How do I find the duplicates in a list and create another list with them?

Question

How can I find the duplicates in a Python list and create another list of the duplicates? The list only contains integers.

Answer 1

I came across this question whilst looking in to something related - and wonder why no-one offered a generator based solution? Solving this problem would be:

>>> print list(getDupes_9([1,2,3,2,1,5,6,5,5,5]))
[1, 2, 5]

I was concerned with scalability, so tested several approaches, including naive items that work well on small lists, but scale horribly as lists get larger (note- would have been better to use timeit, but this is illustrative).

I included @moooeeeep for comparison (it is impressively fast: fastest if the input list is completely random) and an itertools approach that is even faster again for mostly sorted lists... Now includes pandas approach from @firelynx -- slow, but not horribly so, and simple. Note - sort/tee/zip approach is consistently fastest on my machine for large mostly ordered lists, moooeeeep is fastest for shuffled lists, but your mileage may vary.

Advantages

• very quick simple to test for 'any' duplicates using the same code

Assumptions

• Duplicates should be reported once only
• Duplicate order does not need to be preserved
• Duplicate might be anywhere in the list

---

Fastest solution, 1m entries:

def getDupes(c):
        '''sort/tee/izip'''
        a, b = itertools.tee(sorted(c))
        next(b, None)
        r = None
        for k, g in itertools.izip(a, b):
            if k != g: continue
            if k != r:
                yield k
                r = k

---

Approaches tested

import itertools
import time
import random

def getDupes_1(c):
    '''naive'''
    for i in xrange(0, len(c)):
        if c[i] in c[:i]:
            yield c[i]

def getDupes_2(c):
    '''set len change'''
    s = set()
    for i in c:
        l = len(s)
        s.add(i)
        if len(s) == l:
            yield i

def getDupes_3(c):
    '''in dict'''
    d = {}
    for i in c:
        if i in d:
            if d[i]:
                yield i
                d[i] = False
        else:
            d[i] = True

def getDupes_4(c):
    '''in set'''
    s,r = set(),set()
    for i in c:
        if i not in s:
            s.add(i)
        elif i not in r:
            r.add(i)
            yield i

def getDupes_5(c):
    '''sort/adjacent'''
    c = sorted(c)
    r = None
    for i in xrange(1, len(c)):
        if c[i] == c[i - 1]:
            if c[i] != r:
                yield c[i]
                r = c[i]

def getDupes_6(c):
    '''sort/groupby'''
    def multiple(x):
        try:
            x.next()
            x.next()
            return True
        except:
            return False
    for k, g in itertools.ifilter(lambda x: multiple(x[1]), itertools.groupby(sorted(c))):
        yield k

def getDupes_7(c):
    '''sort/zip'''
    c = sorted(c)
    r = None
    for k, g in zip(c[:-1],c[1:]):
        if k == g:
            if k != r:
                yield k
                r = k

def getDupes_8(c):
    '''sort/izip'''
    c = sorted(c)
    r = None
    for k, g in itertools.izip(c[:-1],c[1:]):
        if k == g:
            if k != r:
                yield k
                r = k

def getDupes_9(c):
    '''sort/tee/izip'''
    a, b = itertools.tee(sorted(c))
    next(b, None)
    r = None
    for k, g in itertools.izip(a, b):
        if k != g: continue
        if k != r:
            yield k
            r = k

def getDupes_a(l):
    '''moooeeeep'''
    seen = set()
    seen_add = seen.add
    # adds all elements it doesn't know yet to seen and all other to seen_twice
    for x in l:
        if x in seen or seen_add(x):
            yield x

def getDupes_b(x):
    '''iter*/sorted'''
    x = sorted(x)
    def _matches():
        for k,g in itertools.izip(x[:-1],x[1:]):
            if k == g:
                yield k
    for k, n in itertools.groupby(_matches()):
        yield k

def getDupes_c(a):
    '''pandas'''
    import pandas as pd
    vc = pd.Series(a).value_counts()
    i = vc[vc > 1].index
    for _ in i:
        yield _

def hasDupes(fn,c):
    try:
        if fn(c).next(): return True    # Found a dupe
    except StopIteration:
        pass
    return False

def getDupes(fn,c):
    return list(fn(c))

