Replacing duplicate strings in Python list

Question

I have a Python list and there is exactly one element which is repeated thrice. I want to replace them with unique element.

my_list = [,...,'future_use',...,'future_use',..'future_use',...]

Expected o/p:

[,...,'future_use1',...,'future_use2',..'future_use3',...]

I tried using list comprehension:

count = my_list.count("future_use")
for i in range(1, count+1):
    my_list = [item.replace("future_use","future_use%r" %i,i) for item in my_list]

However, I'm not getting desired output.Can someone point me to the correct procedure?

DSM · Accepted Answer · 2017-10-30 12:55:09Z

6

You could use Counter and defaultdict from collections and count from itertools to make this a little shorter; whether it's clearer depends on how much time you've spent with Python. Patterns like this are used a lot, though, and are worth being familiar with..

(python 3.6 style here, using f-strings)

from collections import Counter, defaultdict
import itertools

seq = ["a", "future_use", "b", "b", "future_use", "c", "future_use"]

counts = Counter(seq)
suffix_counter = defaultdict(lambda: itertools.count(1))
seq2 = [elem if counts[elem] == 1 else elem + f'_{next(suffix_counter[elem])}'
        for elem in seq]

gives me

>>> seq2
['a', 'future_use_1', 'b_1', 'b_2', 'future_use_2', 'c', 'future_use_3']

answered Oct 30, 2017 at 12:55

DSM

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MaximTitarenko · Accepted Answer · 2017-10-30 15:42:44Z

2

Here is the general approach:

my_list = ['foo', 'foo', 'whatever', 'whatever', 'nevermind', 'foo']

new_list = [] 
counts = {}

for el in my_list:
    temp_count = counts.get(el, 0) + 1
    counts[el] = temp_count 

    if temp_count == 1 and my_list.count(el) == 1:
        new_list.append(el)
    else:
        new_list.append(el + str(temp_count))

print(new_list) # ['foo1', 'foo2', 'whatever1', 'whatever2', 'nevermind', 'foo3']

Rather slow, but pretty straightforward.

UPD

Adapting the @DSM's great answer, this can be also implemented as follows:

my_list = ['foo', 'foo', 'whatever', 'whatever', 'nevermind', 'foo']

from collections import Counter

ranges = {k:[str(i + 1) for i in range(v)] for k,v in Counter(my_list).items()}
new_list = [el if ranges[el] == ['1'] else (el + ranges[el].pop(0)) for el in my_list]

print(new_list) # ['foo1', 'foo2', 'whatever1', 'whatever2', 'nevermind', 'foo3']

edited Oct 30, 2017 at 15:42

answered Oct 30, 2017 at 12:49

MaximTitarenko

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1 Comment

Ma0 Over a year ago

for counting the appearances you can also use counter from itertools

Vasilis G. · Accepted Answer · 2017-10-30 12:43:39Z

1

You can try this one:

myList = ['future_use','future_use','future_use']

def makeUnique(myList, word):

    counter = 1

    for i in range(0,len(myList)):
        if myList[i] == word:
            myList[i] = word + str(counter)
            counter = counter + 1

makeUnique(myList, 'future_use')
print(myList)

The result is:

['future_use1', 'future_use2', 'future_use3']

You can use it for every string element on a list.

answered Oct 30, 2017 at 12:43

Vasilis G.

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Comments

MegaIng · Accepted Answer · 2017-10-30 12:45:58Z

0

Try this:

i = 0
for j, item in enumerate(my_list):
    if item == "future_use":
        my_list[j] = "future_use" + str(i)
        i += 1

edited Oct 30, 2017 at 12:45

answered Oct 30, 2017 at 12:41

MegaIng

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4 Comments

Ma0 Over a year ago

i has to be initialized.

ismailsunni Over a year ago

perhaps it should be for for i, item in enumerate(my_list):

Ma0 Over a year ago

@ismailsunni Nope.

MegaIng Over a year ago

@Brad Solomon this is exactly what Vasilis G. did. Same idea but generak.

Iosif Serafeimidis · Accepted Answer · 2017-10-30 13:09:45Z

0

Maybe using a generator here is a bit overkill but its a useful technique to know if you use list comprehensions and generators.

def gen(count):
    for i in range(count):
        yield i+1

my_list = ['mpla', 'future_use','mpla','future_use','future_use','mpla']
count = my_list.count("future_use")
idx = gen(count)
for i in range(count):
    my_list = [item if item!='future_use' 
               else 'future_use' + str(next(idx)) for item in my_list]

print(my_list)

answered Oct 30, 2017 at 13:09

Iosif Serafeimidis

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user2974951 · Accepted Answer · 2017-10-31 04:26:57Z

0

Don't think about more memory efficient, faster, comprehensive and native method.

my_list = ['foo', 'bar', 'foo', 'foobar', 'bar', 'foo', 'foo', 'bar', 'foobar']
keywords = {}
for i, item in enumerate(my_list):
    if item in keywords:
        my_list[i] = item + str(keywords[item])
        keywords[item] += 1
    else:
        keywords[item] = 1
        my_list[i] = item + '1'

edited Oct 31, 2017 at 4:26

answered Oct 30, 2017 at 13:01

user2974951

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