that did exactly the same. Let's do some easy math step-by-step from your original code:

---

if U == 0:
   multiplier = 1
elif U > 0:
   multiplier = U/100.0 + 1
else:
   multiplier = (100 - (U * -1.0))/100.0

---

if U == 0:
   multiplier = 1 + 0                 #sum 0 for convenience
elif U > 0:
   multiplier = 1 + U/100.0           #swap order of addition
else:
   multiplier = (100 + U)) / 100.0    #multiplying by -1 is just changing the sign

---

if U == 0:
   multiplier = 1 + U/100.0    # since U is 0 in this branch, U/100.0 == 0
elif U > 0:
   multiplier = 1 + U/100.0
else:
   multiplier = 1 + U/100.0    # distributive property of division

---

Since the executed code is the same for the three branches, there's no point having a if:

multiplier = 1 + U/100.0

Answer 3

Kevin's answer is technically completely correct, but your code can be simplified a whole lot just from a math standpoint.

You don't need to be performing if-then-else checks on this mathematical operation. Consider the input of "-3": In your code, this will go into the third block, evaluate as (100 - (-3 * -1.0)) / 100.0 = (100 - 3.0) / 100.0 = 0.97.

This is equivalent to the value U going into the second block, evaluated as (-3 / 100.0) + 1 = -0.03 + 1 = 0.97.

Now consider the input of 0: No matter which block it goes in to, the "multiplier" value will come out as 1.

So just have your code be:

multiplier = 1 + (U / 100.0)

and you'll be set.