How can I implement matlabs ``ismember()`` command in Python?

Question

here is my problem: I would like to create a boolean matrix B that contains True everywhere that matrix A has a value contained in vector v. One inconvenient solution would be:

import numpy as np
>>> A = np.array([[0,1,2], [1,2,3], [2,3,4]])
array([[0, 1, 2],
       [1, 2, 3],
       [2, 3, 4]])
>>> v = [1,2]
>>> B = (A==v[0]) + (A==v[1]) # matlab: ``B = ismember(A,v)``
array([[False,  True,  True],
       [ True,  True, False],
       [ True, False, False]], dtype=bool)

Is there maybe a solution that would be more convenient if A and v would have more values?

Cheers!

balpha · Accepted Answer · 2009-08-13 17:19:33Z

4

I don't know much numpy, be here's a raw python one:

>>> A = [[0,1,2], [1,2,3], [2,3,4]]
>>> v = [1,2]
>>> B = [map(lambda val: val in v, a) for a in A]
>>>
>>> B
[[False, True, True], [True, True, False], [True, False, False]]

Edit: As Brooks Moses notes and some simple timing seems to show, this one is probably be better:

>>> B = [ [val in v for val in a] for a in A]

edited Aug 13, 2009 at 17:19

answered Aug 13, 2009 at 16:37

balpha

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3 Comments

Brooks Moses Over a year ago

Naive question: Why the map(lambda...) syntax, rather than just [(val in v) for v in a]? Is there a meaningful difference in this case?

balpha Over a year ago

@Brooks Moses: You're right, I guess there's not, and the double comprehension even seems to be a little faster (I only did some naive timing, though). Edited.

balpha Over a year ago

Actually, for small v and large A, it's a lot faster (factor 2).

Ants Aasma · Accepted Answer · 2009-08-13 18:38:45Z

3

Using numpy primitives:

>>> import numpy as np
>>> A = np.array([[0,1,2], [1,2,3], [2,3,4]])
>>> v = [1,2]
>>> print np.vectorize(lambda x: x in v)(A)
[[False  True  True]
 [ True  True False]
 [ True False False]]

For non-tiny inputs convert v to a set first for a large speedup.

To use numpy.setmember1d:

Auniq, Ainv = np.unique1d(A, return_inverse=True)
result = np.take(np.setmember1d(Auniq, np.unique1d(v)), Ainv).reshape(A.shape)

edited Aug 13, 2009 at 18:38

answered Aug 13, 2009 at 16:44

Ants Aasma

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2 Comments

Alex Martelli Over a year ago

This is broken: see the 2nd row, rightmost column -- what's that True doing there? It corresponds to 3 in A which is NOT in v. Alas, setmember1d does NOT support correctly arrays with duplicates.

Ants Aasma Over a year ago

Corrected, setmember1d documentation could be clearer on this.

Alex Martelli · Accepted Answer · 2009-08-13 18:39:09Z

3

Alas, setmember1d as it exists in numpy is broken when either array has duplicated elements (as A does here). Download this version, call it e.g sem.py somewhere on your sys.path, add to it a first line import numpy as nm, and THEN this finally works:

>>> import sem
>>> print sem.setmember1d(A.reshape(A.size), v).reshape(A.shape)
[[False True True]
 [True True False]
 [True False False]]

Note the difference wrt @Aants' similar answer: this version has the second row of the resulting bool array correct, while his version (using the setmember1d that comes as part of numpy) incorrectly has the second row as all Trues.

answered Aug 13, 2009 at 18:39

Alex Martelli

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Comments

Neil Crighton · Accepted Answer · 2011-05-23 19:20:24Z

2

Since Numpy version 1.4 there's a new function, in1d() that's the equivalent of ismember() in matlab: http://docs.scipy.org/doc/numpy-1.6.0/reference/generated/numpy.in1d.html. But as ars points out, it only returns a 1d array.

answered May 23, 2011 at 19:20

Neil Crighton

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Comments

ars · Accepted Answer · 2009-08-13 16:45:24Z

1

I think the closest you'll get is numpy.ismember1d, but it won't work well with your example. I think your solution (B = (A==v[0]) + (A==v[1])) may actually be the best one.

answered Aug 13, 2009 at 16:45

ars

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Comments

Mark Rushakoff · Accepted Answer · 2009-08-13 17:03:35Z

1

Here's a naive one-liner:

[any (value in item for value in v) for item in A]

Sample output:

>>> A = ( [0,1,2], [1,2,3], [2,3,4] )
>>> v = [1,2]
>>> [any (value in item for value in v) for item in A]
[True, True, True]
>>> v = [1]
>>> [any (value in item for value in v) for item in A]
[True, True, False]

It's a very Pythonic approach, but I'm certain it won't scale well on large arrays or vectors because Python's in operator is a linear search (on lists/tuples, at least).

As Brooks Moses pointed out in the below comment, the output should be a 3x3 matrix. That's why you give sample output in your questions. (Thanks Brooks)

>>> v=[1,2]
>>> [ [item in v for item in row] for row in A]
[[False, True, True], [True, True, False], [True, False, False]]
>>> v=[1]
>>> [ [item in v for item in row] for row in A]
[[False, True, False], [True, False, False], [False, False, False]]

edited Aug 13, 2009 at 17:03

answered Aug 13, 2009 at 16:35

Mark Rushakoff

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1 Comment

Brooks Moses Over a year ago

You've got a good answer to the wrong question, I think -- you want A to be a 3x3 array, and return a 3x3 truth value for each of those 9 elements. Thus, adjusting your answer slightly: [[(item in v) for item in row] for row in A] works fine. I'm also curious why you expect this would be slow.

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