Determining duplicate values in an array

Question

Suppose I have an array

a = np.array([1, 2, 1, 3, 3, 3, 0])

How can I (efficiently, Pythonically) find which elements of a are duplicates (i.e., non-unique values)? In this case the result would be array([1, 3, 3]) or possibly array([1, 3]) if efficient.

I've come up with a few methods that appear to work:

Masking

m = np.zeros_like(a, dtype=bool)
m[np.unique(a, return_index=True)[1]] = True
a[~m]

Set operations

a[~np.in1d(np.arange(len(a)), np.unique(a, return_index=True)[1], assume_unique=True)]

This one is cute but probably illegal (as a isn't actually unique):

np.setxor1d(a, np.unique(a), assume_unique=True)

Histograms

u, i = np.unique(a, return_inverse=True)
u[np.bincount(i) > 1]

Sorting

s = np.sort(a, axis=None)
s[:-1][s[1:] == s[:-1]]

Pandas

s = pd.Series(a)
s[s.duplicated()]

Is there anything I've missed? I'm not necessarily looking for a numpy-only solution, but it has to work with numpy data types and be efficient on medium-sized data sets (up to 10 million in size).

---

Conclusions

Testing with a 10 million size data set (on a 2.8GHz Xeon):

a = np.random.randint(10**7, size=10**7)

The fastest is sorting, at 1.1s. The dubious xor1d is second at 2.6s, followed by masking and Pandas Series.duplicated at 3.1s, bincount at 5.6s, and in1d and senderle's setdiff1d both at 7.3s. Steven's Counter is only a little slower, at 10.5s; trailing behind are Burhan's Counter.most_common at 110s and DSM's Counter subtraction at 360s.

I'm going to use sorting for performance, but I'm accepting Steven's answer because the performance is acceptable and it feels clearer and more Pythonic.

Edit: discovered the Pandas solution. If Pandas is available it's clear and performs well.

Answer 1

If the array is a sorted numpy array, then just do:

a = np.array([1, 2, 2, 3, 4, 5, 5, 6])
rep_el = a[np.diff(a) == 0]

Answer 2

People have already suggested Counter variants, but here's one which doesn't use a listcomp:

>>> from collections import Counter
>>> a = [1, 2, 1, 3, 3, 3, 0]
>>> (Counter(a) - Counter(set(a))).keys()
[1, 3]

[Posted not because it's efficient -- it's not -- but because I think it's cute that you can subtract Counter instances.]

Answer 3

Here's another approach using set operations that I think is a bit more straightforward than the ones you offer:

>>> indices = np.setdiff1d(np.arange(len(a)), np.unique(a, return_index=True)[1])
>>> a[indices]
array([1, 3, 3])

I suppose you're asking for numpy-only solutions, since if that's not the case, it's very difficult to argue with just using a Counter instead. I think you should make that requirement explicit though.

Answer 4

I think this is most clear done outside of numpy. You'll have to time it against your numpy solutions if you are concerned with speed.

>>> import numpy as np
>>> from collections import Counter
>>> a = np.array([1, 2, 1, 3, 3, 3, 0])
>>> [item for item, count in Counter(a).iteritems() if count > 1]
[1, 3]

note: This is similar to Burhan Khalid's answer, but the use of iteritems without subscripting in the condition should be faster.

Answer 5

For Python 2.7+

>>> import numpy
>>> from collections import Counter
>>> n = numpy.array([1,1,2,3,3,3,0])
>>> [x[1] for x in Counter(n).most_common() if x[0] > 1]
[3, 1]

Answer 6

I'm adding my solution to the pile for this 3 year old question because none of the solutions fit what I wanted or used libs besides numpy. This method finds both the indices of duplicates and values for distinct sets of duplicates.

import numpy as np

A = np.array([1,2,3,4,4,4,5,6,6,7,8])

# Record the indices where each unique element occurs.
list_of_dup_inds = [np.where(a == A)[0] for a in np.unique(A)]

# Filter out non-duplicates.
list_of_dup_inds = filter(lambda inds: len(inds) > 1, list_of_dup_inds)

for inds in list_of_dup_inds: print inds, A[inds]
# >> [3 4 5] [4 4 4]
# >> [7 8] [6 6]