Is there a simple way of replacing all negative values in an array with 0?
I'm having a complete block on how to do it using a NumPy array.
E.g.
a = array([1, 2, 3, -4, 5])
I need to return
[1, 2, 3, 0, 5]
a < 0
gives:
[False, False, False, True, False]
This is where I'm stuck - how to use this array to modify the original array.
Here's a way to do it in Python without NumPy. Create a function that returns what you want and use a list comprehension, or the map function.
>>> a = [1, 2, 3, -4, 5]
>>> def zero_if_negative(x):
... if x < 0:
... return 0
... return x
...
>>> [zero_if_negative(x) for x in a]
[1, 2, 3, 0, 5]
>>> map(zero_if_negative, a)
[1, 2, 3, 0, 5]
And yet another possibility:
In [2]: a = array([1, 2, 3, -4, 5])
In [3]: where(a<0, 0, a)
Out[3]: array([1, 2, 3, 0, 5])
Try numpy.clip
:
>>> import numpy
>>> a = numpy.arange(-10, 10)
>>> a
array([-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2,
3, 4, 5, 6, 7, 8, 9])
>>> a.clip(0, 10)
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
You can clip only the bottom half with clip(0)
.
>>> a = numpy.array([1, 2, 3, -4, 5])
>>> a.clip(0)
array([1, 2, 3, 0, 5])
You can clip only the top half with clip(max=n)
. (This is much better than my previous suggestion, which involved passing NaN
to the first parameter and using out
to coerce the type.):
>>> a.clip(max=2)
array([ 1, 2, 2, -4, 2])
Another interesting approach is to use where
:
>>> numpy.where(a <= 2, a, 2)
array([ 1, 2, 2, -4, 2])
Finally, consider aix's answer. I prefer clip
for simple operations because it's self-documenting, but his answer is preferable for more complex operations.
Another minimalist Python solution without using numpy:
[0 if i < 0 else i for i in a]
No need to define any extra functions.
a = [1, 2, 3, -4, -5.23, 6]
[0 if i < 0 else i for i in a]
yields:
[1, 2, 3, 0, 0, 6]
Source: Stackoverflow.com