Dividing two numbers and printing the result and adding one to the result if there is a remainder, without using if-statements or imports or function?

Question

I'm trying to divide number into groups in plain python without importing or the use of if-statements, and get the division into groups + remainder forming one extra group so that 200 / 99 would be 3, And 7 / 3 would be 3, but that 8 / 4 would still be just 2, and 4 / 2 would be 2 etc.. I cannot import anything so it needs to be in plain python.

I tried storing the numbers from inputs from user into variables and dividing them, and then adding one. I also tried // and adding 1 but I cannot get it to work.

Why can't you import anything or use if statements? And when you say you cannot get it to work, what problem do you run into? An error? A different result? Please write that explicitly in your question. — Saaru Lindestøkke, Feb 09 '23 at 22:18

Grismar · Answer 1 · 2023-02-09T22:27:36.030

How about this:

a, b = 200, 99
result, remainder = a // b + int(bool(a % b)), a % b

This computes the result by performing an integer divide a // b and computing the remainder a % b. Converting an integer value to a boolean is False for 0 and True for any other value, and converting that back to an integer gives you the value you want to add to the result. The remainder is computed again to assign it, if you need it.

As user @markransom commented, the conversion to int() isn't even necessary, as bool already 'is' an integer type:

>>> isinstance(True, int)
True

So, this works (although it may be considered a bit less readable):

a, b = 200, 99
result, remainder = a // b + bool(a % b), a % b

If you're using a modern version of Python and really want it to be short, this also works:

result = a // b + bool(remainder := a % b)

This uses the walrus operator to assign the remainder when it is first computed, avoiding having to compute it twice as well.

No need to convert `bool` to `int`, it's already a subclass. — Mark Ransom, Feb 09 '23 at 22:25

tdelaney · Answer 2 · 2023-02-09T22:38:02.833

1

Python boolean operations short-circuit and return the last value evaluated and you can use that to convert a non-zero remainder to 1 for addition to a quotient

def group_me(dividend, divisor):
    quotient, remainder = divmod(dividend, divisor)
    return quotient + (remainder and 1 or 0)

print(group_me(200, 99))
print(group_me(7, 3))
print(group_me(8, 4))

Output

3
3
2

If remainder is non-zero, remainder and 1 short-circuits and returns 1. Otherwise, the or now becomes 0 or 0, which retains its last value 0.

edited Feb 09 '23 at 22:38

answered Feb 09 '23 at 22:30

tdelaney

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bahh this was going to be my answer! – JonSG Feb 09 '23 at 22:32
@JonSG Have a look at the original's top answer, maybe you won't be disappointed about this anymore. – Kelly Bundy Feb 09 '23 at 23:00
@KellyBundy I'm not disappointed at all :-p I was going to also post essentially this as and answer but was in the middle of testing that it worked as I assumed when this got posted. I like the top post as it is a one liner but I would have done it this way for clarity. – JonSG Feb 10 '23 at 00:03
1

@JonSG I thought the "bah" meant you were disappointed you were too slow. Anyway, I find the `-(a // -b)` also clear, ceiling division is simply floor division when viewing the world upside down. – Kelly Bundy Feb 10 '23 at 06:34

score 0 · Answer 3 · answered Feb 09 '23 at 22:52

You could do an if statement mathematically without using if itself:

n, d = 200, 99
x, y = n/d, n//d + 1
result = int((x%1 == 0) * x + (x%1 != 0) * y)

Essentially it is condition * answer 1 + (1 - condition) * answer 2. Then it switch to whichever answer depending on condition being 1 or 0.

Dividing two numbers and printing the result and adding one to the result if there is a remainder, without using if-statements or imports or function?

3 Answers3