删除满足2个条件的2D numpy数组中的元组

Question

So I have a numpy array of tuples and I want to remove all tuples where the first value is less than 0 or the second element is greater than a number, n.
So if n = 10 and we had this array:

[[-1, 5], [3, 11], [-4, 20]]

It would become this:

[[]]

I'm guessing I need to use np.delete and np.where in a smart way?

Thanks in advance.

Answer 1

You can use something like the following:

>>> import numpy as np
>>> A = np.array([[-1, 5], [3, 11], [-4, 20]])
>>> mask = (A[:,0]>0) & (A[:,1] > 10)
>>> A[mask]
array([[ 3, 11]])

The idea is to express your condition using an expression as in mask .

Answer 2

You can solve this using logical indexing. I'm guessing in your example you meant that

[[-1, 5], [3, 11], [-4, 20]]

would become

[]

Since you said that the conditions are:

I want to remove all tuples where the first value is less than 0 or the second element is greater than a number , n.

>>> import numpy as np
>>> arr = np.array([[-1, 5], [3, 11], [-4, 20]])
>>> arr[~((arr[:,0] < 0) | (arr[:,1] > 10))]
array([], shape=(0, 2), dtype=int64)

Basically it's all down to expressing your logical requirement as the combination (through | and & ) of different logical masks.

The ~ takes the inverse of the mask, since you said you wanted to delete elements that match your criteria. A better example of this method working is this:

>>> import numpy as np
>>> arr = np.array([[-1, 5], [1, 1], [3, 11], [-4, 20], [2, 9]])
>>> arr[~((arr[:,0] < 0) | (arr[:,1] > 10))]
array([[1, 1], [2, 9]])