How to check if an object is in an element list of a numpy array?

Question

For example, I have a list l.

l = np.array([[1,2], [3], [2,4]])

I hope to check which elements of the array contain 2.

By using a list comprehension, I can achieve it easily but not efficiently.

result = [2 in element for element in l]

Can I use numpy to get result more efficiently.

Thanks.

Could there be more than one occurance of the search number 2 in any of the lists? — Divakar
– Divakar, Commented Jun 23, 2017 at 7:40
Short answer: No. Look at the type of l: array([[1, 2], [3], [2, 4]], dtype=object). Numpy has no way of operating on object arrays the same way it would floats and ints. — cs95
– cs95, Commented Jun 23, 2017 at 7:41
@liuyihe But the sample in the question is only looking for one number 2. Could you post a more representative sample case? — Divakar
– Divakar, Commented Jun 23, 2017 at 8:16
@Divakar Sorry, I didn't express clearly. In fact you have solved my problem and do not need to worry about my comment above. What I need to do next is just: for item in items: in_eachlist(l, item) and assert len(items) == 1999. Thanks again. — Yihe
– Yihe, Commented Jun 23, 2017 at 11:19

Divakar · Accepted Answer · 2017-06-23 07:50:09Z

1

Here's one approach -

def in_eachlist(l, search_num):
    mask = np.concatenate(l) == search_num
    lens = [len(i) for i in l]
    return np.logical_or.reduceat(mask,np.concatenate(([0], np.cumsum(lens[:-1]))) )

Basically, we are getting a 1D array from the input array of lists and comparing against the search number, giving us a mask. Then, we check if there's any True value within each interval with np.logical_or.reduceat (Thanks to @Daniel F on improvement here as I had used np.add.reduceat earlier and then looked for any sum > 1), giving us the desired output.

Sample run -

In [41]: l
Out[41]: array([[1, 2], [3], [2, 4]], dtype=object)

In [42]: in_eachlist(l,2)
Out[42]: array([ True, False,  True], dtype=bool)

In [43]: in_eachlist(l,3)
Out[43]: array([False,  True, False], dtype=bool)

In [44]: in_eachlist(l,4)
Out[44]: array([False, False,  True], dtype=bool)

edited Jun 23, 2017 at 7:50

answered Jun 23, 2017 at 7:43

Divakar

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4 Comments

Daniel F Over a year ago

Why add.reduceat()>0 when you could do logical_or.reduceat()?

Divakar Over a year ago

@DanielF Thanks, major improvement there! Edited.

Yihe Over a year ago

Thanks! The operation time is reduced from 2.4s to 400ms.

Daniel F Over a year ago

@liuyihe, depending on the output you want after checking the 1999 values you mentioned, it is possible to make this even faster (thinking np.in1d, although @Divakar might have even fancier tricks in mind.)

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