问题
In a numpy array I want to replace all nan and inf into a fixed number. Can I do that in one step to save computation time (arrays are really big)?
a = np.arange(10.0)
a[3] = np.nan
a[5] = np.inf
a[7] = -np.inf
# a: [ 0. 1. 2. nan 4. inf 6. -inf 8. 9.]
a[np.isnan(a)] = -999
a[np.isinf(a)] = -999
# a: [ 0. 1. 2. -999. 4. -999. 6. -999. 8. 9.]
The code above works fine. But I am looking for something like:
a[np.isnan(a) or np.isinf(a)] = -999
Which does not work and I can see why. Just thinking it might be better if every item of a is only checked once.
回答1:
this seems to work:
a[np.isnan(a) | np.isinf(a)] = 2
np.isnan() and np.isinf() in fact return two boolean numpy arrays.
boolean numpy arrays can be combined with bitwise operations such as & and |
回答2:
Numpy comes with its own vectorized version of or:
a[np.logical_or(np.isnan(a), np.isinf(a))] = -999
While the above version is clear understanable, there is a faster one, which is a bit weird:
a[np.isnan(a-a)] = -9999
The idea behind this is, that 'np.inf-np.inf = np.nan`
%timeit a[np.isnan(a-a)] = -999
# 100000 loops, best of 3: 11.7 µs per loop
%timeit a[np.isnan(a) | np.isinf(a)] = -999
# 10000 loops, best of 3: 51.4 µs per loop
%timeit a[np.logical_or(np.isnan(a), np.isinf(a))] = -999
# 10000 loops, best of 3: 51.4 µs per loop
Hence the | and np.logical_or version seem to be internally equivalent
回答3:
You could use np.isfinite which verifies that a number is not infinite nor a NaN:
a[~np.isfinite(a)] = -999
来源：https://stackoverflow.com/questions/45614447/python-combined-masking-in-numpy