How to find duplicate elements in array using for loop in Python?

Question

I have a list with duplicate elements:

 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 find the duplicate elements in the list_a. I don't want to remove the elements from 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 don't know how to increase the value for j.

Answer 1

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]

• Roger Pate: [n for n, i in y.iteritems() if i > 1] instead, and i == 1.
• LeMiz: ...but why the list(y), isn't Counter iterable ?
• YOU: @Roger Pate, thanks, yours is no need to do dict lookup, it could be better.
• Lie Ryan: The only drawbacks of Counter is that it doesn't terminate early if duplicates are found early in a big list and it doesn't work for infinite iterator.
• luistm: My solution: for k, v in x.most_common(): if v > 1: print k

Answer 2

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 3

You could always use a list comprehension:

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

• Alok Singhal: This traverses the list once for each element (Although, OP's code is O(N**2), too).
• Evan Fosmark: Yeah, I understood it's inefficient. If the OP is looking for that, he should go with Roger's answers for sure.
• dMb: I think this is slightly more efficient: [x for i,x in enumerate(list_a) if list_a[i:].count(x) > 1]
• dmitko: this will return a list with duplicates as well as list_a.count(x) > 1 will return True for each occurence of the element. I'd use set() to get unique duplicates

Answer 4

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.

Answer 5

You can use this function to find duplicates:

def get_duplicates(arr):
 dup_arr = arr[:]
 for i in set(arr):
 dup_arr.remove(i) 
 return list(set(dup_arr))

Examples

print get_duplicates([1,2,3,5,6,7,5,2])

[2, 5]

print get_duplicates([1,2,1,3,4,5,4,4,6,7,8,2])

[1, 2, 4]

Answer 6

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 7

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 8

Using only itertools, and works fine on Python 2.5

from itertools import groupby
list_a = sorted([1, 2, 3, 5, 6, 7, 5, 2])
result = dict([(r, len(list(grp))) for r, grp in groupby(list_a)])

Result:

{1: 1, 2: 2, 3: 1, 5: 2, 6: 1, 7: 1}

Answer 9

It looks like you have a list (list_a) potentially including duplicates, which you would rather keep as it is, and build a de-duplicated list tmp based on list_a. In Python 2.7, you can accomplish this with one line:

tmp = list(set(list_a))

Comparing the lengths of tmp and list_a at this point should clarify if there were indeed duplicate items in list_a. This may help simplify things if you want to go into the loop for additional processing.

Answer 10

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):

• Bite code: You should not accept this answer. Yes, it's valid code, but it's not the way you should code in Python. Don't code Python like C/C++, or Java. They are not the same languages, and are not meant to be used the same way.
• Mizipzor: I agree with e-satis, although the the question specifically tries to compare the routine to C/C++ we should try to nudge it in the right direction.

Answer 11

Just quick and dirty,

list_a=[1,2,3,5,6,7,5,2] 
holding_list=[]
for x in list_a:
 if x in holding_list:
 pass
 else:
 holding_list.append(x)
print holding_list

Output [1, 2, 3, 5, 6, 7]

Answer 12

Using numpy:

import numpy as np
count,value = np.histogram(list_a,bins=np.hstack((np.unique(list_a),np.inf)))
print 'duplicate value(s) in list_a: ' + ', '.join([str(v) for v in value[count>1]])

Answer 13

In case of Python3 and if you two lists

def removedup(List1,List2):
 List1_copy = List1[:]
 for i in List1_copy:
 if i in List2:
 List1.remove(i)
List1 = [4,5,6,7]
List2 = [6,7,8,9]
removedup(List1,List2)
print (List1)

Answer 14

Granted, I haven't done tests, but I guess it's going to be hard to beat pandas in speed:

 pd.DataFrame(list_a, columns=["x"]).groupby('x').size().to_dict()

Answer 15

You can use:

b=['E', 'P', 'P', 'E', 'O', 'E']
c={}
for i in b:
 value=0
 for j in b:
 if(i == j):
 value+=1
 c[i]=value
print(c)

Output:

{'E': 3, 'P': 2, 'O': 1}

Answer 16

Find duplicates in the list using loops, conditional logic, logical operators, and list methods

some_list = ['a','b','c','d','e','b','n','n','c','c','h',]
duplicates = [] 
for values in some_list:
 if some_list.count(values) > 1:
 if values not in duplicates:
 duplicates.append(values)
print("Duplicate Values are : ",duplicates) 
 

• Navi: How to get this without using any python library such as count?@Zaheer

Answer 17

Finding the number of repeating elements in a list:

myList = [3, 2, 2, 5, 3, 8, 3, 4, 'a', 'a', 'f', 4, 4, 1, 8, 'D']
listCleaned = set(myList)
for s in listCleaned:
 count = 0
 for i in myList:
 if s == i :
 count += 1
 print(f'total {s} => {count}')

Answer 18

Try like this:

list_a=[1,2,3,5,6,7,5,2]
unique_values = []
duplicates = []
for i in list_a:
 if i not in unique_values:
 unique_values.append(i)
 else:
 found = False
 for x in duplicates:
 if x.get("key") == i:
 found = True
 if found:
 x["occurrence"] += 1
 else:
 duplicates.append({
 "key": i,
 "occurrence": 1
 })

Answer 19

some_string= list(input("Enter any string:\n"))
count={}
dup_count={}
for i in some_string:
 if i not in count:
 count[i]=1
 else:
 count[i]+=1
 dup_count[i]=count[i]
print("Duplicates of given string are below:\n",dup_count)

Answer 20

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.

• Bite code: You must sort before doing that. Use sorted. What's more, you will print the same duplicate several times if there is more than one of the same.
• mthurlin: This will print the same element multiple times if it occurs more than 2 times in the list.
• fortran: Have you guys bothered to read the op's code? It does the exactly the same. @e-satis There's no need to sort, maybe you meant something like [k for k, it in itertools.groupby(sorted(l)) if len(list(it)) > 1] ?