Find and replace duplicates in Array, but replace each nth instance with a different string

Question

I have an array below which consists of repeated strings. I want to find and replace those strings, but each time a match is made I want to change the value of the replace strin

Answer 1

I'd use collections.Counter:

from collections import Counter

numbers = { 
    word: iter([""] if count == 1 else xrange(1, count + 1)) 
    for word, count in Counter(sample).items()
}

result = [
    word + str(next(numbers[word])) 
    for word in sample
]

This doesn't require the list to be sorted or grouped in any way.

This solution uses iterators to generate sequential numbers:

• first, we calculate how many times each word occurs in the list (Counter(sample)).

• then we create a dictionary numbers, which, for each word, contains its "numbering" iterator iter(...). If the word occurs only once count==1, this iterator will return ("yield") an empty string, otherwise it will yield sequential numbers in range from 1 to count [""] if count == 1 else xrange(1, count + 1).

• finally, we iterate over the list once again, and, for each word, pick the next value from its own numbering iterator next(numbers[word]). Since our iterators return numbers, we have to convert them to strings str(...).

Answer 2

groupby is a convenient way to group duplicates:

>>> from itertools import groupby
>>> FinalArray = []
>>> for k, g in groupby(SampleArray):
    # g is an iterator, so get a list of it for further handling
    items = list(g)
    # If only one item, add it unchanged
    if len(items) == 1:
        FinalArray.append(k)
    # Else add index at the end
    else:
        FinalArray.extend([j + str(i) for i, j in enumerate(items, 1)])


>>> FinalArray
['champ', 'king1', 'king2', 'mak1', 'mak2', 'mak3']

Answer 3

One way would be to convert your array into a dictionary like that:

SampleDict = {}
for key in SampleArray:
    if key in SampleDict:
        SampleDict[key][0] = True # means: duplicates
        SampleDict[key][1] += 1 
    else:
        SampleDict[key] = [False, 1] # means: no duplicates

Now you can easily convert that dict back to an array. However if the order in SampleArray is important, then you can do it like that:

for i in range(len(SampleArray)):
    key = SampleArray[i]
    counter = SampleDict[key]
    if index[0]:
        SampleArray[i] = key + str(counter[1])
    counter[1] -= 1

This will give you the reversed order however, i.e.

SampleArray = ['champ', 'king2', 'king1', 'mak3', 'mak2', 'mak1']

But I'm sure you'll be able to tweak it to your needs.

Answer 4

EDIT

Counter and than sorting is simpler:

L = ['champ', 'king', 'king', 'mak', 'mak', 'mak']
counts = Counter(L)
res = []
for word in sorted(counts.keys()):
    if counts[word] == 1:
        res.append(word)
    else:
        res.extend(['{}{}'.format(word, index) for index in 
                   range(1, counts[word] + 1)])

So this

['champ', 'mak', 'king', 'king', 'mak', 'mak']

also gives:

['champ', 'king1', 'king2', 'mak1', 'mak2', 'mak3']

Answer 5

Assuming you want the array sorted :

import collections    
counter = collections.Counter(SampleArray)
res = []
for key in sorted(counter.keys()):
    if counter[key] == 1:
        res.append(key)
    else:
        res.extend([key+str(i) for i in range(1, counter[key]+1)])

>>> res
['champ', 'king1', 'king2', 'mak1', 'mak2', 'mak3']

Answer 6

f = ['champ', 'king', 'king', 'mak', 'mak', 'mak']

fields_out = [x + str(f.count(x) - f[i + 1:].count(x)) for i, x in enumerate(f)]
print(fields_out)

>>['champ1', 'king1', 'king2', 'mak1', 'mak2', 'mak3']

or

fields_out = [(x if i == f.index(x) else x + str(f.count(x) - f[i + 1:].count(x))) for i, x in enumerate(f)]
print(fields_out)

>>['champ', 'king', 'king2', 'mak', 'mak2', 'mak3']