How to create numpy array mask/masked array with multiple conditions?

Question

I know that in Numpy you can mask an array based on a condition. For example, if I want a masked array for all values greater than 0.

arr[arr>0]

But I want to have two conditions for the mask. Intuitively, this would look like:

arr[arr>0 and arr<1]

but the compiler pops an error:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

I know I can use a for loop to solve this:

masked=np.array([i if(i>0 and i<1) for i in mask])

but is there a more elegant solution using something that's built-in for Numpy?

John Zwinck · Accepted Answer · 2023-07-01 09:55:50Z

1

You want this:

arr[(arr > 0) & (arr < 1)]

answered Jul 1, 2023 at 9:55

John Zwinck

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4 Comments

playerJX1 Over a year ago

thanks, I should have tried that. Is there an explanation as to why "and" doesn't work?

John Zwinck Over a year ago

"and" expects to produce a scalar True/False. You want elementwise logic.

playerJX1 Over a year ago

Interesting, now that you mentioned it I found a page that explains this in a similar scenario: stackoverflow.com/questions/8632033/….

hpaulj Over a year ago

and is a Python expression, more like a if/else clause than an operator. True and 1 returns 1, more like 1 if True else False. (except it short circuits)

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