If x is 2.3, then math.floor(x) returns 2.0, the largest integer smaller than or equal to x (as a float.)
How would I get i the largest integer strictly smaller than x (as a integer)?
The best I came up with is:
i = int(math.ceil(x)-1)
Is there a better way?
Note, that if x is 2.0 then math.floor(x) returns 2.0 but I need the largest integer smaller than 2.0, which is 1.
math.ceil(x)-1 is correct and here is the proof.
if x is in Z (the set of integers), then math.ceil(x) = x. Therefore math.ceil(x)-1=x-1, the largest integer smaller than x.
Else we have x in R \ Z and math.ceil(x) is the smallest integer y such that x ≤ y. But then y-1 is an integer smaller than the smallest integer such that x ≤ y, therefore x > y-1 and by construction y-1 is the largest such integer smaller than x.
It's simple enough that I wouldn't bother with those if-else. But to avoid computation errors with floats I would do the -1 outside the int conversion.
int(math.ceil(x))-1
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