How to create a Boolean array from an array of indices in numpy?

Question

Given I have an multidimensional array of indices, how do I create a Boolean array from these? For the 1D case it would look like this:

a = [1,5,6]
b = somefunction(total_array_length=10, a)
>>> [False, True, False, False, False, True, True, False, False, False]

For the 2D case it would look like this:

a = [[1,3],[4,2]]
b = somefunction(total_array_length=5, a)
>>> [[False, True, False, True, False], [False, False, True, False, True]]

I want to use this to create a mask for an array. I have a multi dimensional array of dimension 8 and for the last axis I can find the indices of the elements I want to keep. In other words, I have an 8D array in which the last axis consists all indices I want to keep in the original array. Does anyone know how to do this?

In the function above the total_array_length would be equal to the length of the original array.

So how would I do this for an array of shape (23,5,76,32,1,3,8,9) with indices array of shape (23,5,76,32,1,3,8,4)? Note 4<9 but other than that it has the same dimensions.

a.shape = (23,5,76,32,1,3,8,4) 
b = somefunction(total_array_length=9, a)
b.shape =(23,5,76,32,1,3,8,9)

It's not clear what total_array_length is and why the shape of b does not match the shape of a in either example. Also it's unclear what you are trying to say about the last axis of your 8d array. It doesn't sound analogous to your examples. — Imran
– Imran, Commented Dec 24, 2017 at 6:08
Please just try writing the code and post it if you are stuck. Don't ask us to start from scratch. — John Zwinck
– John Zwinck, Commented Dec 24, 2017 at 7:02

hpaulj · Accepted Answer · 2017-12-24 07:35:10Z

9

For the first case:

In [23]: a = [1,5,6]
In [24]: b = np.zeros(10, dtype=bool)
In [25]: b[a] = True
In [26]: b
Out[26]: array([False,  True, False, False, False,  True,  True, False, False, False], dtype=bool)

An equally simple, and possibly as fast, list version:

In [27]: [True if i in a else False for i in range(10)]
Out[27]: [False, True, False, False, False, True, True, False, False, False]

And for the list of lists just nest that list comprehension:

In [34]: [[True if i in a1 else False for i in range(5)] for a1 in a]
Out[34]: [[False, True, False, True, False], [False, False, True, False, True]]

An array version is of this second case is:

In [41]: a = [[1,3],[4,2]]
In [42]: b = np.zeros((len(a),5), bool)
In [43]: b[[[0],[1]],a]
Out[43]: 
array([[False, False],
       [False, False]], dtype=bool)
In [44]: b[[[0],[1]],a]=True
In [45]: b
Out[45]: 
array([[False,  True, False,  True, False],
       [False, False,  True, False,  True]], dtype=bool)

This, though, will only work if the sublists of a are all the same length. If they differ, I think we'll have to work with a flattened version of this. In effect turn the 2d case into the original 1d.

For a ragged a, the list version is still easy:

In [49]: a = [[1,2,3],[4,2],[1]]
In [50]: [[True if i in a1 else False for i in range(5)] for a1 in a]
Out[50]: 
[[False, True, True, True, False],
 [False, False, True, False, True],
 [False, True, False, False, False]]

But an array version is a bit more involved:

Construct 2 arrays of indices, for rows and columns:

In [53]: a0 = np.repeat(np.arange(3),[len(i) for i in a])
In [54]: a0
Out[54]: array([0, 0, 0, 1, 1, 2])
In [55]: a1 = np.hstack(a)
In [56]: a1
Out[56]: array([1, 2, 3, 4, 2, 1])

Get the equivalent raveled (1d) indexing:

In [57]: np.ravel_multi_index((a0,a1),(3,5))
Out[57]: array([ 1,  2,  3,  9,  7, 11], dtype=int32)

Apply it to a 2d array via flat:

In [58]: b = np.zeros((3,5),bool)
In [59]: b.flat[Out[57]] = True
In [60]: b
Out[60]: 
array([[False,  True,  True,  True, False],
       [False, False,  True, False,  True],
       [False,  True, False, False, False]], dtype=bool)

edited Dec 24, 2017 at 7:35

answered Dec 24, 2017 at 7:17

hpaulj

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3 Comments

Perm. Questiin Over a year ago

How would I do this for an array of shape (23,5,76,32,1,3,8,9) with indices array of shape (23,5,76,32,1,3,8,4)? Note 4<9 but other than that it has the same dimensions.

hpaulj Over a year ago

Try reshaping the arrays into (n,4) and (n,9). Do the mapping (eg as 44). Then reshape back.

Gregory Morse Over a year ago

True if i in a1 else False is the same as i in a1 and it would be faster and more readable to use the latter.

kuzand · Accepted Answer · 2020-02-19 07:28:34Z

2

For 1D:

n = 10
inds_all = np.arange(n)
inds = [1, 5, 6]

b = np.isin(inds_all, inds)
# array([False, True, False, False, False, True, True, False, False, False])

For 2D it's a little bit trickier, especially when the list of list-indices is of uneven sizes:

n = 5
inds_all = np.arange(len(inds)*n).reshape(len(inds), n)
inds = [[1,3], [4,2], [3]]
inds = np.concatenate(inds) + np.repeat(inds_all[:,0], list(map(len, inds)))

b = np.isin(inds_all, inds)
# array([[False,  True, False,  True, False],
#        [False, False,  True, False,  True],
#        [False, False, False,  True, False]])

edited Feb 19, 2020 at 7:28

answered Feb 19, 2020 at 0:02

kuzand

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