STABLE = True
if STABLE:
    print 'Finding FIRST then ALL duplicates, single dupe of "nth" placed element in 1m element array'
else:
    print 'Finding FIRST then ALL duplicates, single dupe of "n" included in randomised 1m element array'
for location in (50,250000,500000,750000,999999):
    for test in (getDupes_2, getDupes_3, getDupes_4, getDupes_5, getDupes_6,
                 getDupes_8, getDupes_9, getDupes_a, getDupes_b, getDupes_c):
        print 'Test %-15s:%10d - '%(test.__doc__ or test.__name__,location),
        deltas = []
        for FIRST in (True,False):
            for i in xrange(0, 5):
                c = range(0,1000000)
                if STABLE:
                    c[0] = location
                else:
                    c.append(location)
                    random.shuffle(c)
                start = time.time()
                if FIRST:
                    print '.' if location == test(c).next() else '!',
                else:
                    print '.' if [location] == list(test(c)) else '!',
                deltas.append(time.time()-start)
            print ' -- %0.3f  '%(sum(deltas)/len(deltas)),
        print
    print

The results for the 'all dupes' test were consistent, finding "first" duplicate then "all" duplicates in this array:

Finding FIRST then ALL duplicates, single dupe of "nth" placed element in 1m element array
Test set len change :    500000 -  . . . . .  -- 0.264   . . . . .  -- 0.402  
Test in dict        :    500000 -  . . . . .  -- 0.163   . . . . .  -- 0.250  
Test in set         :    500000 -  . . . . .  -- 0.163   . . . . .  -- 0.249  
Test sort/adjacent  :    500000 -  . . . . .  -- 0.159   . . . . .  -- 0.229  
Test sort/groupby   :    500000 -  . . . . .  -- 0.860   . . . . .  -- 1.286  
Test sort/izip      :    500000 -  . . . . .  -- 0.165   . . . . .  -- 0.229  
Test sort/tee/izip  :    500000 -  . . . . .  -- 0.145   . . . . .  -- 0.206  *
Test moooeeeep      :    500000 -  . . . . .  -- 0.149   . . . . .  -- 0.232  
Test iter*/sorted   :    500000 -  . . . . .  -- 0.160   . . . . .  -- 0.221  
Test pandas         :    500000 -  . . . . .  -- 0.493   . . . . .  -- 0.499  

When the lists are shuffled first, the price of the sort becomes apparent - the efficiency drops noticeably and the @moooeeeep approach dominates, with set & dict approaches being similar but lessor performers:

Finding FIRST then ALL duplicates, single dupe of "n" included in randomised 1m element array
Test set len change :    500000 -  . . . . .  -- 0.321   . . . . .  -- 0.473  
Test in dict        :    500000 -  . . . . .  -- 0.285   . . . . .  -- 0.360  
Test in set         :    500000 -  . . . . .  -- 0.309   . . . . .  -- 0.365  
Test sort/adjacent  :    500000 -  . . . . .  -- 0.756   . . . . .  -- 0.823  
Test sort/groupby   :    500000 -  . . . . .  -- 1.459   . . . . .  -- 1.896  
Test sort/izip      :    500000 -  . . . . .  -- 0.786   . . . . .  -- 0.845  
Test sort/tee/izip  :    500000 -  . . . . .  -- 0.743   . . . . .  -- 0.804  
Test moooeeeep      :    500000 -  . . . . .  -- 0.234   . . . . .  -- 0.311  *
Test iter*/sorted   :    500000 -  . . . . .  -- 0.776   . . . . .  -- 0.840  
Test pandas         :    500000 -  . . . . .  -- 0.539   . . . . .  -- 0.540