How to create a numpy masked array using lower dim array as a mask?

Question

Assume

a = [
        [1, 2, 3],
        [4, 5, 6],
        [7, 8, 9],
    ]

mask = [1, 0, 1]

I want

a[mask] == [
           [1, 2, 3],
           [False, False, False],
           [7, 8, 9],
           ]

or equivalent.

Meaning, I want to access a with mask, where mask is of lower dimension, and have auto broadcasting. I want to use that in the constructor of np.ma.array in the mask= argument.

What's with this mix of numbers and booleans?

hpaulj
– hpaulj

2020-04-26 18:46:10 +00:00
Commented Apr 26, 2020 at 18:46 — hpaulj
– hpaulj, Commented Apr 26, 2020 at 18:46
numpy.org/devdocs/reference/maskedarray.generic.html

Gulzar
– Gulzar

2020-04-26 18:47:31 +00:00
Commented Apr 26, 2020 at 18:47 — Gulzar
– Gulzar, Commented Apr 26, 2020 at 18:47

orlp · Accepted Answer · 2020-04-26 19:02:13Z

2

This should work. Note that your mask has the opposite meaning of np.ma.masked_array, where 1 means removed, so I inverted your mask:

>>> a = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
>>> mask = ~np.array([1, 0, 1], dtype=np.bool)  # Note - inverted mask.
>>> masked_a = np.ma.masked_array(
...     a,
...     np.repeat(mask, a.shape[1]).reshape(a.shape)
... )
>>> masked_a
masked_array(
  data=[[1, 2, 3],
        [--, --, --],
        [7, 8, 9]],
  mask=[[False, False, False],
        [ True,  True,  True],
        [False, False, False]],
  fill_value=999999)

answered Apr 26, 2020 at 19:02

orlp

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