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.
You are halfway there. Try:
In [4]: a[a < 0] = 0
In [5]: a
Out[5]: 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.
numpy? Did you try the clip method of a? The built-in function numpy.clip gives me the same error, but the method does not.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]
for applies it to every element of the list, putting it after, means only if the condition is met does it go into the resulting list.
i < 0 ? 0 : i in C) inside a list comprehension. Put brackets in to make it clearer [(0 if i < 0 else i) for i in a]. Putting the if after is using the filter part of the list expression construct. [(i) for i in a if i < 0] will only return a list of items that are less than zero.
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])
Benchmark using numpy:
%%timeit
a = np.random.random(1000) - 0.5
b = np.maximum(a,0)
# 18.2 µs ± 204 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%%timeit
a = np.random.random(1000) - 0.5
a[a < 0] = 0
# 19.6 µs ± 304 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%%timeit
a = np.random.random(1000) - 0.5
b = np.where(a<0, 0, a)
# 21.1 µs ± 134 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%%timeit
a = np.random.random(1000) - 0.5
b = a.clip(0)
# 37.7 µs ± 124 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Supprisingly, np.maximum beat @NPE answer.
Caveat:
os[os < 0] = 0 is faster than np.where() but not supported by numba. But whatever, np.maximum() is the fastest that I found.
np.maximum() is different from np.max() and np.amax(). np.maximum() can compare vector with single value.
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]
