96

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.

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6
  • 2
    Could you explain why the sorting solution works? I tried it out but for some reason I don't really get it. Commented Jul 18, 2013 at 16:06
  • 2
    @Markus if you sort an array, any duplicate values are adjacent. You then use a boolean mask to take only those items that are equal to the previous item. Commented Jul 18, 2013 at 16:53
  • 1
    Shouldn't it be s[:-1][ s[1:] == s[:-1] ]? I get an IndexError otherwise, the boolean mask being one element shorter than the s-array.... Commented Nov 13, 2017 at 16:22
  • @snake_charmer I think earlier versions of numpy were more forgiving in this regard. I'll fix it, thanks. Commented Nov 13, 2017 at 17:43
  • pandas seems to have improved the performance of some underlying methods. On my machine, pandas is only 29% slower than the sorting method. The method proposed by Mad Physicist is 17% slower than sorting. Commented Oct 4, 2019 at 11:14

11 Answers 11

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94

As of numpy version 1.9.0, np.unique has an argument return_counts which greatly simplifies your task:

u, c = np.unique(a, return_counts=True)
dup = u[c > 1]

This is similar to using Counter, except you get a pair of arrays instead of a mapping. I'd be curious to see how they perform relative to each other.

It's probably worth mentioning that even though np.unique is quite fast in practice due to its numpyness, it has worse algorithmic complexity than the Counter solution. np.unique is sort-based, so runs asymptotically in O(n log n) time. Counter is hash-based, so has O(n) complexity. This will not matter much for anything but the largest datasets.

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9 Comments

...and on a very large dataset Counter's per-object space overhead can be prohibitive.
@TLW. Presumably the dataset is already in memory. Counter's hashtable is just an array with maybe some additional storage for buckets. Objects are stored by reference, not copy.
@MadPhysicist - You should look at the size of Python objects. The per-object overhead may surprise you.
numpy arrays don't have said overhead because they directly store the data - it's one of the reasons why numpy arrays can be quirky to work with, but also means that e.g. a billion-element uint8 array takes ~1GB instead of the ~8B it would be if the elements were Python objects.
@TLW. Gotcha. Converting a packed numpy array is indeed going to be expensive. I completely forgot that this or exclusively about numpy and assumed the data would be in memory as python objects ahead of time. I'm familiar with the python integer format and interning.
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38

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).items() if count > 1]
[1, 3]

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

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1 Comment

Note: Counter(a).items() has to be used in python 3
12

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.]

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1 Comment

More efficient not to recompute set: c = Counter(a); result = (c - Counter(c.keys())).keys()
7

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]
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1 Comment

shouldn't x[0] > 1 be x[1] > 1? the latter x represents the frequency.
6

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.

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3 Comments

I see it as a wart on this approach is that the 3 is repeated while the 1 is not. It would be nice to have it one way or the other. (This is not a criticism of your answer so much as of the original approach by the OP.)
@StevenRumbalski, yeah, I see what you mean. My sense is that the repeated 3 makes sense if what's really needed is a mask rather than a list of items; if what's needed is a list of items, then I agree that not having repeated items is better.
I'm not opposed to using Counter, but I am concerned about efficiency and compatibility.
5

If a is made up of small integers you can use numpy.bincount directly:

import numpy as np

a = np.array([3, 2, 2, 0, 4, 3])
counts = np.bincount(a)
print np.where(counts > 1)[0]
# array([2, 3])

This is very similar your "histogram" method, which is the one I would use if a was not made up of small integers.

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Comments

5

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]
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1 Comment

a[1:][np.diff(a) == 0], no?
3

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]
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1 Comment

Three years later still, and you can use the return_counts argument to unique for this too. See my answer.
3 
>>> import numpy as np

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

>>> uniques, uniq_idx, counts = np.unique(a,return_index=True,return_counts=True)
>>> duplicates = a[ uniq_idx[counts>=2] ]  # <--- Get duplicates

If you also want to get the orphans:

>>> orphans = a[ uniq_idx[counts==1] ]
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Comments

1

Combination of Pandas and Numpy (Utilizing value_counts():

import pandas as pd
import numpy as np

arr=np.array(('a','b','b','c','a'))
pd.Series(arr).value_counts()

OUTPUT:

a    2
b    2
c    1
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Comments

1

Using the subtract method of a counter circumvents creating a second counter like in DSM's answer and to get only the positive counts (ie: duplicates) use the unary + operator on the resulting Counter

a = [1, 2, 1, 3, 3, 3, 0]

c = Counter(a)
c.subtract(c.keys())
dupes = (+c).keys()

I found this to be the best performing solution in my testing

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2 Comments

What is d in this case?
@MadPhysicist Thank you for spotting my mistake I have edited the answer.

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