Numpy: How to check if array contains certain numbers?

Question

For example: I have a = np.array([123, 412, 444]) and b = np.array([123, 321])

I want to know if a contains all the elements in b. Is there a simple operation for this? In this case that would not be true.

Do you mean unique elements? a = [3,3,4], b = [3,4,5], are all elements of a contained in b? — Akavall
– Akavall, Commented May 12, 2012 at 16:56
No I meant that all the elements in b has to be in a. The answers below where what I was looking for. Thanks — luffe
– luffe, Commented May 12, 2012 at 17:03

the wolf · Accepted Answer · 2012-05-12 17:02:50Z

14

You could alway use a set:

>>> a = numpy.array([123, 412, 444])
>>> b = numpy.array([123, 321])
>>> set(b) in set(a)
False

Or with newer versions of numpy:

>>> numpy.in1d(b,a)
array([ True, False], dtype=bool)

If you want just 'the answer' rather than an array:

>>> numpy.in1d(b,a).all()
False

Or (least desirable):

>>> numpy.array([x in a for x in b]) 
array([ True, False], dtype=bool)

Looping is slowish on numpy arrays and should be avoided.

edited May 12, 2012 at 17:02

answered May 12, 2012 at 16:56

the wolf

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1 Comment

TryPyPy Over a year ago

set(b) in set(a) probably doesn't do what you want, that would be set.is[super|sub]set.

petr · Accepted Answer · 2015-12-16 11:37:47Z

11

You can use set difference to determine what you are looking for. Numpy has a built-in function called numpy.setdiff1d(ar1, ar2):

Return the sorted, unique values in ar1 that are not in ar2.

Example for your case:

>>> a = np.array([123, 412, 444])
>>> b = np.array([123, 321])
>>> diff = np.setdiff1d(b, a)
>>> print diff
array([321])
>>> if diff.size:
>>>    print "Not passed"

So for your case, you would do a set difference you would subtract a from b and obtain an array with elements in b which are not in a. Then you can check if that was empty or not. As you can see, the output is 312, which is an entry present in a but not in b; the length of it is now larger then zero, therefore there were elements in b which were not present in a.

edited Dec 16, 2015 at 11:37

answered May 12, 2012 at 16:50

petr

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3 Comments

petr Over a year ago

@luffe if I answered your question, please mark it as an answer.

jamylak Over a year ago

@petr better to check if diff instead

max29 Over a year ago

@petr, do you mean 321 instead of 312 for array elements which are in b but not in a?

Markus Dutschke · Accepted Answer · 2021-05-27 13:05:16Z

5

that means you want to check if each element of b is contained in a. in1d does that:

from numpy import array, in1d
a = array([123, 412, 444])
b = array([123, 321])
print in1d(b, a).all()

Update from 2021: nowadays np.isin is recommended

edited May 27, 2021 at 13:05

Markus Dutschke

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answered May 12, 2012 at 16:54

mata

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Comments

cobie · Accepted Answer · 2012-05-12 16:50:52Z

-1

you could do:

 a = an_array
 b = another_array
 for i in b:
    if i not in a:
        return False
return True

answered May 12, 2012 at 16:50

cobie

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3 Comments

rubik Over a year ago

That is not valid Python. Also, a Python loop on a numpy array is very slow.

cobie Over a year ago

how exactly is that invalid python?

rubik Over a year ago

Obviously you have not run it in the interpreter. If you had done so you would have discovered your code raises a SyntaxError: 'return' outside function.

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