How to find duplicate elements in array using for loop in python like c/c++?

Question

i have a list with duplicate elements:

In python:

 list_a=[1,2,3,5,6,7,5,2]

 tmp=[]

 for i in list_a:
     if tmp.__contains__(i):
         print i
     else:
         tmp.append(i)

i have used the above code to found the duplicate elements in the list_a. i dont want to remove the elements form list.

But i want to use for loop here: Normally C/C++ we use like this i guess:

 for (int i=0;i<=list_a.length;i++)
     for (int j=i+1;j<=list_a.length;j++)
         if (list_a[i]==list_a[j])
             print list_a[i]

how do we use like this in Python ?

for i in list_a:
    for j in list_a[1:]:
    ....

i tried the above code .But it gets solution wrong.i dont know how to increase the value for j.pls help...

Answer 1

Use the in operator instead of calling __contains__ directly.

What you have almost works (but is O(n**2)):

for i in xrange(len(list_a)):
  for j in xrange(i + 1, len(list_a)):
    if list_a[i] == list_a[j]:
      print "duplicate:", list_a[i]

But it's far easier to use a set (roughly O(n) due to the hash table):

seen = set()
for n in list_a:
  if n in seen:
    print "duplicate:", n
  else:
    seen.add(n)

Or a dict, if you want to track locations of duplicates (also O(n)):

import collections
items = collections.defaultdict(list)
for i, item in enumerate(list_a):
  items[item].append(i)
for item, locs in items.iteritems():
  if len(locs) > 1:
    print "duplicates of", item, "at", locs

Or even just detect a duplicate somewhere (also O(n)):

if len(set(list_a)) != len(list_a):
  print "duplicate"

Answer 2

If you're looking for one-to-one mapping between your nested loops and Python, this is what you want:

n = len(list_a)
for i in range(n):
    for j in range(i+1, n):
        if list_a[i] == list_a[j]:
            print list_a[i]

The code above is not "Pythonic". I would do it something like this:

seen = set()
for i in list_a:
   if i in seen:
       print i
   else:
       seen.add(i)

Also, don't use __contains__, rather, use in (as above).

Answer 3

You could just "translate" it line by line.

c++

for (int i=0;i<=list_a.length;i++)
    for (int j=i+1;j<=list_a.length;j++)
        if (list_a[i]==list_a[j])
            print list_a[i]

Python

for i in range(0, len(list_a)):
    for j in range(i + 1, len(list_a))
        if list_a[i] == list_a[j]:
            print list_a[i]

c++ for loop:

for(int x = start; x < end; ++x)

Python equivalent:

for x in range(start, end):

Answer 4

You could always use a list comprehension:

dups = [x for x in list_a if list_a.count(x) > 1]

Answer 5

The following requires the elements of your list to be hashable (not just implementing __eq__ ). I find it more pythonic to use a defaultdict (and you have the number of repetitions for free):

import collections
l = [1, 2, 4, 1, 3, 3]
d = collections.defaultdict(int)
for x in l:
   d[x] += 1
print [k for k, v in d.iteritems() if v > 1]
# prints [1, 3]

Answer 6

Just for information, In python 2.7+, we can use Counter

import collections

x=[1, 2, 3, 5, 6, 7, 5, 2]

>>> x
[1, 2, 3, 5, 6, 7, 5, 2]

>>> y=collections.Counter(x)
>>> y
Counter({2: 2, 5: 2, 1: 1, 3: 1, 6: 1, 7: 1})

Unique List

>>> list(y)
[1, 2, 3, 5, 6, 7]

Items found more than 1 time

>>> [i for i in y if y[i]>1]
[2, 5]

Items found only one time

>>> [i for i in y if y[i]==1]
[1, 3, 6, 7]

Answer 7

A little bit more Pythonic implementation (not the most, of course), but in the spirit of your C code could be:

for i, elem in enumerate(seq):
    if elem in seq[i+1:]:
        print elem

Edit: yes, it prints the elements more than once if there're more than 2 repetitions, but that's what the op's C pseudo code does too.

Answer 8

Before Python 2.3, use dict() :

>>> lst = [1, 2, 3, 5, 6, 7, 5, 2]
>>> stats = {}
>>> for x in lst : # count occurrences of each letter:
...     stats[x] = stats.get(x, 0) + 1 
>>> print stats
{1: 1, 2: 2, 3: 1, 5: 2, 6: 1, 7: 1} # filter letters appearing more than once:
>>> duplicates = [dup for (dup, i) in stats.items() if i > 1] 
>>> print duplicates

So a function :

def getDuplicates(iterable):
    """
       Take an iterable and return a generator yielding its duplicate items.
       Items must be hashable.

       e.g :

       >>> sorted(list(getDuplicates([1, 2, 3, 5, 6, 7, 5, 2])))
       [2, 5]
    """
    stats = {}
    for x in iterable : 
        stats[x] = stats.get(x, 0) + 1
    return (dup for (dup, i) in stats.items() if i > 1)

With Python 2.3 comes set(), and it's even a built-in after than :

def getDuplicates(iterable):
    """
       Take an iterable and return a generator yielding its duplicate items.
       Items must be hashable.

       e.g :

       >>> sorted(list(getDuplicates([1, 2, 3, 5, 6, 7, 5, 2])))
       [2, 5]
    """
    try: # try using built-in set
        found = set() 
    except NameError: # fallback on the sets module
        from sets import Set
        found = Set()

    for x in iterable:
        if x in found : # set is a collection that can't contain duplicate
            yield x
        found.add(x) # duplicate won't be added anyway

With Python 2.7 and above, you have the collections module providing the very same function than the dict one, and we can make it shorter (and faster, it's probably C under the hood) than solution 1 :

import collections

def getDuplicates(iterable):
    """
       Take an iterable and return a generator yielding its duplicate items.
       Items must be hashable.

       e.g :

       >>> sorted(list(getDuplicates([1, 2, 3, 5, 6, 7, 5, 2])))
       [2, 5]
    """
    return (dup for (dup, i) in collections.counter(iterable).items() if i > 1)

I'd stick with solution 2.