Map function with array based on integers on 2D numpy array

Question

>>> a = 0.001 * np.random.randint(0, 1000, size=(5))
>>> a
array([0.524, 0.311, 0.603, 0.957, 0.923])

>>> b = np.random.randint(0, 2, size=(3, 5))
>>> b
array([[1, 1, 1, 0, 1],
       [0, 0, 1, 0, 0],
       [0, 0, 1, 0, 1]])

I want to map over each row of the 2D array b, and if the value at row[i] == 1, set it to a[i], if the value at row[i] == 0, I want to set it to 1 - a[i].

For example, for b[1] = row = [0, 0, 1, 0, 0], it should become:

[1 - 0.524, 1 - 0.311, 0.603, 1 - 0.957, 1 - 0.923]

i.e.

[0.476, 0.689, 0.603, 0.043, 0.077]

How can I achieve this for all rows of the 2D matrix without resorting to for loops? (i.e. leverage efficiency of numpy)

What's the shape of the result supposed to be? Show the full working iterative code. — hpaulj
– hpaulj, Commented Jul 4, 2022 at 2:43

Mad Physicist · Accepted Answer · 2022-07-04 04:13:38Z

1

There are a couple of ways of doing this. I would probably use the where and out arguments to np.subtract:

np.subtract(np.ones(len(b)), a, out=np.broadcast_to(a, b.shape).copy(), where=b.astype(bool))

Going with @hpaulj's solution to use the 3-arg version of np.where is probably much cleaner in this case:

np.where(b, a, 1 - a)

edited Jul 4, 2022 at 4:13

answered Jul 4, 2022 at 1:52

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4 Comments

herophant Over a year ago

When I do this, I get non-broadcastable output operand with shape (5,) doesn't match the broadcast shape (3,5) as an error, unfortunately, is there any way to perform it on a 2D array?

hpaulj Over a year ago

The out needs to be (3,5) the shape of b. For example a[None].repeat(3, axis=0)

hpaulj Over a year ago

Here the np.where function is a cleaner solution.

Mad Physicist Over a year ago

@herophant. Sorry, I forgot to broadcast the ones correctly.

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