Get first index in array where relation is true

Question

How can I get the last index of the element in a where b > a when a and b have different length using numpy.

For instance, for the following values:

>>> a = np.asarray([10, 20, 30, 40])
>>> b = np.asarray([12, 25])

I would expect a result of [0, 1] (0.. because 12 > 10 -> index 0 in a; 1.. because 25 > 20 -> index 1 in a). Obviously, the length of the result vector should equal the length of b (and the values of the result list should be less than the length of a (as they refer to the indices in a)).

Another test is for b = np.asarray([12, 25, 31, 9, 99]) (same a as above), the result should be array([ 0, 1, 2, -1, 3]).

@Kasramvd: You're right. I didn't describe this properly. Let me rectify the question... — orange
– orange, Commented Jun 28, 2018 at 6:35

Nils Werner · Accepted Answer · 2018-06-28 08:08:25Z

3

A vectorized solution:

Remember that you can compare all elements in b with all elements in a using broadcasting:

b[:, None] > a
# array([[ True, False, False, False],     # b[0] > a[:]
#        [ True,  True, False, False]])    # b[1] > a[:]

And now find the index of the last True value in each row, which equals to the first False value in each row, minus 1

np.argmin((b[:, None] > a), axis=1) - 1
# array([0, 1])

Note that there might be an ambiguity as to what a returned value of -1 means. It could mean

b[x] was larger than all elements in a, or
b[x] was not larger than any element in a

In our data, this means

a = np.asarray([10, 20, 30, 40])
b = np.asarray([9, 12, 25, 39, 40, 41, 50])

mask = b[:, None] > a
# array([[False, False, False, False],   # 9 is smaller than a[:], case 2
#        [ True, False, False, False],
#        [ True, False, False, False],
#        [ True,  True,  True, False],
#        [ True,  True,  True, False],
#        [ True,  True,  True,  True],   # 41 is larger than a[:], case 1
#        [ True,  True,  True,  True]])  # 50 is larger than a[:], case 1

So for case 1 we need to find rows with all True values:

is_max = np.all(mask, axis=1)

And for case 2 we need to find rows with all False values:

none_found = np.all(~mask, axis=1)

This means we can use the is_max to find and replace all case 1 -1 values with a positive index

mask = b[:, None] > a
is_max = np.all(mask, axis=1)
# array([False, False, False, False, False,  True,  True])
idx = np.argmin(mask, axis=1) - 1
# array([-1,  0,  0,  2,  2, -1, -1])
idx[is_max] = len(a) - 1
# array([-1,  0,  0,  2,  2,  3,  3])

However be aware that the index -1 has a meaning: Just like 3 it already means "the last element". So if you want to use idx for indexing, keeping -1 as an invalid value marker may cause trouble down the line.

edited Jun 28, 2018 at 8:08

answered Jun 28, 2018 at 6:53

Nils Werner

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6 Comments

orange Over a year ago

Awesome ;-)! Thanks.

orange Over a year ago

Any idea how I can handle inputs (b) that are smaller than all values in a, i.e. for which there is no index? For b = np.asarray([12, 25, 31, 9]) the result is array([0, 1, 2, 3]) which is the same as if the last value in b (9) was > 40.

orange Over a year ago

Your last edit provided what I need. The index is now -1 which isn't ambiguous any more.

Nils Werner Over a year ago

the index -1 is "the last index". Maybe you should mark too small elements differently

Nils Werner Over a year ago

I have added some explanation what -1 means, and a way of detecting what condition caused a -1 to be returned.

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ᴀʀᴍᴀɴ · Accepted Answer · 2018-06-28 06:19:53Z

0

Works even a has shorter length than b , first choose shorter list length then check if its has smaller numbers element wise :

[i for i in range(min(len(a),len(b))) if min(a, b, key=len)[i] > max(a, b, key=len)[i]]
# [0, 1]

answered Jun 28, 2018 at 6:19

ᴀʀᴍᴀɴ

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Comments

Sunitha · Accepted Answer · 2018-06-28 06:49:14Z

0

You can zip a and b to combine them and then enumerate to iterate it with its index

[i for i,(x,y) in enumerate(zip(a,b)) if y>x]
# [0, 1]

answered Jun 28, 2018 at 6:49

Sunitha

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RohithS98 · Accepted Answer · 2018-06-28 06:53:59Z

0

np.asarray([i for i in range(len(b)) if b[i]>a[i]])

This should give you the answer. Also the length does not have to be same as that of either a or b.

edited Jun 28, 2018 at 6:53

answered Jun 28, 2018 at 6:14

RohithS98

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