根据一维条件删除ndarray中的元素

Question

In a Numpy ndarray, how do I remove elements in a dimension based on condition in a different dimension?

I have:

[[[1 3]
  [1 4]]

 [[2 6]
  [2 8]]

 [[3 5]
  [3 5]]]

I want to remove based on condition x[:,:,1] < 7

Desired output ( [:,1,:] removed):

[[[1 3]
  [1 4]]

 [[3 5]
  [3 5]]]

EDIT: fixed typo

Answer 1

This may work:

x[np.where(np.all(x[..., 1] < 7, axis=1)), ...]

yields

array([[[[1, 3],
         [1, 4]],

        [[3, 5],
         [3, 5]]]])

You do get an extra dimension, but that's easy to remove:

np.squeeze(x[np.where(np.all(x[..., 1] < 7, axis=1)), ...])

Briefly how it works:

First the condition: x[..., 1] < 7 .
Then test if the condition is valid for all elements along the specific axis: np.all(x[..., 1] < 7, axis=1) .
Then, use where to grab the indices instead of an array of booleans: np.where(np.all(x[..., 1] < 7, axis=1)) .
And insert those indices into the relevant dimension: x[np.where(np.all(x[..., 1] < 7, axis=1)), ...] .

Answer 2

As your desired output, you filter x on axis=0. Therefore, you may try this way

m = (x[:,:,1] < 7).all(1)
x_out = x[m,:,:]

Or simply

x_out = x[m]   

Out[70]:
array([[[1, 3],
        [1, 4]],

       [[3, 5],
        [3, 5]]])