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]

Answer 1

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]

Answer 2

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]

Answer 3

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

Answer 4

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