python - Implementing ceil function without using if-else

Question

I just wanted to know that is there a way of implementing ceil function without using if-else? With if-else (for a/b) it can be implemented as:

if a%b == 0:
    return(a/b)
else:
    return(a//b + 1)

Answer 1

最简单的是。

a//b + bool(a%b)

而只是为了安全，

b and (a//b + bool(a%b))

干杯。

Answer 2

如果它们是整数，这样应该可以工作（我猜你有一个有理数表示）：

a/b + (a%b!=0)

否则，请替换a/b为int(a/b)，或者更好，如下所示a//b。

Answer 3

-(-a//b)

Perhaps the simplest?

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*Edited per @Gilles's comment:

for an integer n,

floor(x)=n for x in [n, n+1)
ceil(y)=n+1 for y in (n, n+1]

So, floor(-y)=-n-1 for -y in [-n-1, -n),

and ceil(y)=-floor(-y)=n+1 for y in (n, n+1]

In Python, floor(a/b) = a//b. Thus ceil(a/b) = -(-a//b)