Is there a better way of replacing duplicates in a list (Python)

Question

Given a list:

• l1: ['a', 'b', 'c', 'a', 'a', 'b']
• output: ['a', 'b', 'c', 'a_1', 'a_2', 'b_1' ]

I created the following code to get the output. It's messyyy..

for index in range(len(l1)):
    counter = 1
    list_of_duplicates_for_item = [dup_index for dup_index, item in enumerate(l1) if item == l1[index] and l1.count(l1[index]) > 1]
    for dup_index in list_of_duplicates_for_item[1:]: 
        l1[dup_index] = l1[dup_index] + '_' + str(counter)
        counter = counter + 1

Is there a more pythonic way of doing this? I couldn't find anything on the web.

Answer 1

I would do something like this:

a1 = ['a', 'b', 'c', 'a', 'a', 'b']
a2 = []

d = {}

for i in a1:

    d.setdefault(i, -1)
    d[i] += 1

    if d[i] >= 1:
        a2.append('%s_%d' % (i, d[i]))
    else:
        a2.append(i)

print a2

Answer 2

I think the output you're asking for is messy itself, and so there is no clean way of creating it.

How do you intend to use this new list? Would a dictionary of counts like the following work instead?

{'a':3, 'b':2, 'c':1}

If so, I would recommend:

from collections import defaultdict
d = defaultdict(int) # values default to 0
for key in l1:
    d[key] += 1

Answer 3

In Python, generating a new list is usually much easier than changing an existing list. We have generators to do this efficiently. A dict can keep count of occurrences.

l = ['a', 'b', 'c', 'a', 'a', 'b']

def rename_duplicates( old ):
    seen = {}
    for x in old:
        if x in seen:
            seen[x] += 1
            yield "%s_%d" % (x, seen[x])
        else:
            seen[x] = 0
            yield x

print list(rename_duplicates(l))

Answer 4

Based on your comment to @mathmike, if your ultimate goal is to create a dictionary from a list with duplicate keys, I would use a defaultdict from the `collections Lib.

>>> from collections import defaultdict
>>> multidict = defaultdict(list)
>>> multidict['a'].append(1)
>>> multidict['b'].append(2)
>>> multidict['a'].append(11)
>>> multidict
defaultdict(<type 'list'>, {'a': [1, 11], 'b': [2]})