I want to take a floating-point number and round it down to the nearest integer. However, if it's not a whole, I always want to round down the variable, regardless of how close it is to the next integer up. Is there a way to do this?
This question is related to
python
floating-point
integer
rounding
If you don't want to import math, you could use:
int(round(x))
Here's a piece of documentation:
>>> help(round)
Help on built-in function round in module __builtin__:
round(...)
round(number[, ndigits]) -> floating point number
Round a number to a given precision in decimal digits (default 0 digits).
This always returns a floating point number. Precision may be negative.
I used this code where you subtract 0.5 from the number and when you round it, it is the original number rounded down.
round(a-0.5)
I think you need a floor function :
One of these should work:
import math
math.trunc(1.5)
> 1
math.trunc(-1.5)
> -1
math.floor(1.5)
> 1
math.floor(-1.5)
> -2
a lot of people say to use int(x)
, and this works ok for most cases, but there is a little problem. If OP's result is:
x = 1.9999999999999999
it will round to
x = 2
after the 16th 9 it will round. This is not a big deal if you are sure you will never come across such thing. But it's something to keep in mind.
Simple
print int(x)
will work as well.
It may be very simple, but couldn't you just round it up then minus 1? For example:
number=1.5
round(number)-1
> 1
x//1
The //
operator returns the floor of the division. Since dividing by 1 doesn't change your number, this is equivalent to floor but no import is needed.
Notes:
To get floating point result simply use:
round(x-0.5)
It works for negative numbers as well.
Just make round(x-0.5) this will always return the next rounded down Integer value of your Float. You can also easily round up by do round(x+0.5)
If you working with numpy, you can use the following solution which also works with negative numbers (it's also working on arrays)
import numpy as np
def round_down(num):
if num < 0:
return -np.ceil(abs(num))
else:
return np.int32(num)
round_down = np.vectorize(round_down)
round_down([-1.1, -1.5, -1.6, 0, 1.1, 1.5, 1.6])
> array([-2., -2., -2., 0., 1., 1., 1.])
I think it will also work if you just use the math
module instead of numpy
module.
Don't know if you solved this, but I just stumble upon this question. If you want to get rid of decimal points, you could use int(x) and it will eliminate all decimal digits. Theres no need to use round(x).
Source: Stackoverflow.com