Replace all elements of Python NumPy Array that are greater than some value
Since you actually want a different array which is arr
where arr < 255
, and 255
otherwise, this can be done simply:
result = np.minimum(arr, 255)
More generally, for a lower and/or upper bound:
result = np.clip(arr, 0, 255)
If you just want to access the values over 255, or something more complicated, @mtitan8's answer is more general, but np.clip
and np.minimum
(or np.maximum
) are nicer and much faster for your case:
In [292]: timeit np.minimum(a, 255)
100000 loops, best of 3: 19.6 µs per loop
In [293]: %%timeit
.....: c = np.copy(a)
.....: c[a>255] = 255
.....:
10000 loops, best of 3: 86.6 µs per loop
If you want to do it in-place (i.e., modify arr
instead of creating result
) you can use the out
parameter of np.minimum
:
np.minimum(arr, 255, out=arr)
or
np.clip(arr, 0, 255, arr)
(the out=
name is optional since the arguments in the same order as the function's definition.)
For in-place modification, the boolean indexing speeds up a lot (without having to make and then modify the copy separately), but is still not as fast as minimum
:
In [328]: %%timeit
.....: a = np.random.randint(0, 300, (100,100))
.....: np.minimum(a, 255, a)
.....:
100000 loops, best of 3: 303 µs per loop
In [329]: %%timeit
.....: a = np.random.randint(0, 300, (100,100))
.....: a[a>255] = 255
.....:
100000 loops, best of 3: 356 µs per loop
For comparison, if you wanted to restrict your values with a minimum as well as a maximum, without clip
you would have to do this twice, with something like
np.minimum(a, 255, a)
np.maximum(a, 0, a)
or,
a[a>255] = 255
a[a<0] = 0
I think both the fastest and most concise way to do this is to use NumPy's built-in Fancy indexing. If you have an ndarray
named arr
, you can replace all elements >255
with a value x
as follows:
arr[arr > 255] = x
I ran this on my machine with a 500 x 500 random matrix, replacing all values >0.5 with 5, and it took an average of 7.59ms.
In [1]: import numpy as np
In [2]: A = np.random.rand(500, 500)
In [3]: timeit A[A > 0.5] = 5
100 loops, best of 3: 7.59 ms per loop