Is there a better way to do this?
if a > 1:
a = 1
if a < 0:
a = 0
I was thinking of using a function because I have a lot of these in my code, Was curious if there was a shorter comprehensible way to do it.
limit-number.pydef limit(num, minimum=1, maximum=255): """Limits input 'num' between minimum and maximum values. Default minimum value is 1 and maximum value is 255.
clamp() method clamps all the input elements into the range [ min, max ] and return a resulting tensor. inp: This is input tensor. min: This is a number and specifies the lower-bound of the range to which input to be clamped. max: This is a number and specifies the upper-bound of the range to which input to be clamped.
What you describe is usually called clipping. There are several ways to do clipping.
numpy
)You can use numpy.clip
for that:
numpy.clip(a, a_min, a_max, out=None)
So:
import numpy
numpy.clip(x,0,1)
Although since function calls in Python are expensive, and numpy
usually processes data in batch, for a single value it will result in computational overhead.
For example:
>>> x = -2.5
>>> numpy.clip(x,0,1)
0.0
>>> x = 2.5
>>> numpy.clip(x,0,1)
1.0
>>> x = 0.5
>>> numpy.clip(x,0,1)
0.5
Usually you use numpy
to perform operations on (large) matrices, for instance if you have to process a 1000x1000 matrix, than using numpy
will definitely pay off.
A pure python approach can be obtained with:
max(0,min(1,x))
But here you have two calls, as a result, it will be slower than using if
statements.
Finally if you stick with the if
-code, you can optimize it a tiny bit by using an elif
:
if x < 0:
x = 0
elif x > 1:
x = 1
Or a more generic function:
def clip(x,min,max):
if x < min:
return min
elif x > max:
return max
return x
I'd always use something like max(0, min(1, x))
.
If x is more than 1, min(1, x)
will be 1, and max(0, 1)
is still 1.
If x is less than 0, min(1, x)
will be x
, and max(0, x)
will be 0.
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