How to delete a row based on a condition from a numpy array?

Question

From the following array :

test = np.array([[1,2,'a'],[4,5,6],[7,'a',9],[10,11,12]])

How can I delete the rows that contain 'a' ? Expected result :

array([[ 4,  5,  6],
   [10, 11, 12]])

Answer 1

Note, numpy supports vectorized comparisons:

>>> test
array([[1, 2, 'a'],
       [4, 5, 6],
       [7, 'a', 9],
       [10, 11, 12]], dtype=object)
>>> test == 'a'
array([[False, False,  True],
       [False, False, False],
       [False,  True, False],
       [False, False, False]], dtype=bool)

Now, you want the rows where all are not equalt to 'a':

>>> (test != 'a').all(axis=1)
array([False,  True, False,  True], dtype=bool)

So, simply select the rows with the mask:

>>> row_mask = (test != 'a').all(axis=1)
>>> test[row_mask,:]
array([[4, 5, 6],
       [10, 11, 12]], dtype=object)

Answer 2

Also, like this maybe? (Inspired from one of my another answers )

In [100]: mask = ~(test == 'a')

In [101]: mask
Out[101]: 
array([[ True,  True, False],
       [ True,  True,  True],
       [ True, False,  True],
       [ True,  True,  True]], dtype=bool)

In [102]: test[np.all(mask, axis=1), :]
Out[102]: 
array([['4', '5', '6'],
       ['10', '11', '12']],
      dtype='<U21')

But, please note that here we're not deleting any rows from the original array. We're just slicing out the rows which doesn't have the alphabet a.