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up vote 0 down vote favorite

I have a 2D numpy array and want to replace each NaN in each row by the corresponding value from a 1D array. For example, this matrix:

[[1.  2. NaN]
 [4.  5.  6.]
 [NaN NaN 9.]]

using vector [3. 7. 8.] would be converted to:

[[1. 2. 8.]
 [4. 5. 6.]
 [3. 7. 9.]]

How to do it without iterating over indices?

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up vote 1 down vote accepted

Use numpy.where and broadcasting:

>>> a = np.array([[1.,  2., np.nan],
                  [4.,  5.,  6.],
                  [np.nan, np.nan, 9.]])
>>> v = np.array([3, 7, 8])
>>> np.where(np.isnan(a), v, a)
array([[ 1.,  2.,  8.],
       [ 4.,  5.,  6.],
       [ 3.,  7.,  9.]])

numpy.isnan() gives you an array of booleans with NaN having the value True and False otherwise.

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