Select elements of numpy array via boolean mask array

You probably want something like this:

>>> a = np.array([True, True, True, False, False])>>> b = np.array([[1,2,3,4,5], [1,2,3,4,5]])>>> b[:,a]array([[1, 2, 3],       [1, 2, 3]])

Note that for this kind of indexing to work, it needs to be an ndarray, like you were using, not a list, or it'll interpret the False and True as 0 and 1 and give you those columns:

>>> b[:,[True, True, True, False, False]]   array([[2, 2, 2, 1, 1],       [2, 2, 2, 1, 1]])

You can use numpy.ma module and use np.ma.masked_array function to do so.

>>> x = np.array([1, 2, 3, -1, 5])                                                >>> mx = ma.masked_array(x, mask=[0, 0, 0, 1, 0])masked_array(data=[1, 2, 3, --, 5], mask=[False, False,  False, True, False], fill_value=999999)

Hope I'm not too late! Here's your array:

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

Let's create an array of zeros of the same shape as X:

mask = np.zeros_like(X)# array([[0, 0, 0, 0, 0],#        [0, 0, 0, 0, 0]])

Then, specify the columns that you want to mask out or hide with a 1. In this case, we want the last 2 columns to be masked out.

mask[:, -2:] = 1# array([[0, 0, 0, 1, 1],#        [0, 0, 0, 1, 1]])

Create a masked array:

X_masked = np.ma.masked_array(X, mask)# masked_array(data=[[1, 2, 3, --, --],#                    [1, 2, 3, --, --]],#              mask=[[False, False, False,  True,  True],#                    [False, False, False,  True,  True]],#              fill_value=999999)

We can then do whatever we want with X_masked, like taking the sum of each column (along axis=0):

np.sum(X_masked, axis=0)# masked_array(data=[2, 4, 6, --, --],#              mask=[False, False],#              fill_value=1e+20)

Great thing about this is that X_masked is just a view of X, not a copy.

X_masked.base is X# True