Test if a numpy array is a member of a list of numpy arrays, and remove it from the list

Question

When testing if a numpy array c is member of a list of numpy arrays CNTS:

import numpy as np

c = np.array([[[ 75, 763]],
              [[ 57, 763]],
              [[ 57, 749]],
              [[ 75, 749]]])

CNTS = [np.array([[[  78, 1202]],
                  [[  63, 1202]],
                  [[  63, 1187]],
                  [[  78, 1187]]]),
        np.array([[[ 75, 763]],
                  [[ 57, 763]],
                  [[ 57, 749]],
                  [[ 75, 749]]]),
        np.array([[[ 72, 742]],
                  [[ 58, 742]],
                  [[ 57, 741]],
                  [[ 57, 727]],
                  [[ 58, 726]],
                  [[ 72, 726]]]),
        np.array([[[ 66, 194]],
                  [[ 51, 194]],
                  [[ 51, 179]],
                  [[ 66, 179]]])]

print(c in CNTS)

I get:

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

However, the answer is rather clear: c is exactly CNTS[1], so c in CNTS should return True!

How to correctly test if a numpy array is member of a list of numpy arrays?

The same problem happens when removing:

CNTS.remove(c)

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

Application: test if an opencv contour (numpy array) is member of a list of contours, see for example Remove an opencv contour from a list of contours.

@DavidG the real problem (for the application) is not only test membership, but also remove a numpy array from a list of numpy arrays, I added more details to the question. It's not really a duplicate. — Basj
– Basj, Commented Oct 30, 2018 at 13:29
I think you can use bool(sum((map(lambda x: np.array_equal(x,c), CNTS)))) — Rudolf Morkovskyi
– Rudolf Morkovskyi, Commented Oct 30, 2018 at 13:30
@RudolfMorkovskyi. You could just use any, which would also do short-circuiting — Mad Physicist
– Mad Physicist, Commented Oct 30, 2018 at 13:31
@MadPhysicist No, those arrays are contours, i.e. list of points, it can be a rectangle (4 points), pentagon (5 points), etc. — Basj
– Basj, Commented Oct 30, 2018 at 13:39

Mad Physicist · Accepted Answer · 2018-10-30 13:56:17Z

6

You are getting the error because in essentially invokes bool(c == x) on every element x of CNTS. It's the __bool__ conversion that is raising the error:

>>> c == CNTS[1]
array([[[ True,  True]],
       [[ True,  True]],
       [[ True,  True]],
       [[ True,  True]]])

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

The same applies for removal, since it tests for equality with each element.

Containment

The solution is to use np.array_equal or apply the all method to each comparison:

any(np.array_equal(c, x) for x in CNTS)

OR

any((c == x).all() for x in CNTS)

Removal

To perform the removal, you are more interested in the index of the element than its existence. The fastest way I can think of is to iterate over the indices, using the elements of CNTS as comparison keys:

index = next((i for i, x in enumerate(CNTS) if (c == x).all()), -1)

This option short circuits quite nicely, and returns -1 as the default index rather than raising a StopIteration. You can remove the argument -1 to next if you prefer the error. If you prefer, you can replace (c == x).all() with np.array_equal(c, x).

Now you can remove as usual:

del CNTS[index]

edited Oct 30, 2018 at 13:56

answered Oct 30, 2018 at 13:29

Mad Physicist

116k29 gold badges202 silver badges292 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Basj Over a year ago

Thank you @MadPhysicist. As I mentioned in the edited question, the goal is to test membership to finally remove it (if present in the list of numpy arrays). How would you do the latter?

Mad Physicist Over a year ago

@Basj. Updated to include removal

Basj Over a year ago

Weird indeed (I upvoted all the 3 answers) @MadPhysicist!

jkr · Accepted Answer · 2018-10-30 13:40:50Z

3

This solution could work for this case:

def arrayisin(array, list_of_arrays):
    for a in list_of_arrays:
        if np.array_equal(array, a):
            return True
    return False

This function iterates over a list of arrays and tests the equality against some other array. So the usage would be:

>>> arrayisin(c, CNTS)
True

To remove the array from the list, you can get the index of the array and then use list.pop. In the function get_index, we enumerate the list of arrays, meaning we zip the indices of the list and the contents of the list. If there is a match, we return the index of the match.

def get_index(array, list_of_arrays):
    for j, a in enumerate(list_of_arrays):
        if np.array_equal(array, a):
            return j
    return None

idx = get_index(c, CNTS)  # 1
CNTS.pop(idx)

Please see the python data structures tutorial for the documentation of list.pop https://docs.python.org/3/tutorial/datastructures.html

edited Oct 30, 2018 at 13:40

answered Oct 30, 2018 at 13:29

jkr

19.6k5 gold badges48 silver badges78 bronze badges

1 Comment

Basj Over a year ago

Thank you @Jakub. As I mentioned in the question, the goal is to test membership to finally remove it (if present in the list of numpy arrays). How would you do that?

Khalil Al Hooti · Accepted Answer · 2018-10-30 14:06:25Z

1

Use del to delete the index of the list you want to remove.

del CNTS[int(np.where(list(np.array_equal(row, c) for row in CNTS))[0])]

CNTS

[array([[[  78, 1202]],

        [[  63, 1202]],

        [[  63, 1187]],

        [[  78, 1187]]]), array([[[ 72, 742]],

        [[ 58, 742]],

        [[ 57, 741]],

        [[ 57, 727]],

        [[ 58, 726]],

        [[ 72, 726]]]), array([[[ 66, 194]],

        [[ 51, 194]],

        [[ 51, 179]],

        [[ 66, 179]]])]

edited Oct 30, 2018 at 14:06

answered Oct 30, 2018 at 13:42

Khalil Al Hooti

4,6065 gold badges28 silver badges51 bronze badges

6 Comments

Mad Physicist Over a year ago

This does not short circuit, and creates an extra numpy array, just to call np.where...

Khalil Al Hooti Over a year ago

Thanks @MadPhysicist, answer modified

Mad Physicist Over a year ago

I tried running your updated code and it does not work as advertised. np.where will wrap any input in an array. Generators do not get expanded, so it gives you index 0 no matter what (since a generator object is truthy).

Mad Physicist Over a year ago

I would recommend going back to your original answer with list. While not optimal, it worked perfectly.

Mad Physicist Over a year ago

int(np.where(list(np.array_equal(row, c) for row in CNTS))[0]) -> 1, int(np.where((np.array_equal(row, c) for row in CNTS))[0]) -> 0. I'm using numpy 1.14.2. Perhaps in version 15, generators are expanded?

|

Collectives™ on Stack Overflow

Test if a numpy array is a member of a list of numpy arrays, and remove it from the list

3 Answers 3

3 Comments

1 Comment

6 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

1 Comment

6 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related