Filtering numpy array of tuples yields a ValueError

Question

I have a numpy array of tuples and wish to filter them, but nothing seems to work

 possible_points = np.random.normal((0,0),0.01,size=(100,2))
 possible_points = possible_points[possible_points > (0,0) and possible_points < (1,1)]

This yields an error

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

I suspect the problem is somehow related with the following. When I do, for example,

possible_points = [possible_points > (0,0)]

possible_points becomes something like

[array([[False, False],
       [ True, False],
       [ True, False],
       [ True,  True],
       [ True,  True],
       [ True,  True],
       [ True,  True],
       [False, False],
       [ True, False],
    ...

From my understanding, it should return True if possible_points > (0,0) and False otherwise, but it's evaluating each tuple's coordinate individually. There are even some points that become [True, False]. If I do

possible_points = possible_points[possible_points > (0,0)]

I stop having tuples, and have something like

array([3.36687201e-03, 7.97227921e-04, 3.44875252e-03, 8.79364101e-03,....])

And I suspect this is what yields the ValueErros.

Can anyone help me understand what I'm doing wrong and help me achieve my goal?

@MichaelSzczesny, I wish to have the elements of possible points that are bigger than (0,0) and smaller than (1,1) — João Dionísio
– João Dionísio, Commented Oct 31, 2020 at 13:01

Dani Mesejo · Accepted Answer · 2020-10-31 13:01:30Z

2

IIUC, you should use the & operator together with all across the columns (axis=1), like this:

import numpy as np

# set a seed for reproducibility
np.random.seed(42)

possible_points = np.random.normal((0, 0), 0.01, size=(100, 2))

# create the filtering mask
mask = ((possible_points > (0, 0)) & (possible_points < (1, 1))).all(axis=1)

# filter
possible_points = possible_points[mask]

print(possible_points)

Output

[[0.00647689 0.0152303 ]
 [0.01579213 0.00767435]
 [0.00738467 0.00171368]
 [0.01031    0.0093128 ]
 [0.00331263 0.00975545]
 [0.00812526 0.0135624 ]
 [0.00361396 0.01538037]
 [0.00915402 0.00328751]
 [0.00097078 0.00968645]
 [0.0029612  0.00261055]
 [0.01886186 0.00174578]
 [0.0006023  0.02463242]
 [0.01142823 0.00751933]
 [0.00586857 0.02190456]
 [0.0022746  0.01307143]
 [0.00259883 0.00781823]
 [0.00521942 0.00296985]
 [0.00250493 0.00346448]
 [0.01865775 0.00473833]
 [0.00963376 0.00412781]
 [0.0082206  0.01896793]
 [0.00276691 0.00827183]
 [0.00013002 0.01453534]
 [0.00173181 0.00385317]]

answered Oct 31, 2020 at 13:01

Dani Mesejo

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1 Comment

João Dionísio Over a year ago

Yes, this does the trick, thank you! I'll accept the answer when I can. I'll also try to investigate why that makes sense.

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