How to mask one ndarray using a another ndarray

Question

I have an items array which has shape (n, 3), and a counts array which has the same shape (n, 3). How can I make every C in items to have count = 0 without resorting to loops?

items = np.array([['B', 'A', 'C'],
    ['B', 'B', 'C'],
    ['A', 'B', 'A'],
    ['C', 'C', 'C'],
    ['B', 'B', 'B']])

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

Expected output:

>>> counts
np.array([[1, 3, 0],
    [4, 2, 0],
    [2, 2, 1],
    [0, 0, 0],
    [1, 2, 1]])

Chan Kha Vu · Accepted Answer · 2019-07-23 01:27:19Z

3

What you're looking for is:

counts[items == 'C'] = 0

A more explicit way to express it is:

c_indices = np.where(items == 'C')
counts[c_indices] = 0

answered Jul 23, 2019 at 1:27

Chan Kha Vu

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2 Comments

Mad Physicist Over a year ago

There is no good reason to use where for this. It is longer, slower, consumes more memory, and in my opinion less legible. +1 for how to construct the mask though.

Jenny Over a year ago

I can't believe it's that simple. Oh well

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