替换数字数组中的负值

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.

You are halfway there. Try:

In [4]: a[a < 0] = 0


In [5]: a
Out[5]: array([1, 2, 3, 0, 5])

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]

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]

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:

1. os[os < 0] = 0 is faster than np.where() but not supported by numba. But whatever, np.maximum() is the fastest that I found.

2. np.maximum() is different from np.max() and np.amax(). np.maximum() can compare vector with single value.