mask a 2D numpy array based on values in one column

Question

Suppose I have the following numpy array:

a = [[1, 5, 6],
     [2, 4, 1],
     [3, 1, 5]]

I want to mask all the rows which have 1

Answer 1

You can create the desired mask by

mask = numpy.repeat(a[:,0]==1, a.shape[1])

and the masked array by

masked_a = numpy.ma.array(a, mask=numpy.repeat(a[:,0]==1, a.shape[1]))

Answer 2

You could simply create an empty mask and then use numpy-broadcasting (like @eumiro showed) but using the element- and bitwise "or" operator |:

>>> a = np.array([[1, 5, 6], [2, 4, 1], [3, 1, 5]])

>>> mask = np.zeros(a.shape, bool) | (a[:, 0] == 1)[:, None]

>>> np.ma.array(a, mask=mask)
masked_array(data =
 [[-- -- --]
 [2 4 1]
 [3 1 5]],
             mask =
 [[ True  True  True]
 [False False False]
 [False False False]],
       fill_value = 999999)

A bit further explanation:

>>> # select first column
>>> a[:, 0]  
array([1, 2, 3])

>>> # where the first column is 1
>>> a[:, 0] == 1  
array([ True, False, False], dtype=bool)

>>> # added dimension so that it correctly broadcasts to the empty mask
>>> (a[:, 0] == 1)[:, None]  
array([[ True],
       [False],
       [False]], dtype=bool)

>>> # create the final mask
>>> np.zeros(a.shape, bool) | (a[:, 0] == 1)[:, None]  
array([[ True,  True,  True],
       [False, False, False],
       [False, False, False]], dtype=bool)

One further advantage of this approach is that it doesn't need to use potentially expensive multiplications or np.repeat so it should be quite fast.

Answer 3

import numpy as np

a = np.array([[1, 5, 6],
              [2, 4, 1],
              [3, 1, 5]])

np.ma.MaskedArray(a, mask=(np.ones_like(a)*(a[:,0]==1)).T)

# Returns: 
masked_array(data =
 [[-- -- --]
 [2 4 1]
 [3 1 5]],
             mask =
 [[ True  True  True]
 [False False False]
 [False False False]])