List indexes of duplicate values in a list with Python

Question

I am trying to modify this definition that lists duplicate items so that it lists indexes of duplicate values. Also, I would like it to list ALL of the duplicates that means the resultant for a = [1,2,3,2,1,5,6,5,5,5] would be duplicate_indexes = [3,4,7,8,9] Here's the definition:

def list_duplicates(seq):
    seen = set()
    seen_add = seen.add
    # adds all elements it doesn't know yet to seen and all other to seen_twice
    seen_twice = set( x for x in seq if x in seen or seen_add(x) )
    # turn the set into a list (as requested)
    return list( seen_twice )

a = [1,2,3,2,1,5,6,5,5,5]
list_duplicates(a) # yields [1, 2, 5]

Karthik · Accepted Answer · 2014-05-14 04:27:32Z

17

List comprehension to print the index of duplicates. It slices the list till the selected index and return the index value if the item is already present in the sliced list

a= [1, 2, 3, 2, 1, 5, 6, 5, 5, 5]
result=[idx for idx, item in enumerate(a) if item in a[:idx]]
print result #[3, 4, 7, 8, 9]

answered May 14, 2014 at 4:27

Karthik

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1 Comment

MarnixKlooster ReinstateMonica Over a year ago

+1 for being both the shortest and most clearly expressing the specification, compared to the other answers.

thefourtheye · Accepted Answer · 2014-05-14 04:12:40Z

8

a, seen, result = [1, 2, 3, 2, 1, 5, 6, 5, 5, 5], set(), []
for idx, item in enumerate(a):
    if item not in seen:
        seen.add(item)          # First time seeing the element
    else:
        result.append(idx)      # Already seen, add the index to the result
print result
# [3, 4, 7, 8, 9]

Edit: You can just use list comprehension in that function, like this

def list_duplicates(seq):
    seen = set()
    seen_add = seen.add
    return [idx for idx,item in enumerate(seq) if item in seen or seen_add(item)]

print list_duplicates([1, 2, 3, 2, 1, 5, 6, 5, 5, 5])
# [3, 4, 7, 8, 9]

edited May 14, 2014 at 4:12

answered May 14, 2014 at 4:06

thefourtheye

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1 Comment

wwii Over a year ago

You are using a set for seen to make the membership test fast?

jspurim · Accepted Answer · 2014-05-14 04:38:54Z

1

def list_duplicates_index(seq):
    return [i for (i,x) in enumerate(a) if x in list_duplicates(a)]

edited May 14, 2014 at 4:38

answered May 14, 2014 at 4:14

jspurim

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Comments

user3552506 · Accepted Answer · 2014-05-14 04:49:42Z

1

def list_duplicates(seq):
    d = {}
    for i in seq:
        if i in d:
            d[i] += 1
        else:
            d[i] = 1
    dups = []
    for i in d:
        if d[i] > 1:
            dups.append(i)
    lst = []
    for i in dups:
        l = []
        for index in range(len(seq)):
            if seq[index] == i:
                l.append(index)
        lst.append(l[1:])
    new = []
    for i in lst:
        for index in i:
            new.append(index)   
    return new

answered May 14, 2014 at 4:49

user3552506

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