Identify duplicate values in a list in Python

Question

Is it possible to get which values are duplicates in a list using python?

I have a list of items:

    mylist = [20, 30, 25, 20]

I know the best way of removing the duplicates is set(mylist), but is it possible to know what values are being duplicated? As you can see, in this list the duplicates are the first and last values. [0, 3].

Is it possible to get this result or something similar in python? I'm trying to avoid making a ridiculously big if elif conditional statement.

Answer 1

You could identify these items using the iteration_utilities library:

from iteration_utilities import duplicates
list(duplicates(mylist))

Output: [20]

Note that if 20 appeared 3 times in your original list, the output would instead be [20, 20].

Answer 2

def checkduplicate(lists): 
 a = []
 for i in lists:
    if  i in a:
        pass   
    else:
        a.append(i)
 return i          
            
print(checkduplicate([1,9,78,989,2,2,3,6,8]))

Answer 3

It looks like you want the indices of the duplicates. Here is some short code that will find those in O(n) time, without using any packages:

dups = {}
[dups.setdefault(v, []).append(i) for i, v in enumerate(mylist)]
dups = {k: v for k, v in dups.items() if len(v) > 1}
# dups now has keys for all the duplicate values
# and a list of matching indices for each

# The second line produces an unused list. 
# It could be replaced with this:
for i, v in enumerate(mylist):
    dups.setdefault(v, []).append(i)

Answer 4

mylist = [20, 30, 25, 20]

kl = {i: mylist.count(i) for i in mylist if mylist.count(i) > 1 }

print(kl)

Answer 5

You can print duplicate and Unqiue using below logic using list.

def dup(x):
    duplicate = []
    unique = []
    for i in x:
        if i in unique:
            duplicate.append(i)
        else:
            unique.append(i)
    print("Duplicate values: ",duplicate)
    print("Unique Values: ",unique)

list1 = [1, 2, 1, 3, 2, 5]
dup(list1)

Answer 6

The following code will fetch you desired results with duplicate items and their index values.

  for i in set(mylist):
    if mylist.count(i) > 1:
         print(i, mylist.index(i))

Answer 7

I tried below code to find duplicate values from list

1) create a set of duplicate list

2) Iterated through set by looking in duplicate list.

glist=[1, 2, 3, "one", 5, 6, 1, "one"]
x=set(glist)
dup=[]
for c in x:
    if(glist.count(c)>1):
        dup.append(c)
print(dup)

OUTPUT

[1, 'one']

Now get the all index for duplicate element

glist=[1, 2, 3, "one", 5, 6, 1, "one"]
x=set(glist)
dup=[]
for c in x:
    if(glist.count(c)>1):
        indices = [i for i, x in enumerate(glist) if x == c]
        dup.append((c,indices))
print(dup)

OUTPUT

[(1, [0, 6]), ('one', [3, 7])]

Hope this helps someone

Answer 8

simplest way without any intermediate list using list.index():

z = ['a', 'b', 'a', 'c', 'b', 'a', ]
[z[i] for i in range(len(z)) if i == z.index(z[i])]
>>>['a', 'b', 'c']

and you can also list the duplicates itself (may contain duplicates again as in the example):

[z[i] for i in range(len(z)) if not i == z.index(z[i])]
>>>['a', 'b', 'a']

or their index:

[i for i in range(len(z)) if not i == z.index(z[i])]
>>>[2, 4, 5]

or the duplicates as a list of 2-tuples of their index (referenced to their first occurrence only), what is the answer to the original question!!!:

[(i,z.index(z[i])) for i in range(len(z)) if not i == z.index(z[i])]
>>>[(2, 0), (4, 1), (5, 0)]

or this together with the item itself:

[(i,z.index(z[i]),z[i]) for i in range(len(z)) if not i == z.index(z[i])]
>>>[(2, 0, 'a'), (4, 1, 'b'), (5, 0, 'a')]

or any other combination of elements and indices....

Answer 9

You can use list compression and set to reduce the complexity.

my_list = [3, 5, 2, 1, 4, 4, 1]
opt = [item for item in set(my_list) if my_list.count(item) > 1]

Answer 10

m = len(mylist)
for index,value in enumerate(mylist):
        for i in xrange(1,m):
                if(index != i):
                    if (L[i] == L[index]):
                        print "Location %d and location %d has same list-entry:  %r" % (index,i,value)

This has some redundancy that can be improved however.

Answer 11

That's the simplest way I can think for finding duplicates in a list:

my_list = [3, 5, 2, 1, 4, 4, 1]

my_list.sort()
for i in range(0,len(my_list)-1):
               if my_list[i] == my_list[i+1]:
                   print str(my_list[i]) + ' is a duplicate'

Answer 12

The following list comprehension will yield the duplicate values:

[x for x in mylist if mylist.count(x) >= 2]

Answer 13

These answers are O(n), so a little more code than using mylist.count() but much more efficient as mylist gets longer

If you just want to know the duplicates, use collections.Counter

from collections import Counter
mylist = [20, 30, 25, 20]
[k for k,v in Counter(mylist).items() if v>1]

If you need to know the indices,

from collections import defaultdict
D = defaultdict(list)
for i,item in enumerate(mylist):
    D[item].append(i)
D = {k:v for k,v in D.items() if len(v)>1}

Answer 14

You should sort the list:

mylist.sort()

After this, iterate through it like this:

doubles = []
for i, elem in enumerate(mylist):
    if i != 0:
        if elem == old:
            doubles.append(elem)
            old = None
            continue
    old = elem

Answer 15

Here's a list comprehension that does what you want. As @Codemonkey says, the list starts at index 0, so the indices of the duplicates are 0 and 3.

>>> [i for i, x in enumerate(mylist) if mylist.count(x) > 1]
[0, 3]