Replace NaNs in masked numpy array

Question

var[np.isnan(var)] = 0.0

var.shape
(50, 360, 720)

I want to replace NaNs in the array with 0.0. However, I get an error: *** IndexError: Index cannot be multidimensional

var[0]

masked_array(data =
 [[-- -- -- ..., -- -- --]
 [-- -- -- ..., -- -- --]
 [-- -- -- ..., -- -- --]
 ..., 
 [-- -- -- ..., -- -- --]
 [-- -- -- ..., -- -- --]
 [-- -- -- ..., -- -- --]],
             mask =
 [[ True  True  True ...,  True  True  True]
 [ True  True  True ...,  True  True  True]
 [ True  True  True ...,  True  True  True]
 ..., 
 [ True  True  True ...,  True  True  True]
 [ True  True  True ...,  True  True  True]
 [ True  True  True ...,  True  True  True]],
       fill_value = nan)

How to fix it?

Eli Bendersky · Accepted Answer · 2017-01-05 00:12:01Z

1

nan_to_num is the right way to do it. Here's a 2D example:

In [25]: arr = np.array([[10, 20], [np.nan, 30], [np.nan, -10]])

In [26]: arr
Out[26]: 
array([[ 10.,  20.],
       [ nan,  30.],
       [ nan, -10.]])

In [27]: np.nan_to_num(arr)
Out[27]: 
array([[ 10.,  20.],
       [  0.,  30.],
       [  0., -10.]])

I believe nan_to_num should work on masked arrays as well. For example:

In [33]: mx = ma.masked_array([np.nan, 2, 3, 4], mask=[0, 0, 1, 0])

In [34]: mx.compressed()
Out[34]: array([ nan,   2.,   4.])

And now:

In [36]: np.nan_to_num(ma.masked_array([np.nan, 2, 3, 4], mask=[0, 0, 1, 0])).compressed()
Out[36]: array([ 0.,  2.,  4.])

edited Jan 5, 2017 at 0:12

answered Jan 4, 2017 at 23:59

Eli Bendersky

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3 Comments

Eli Bendersky Over a year ago

@user308827: it really shouldn't. what makes you think so?

user308827 Over a year ago

In my full dataset, it is converting everything to 0.0

Eli Bendersky Over a year ago

@user308827: see my update to the answer. Maybe something is wrong with your dataset (all NaNs?) -- check it step by step

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