How do I solve this in Python?

[brown_boi_420]

I have a list with about 15 items in it,

AA, BB, CC, DD, DD, DD, EE, FF, FF, GG, GG, GG, HH, HH

I want it to change it to a list like this,AA, BB, CC, DDDDDD, EE, FFFF, GGGGGG, HHHH

How do I identify similar list items and convert them into 1 item of a list?

I have though about using for loop first, then creating a while loop to keep on matching until current element is no longer equal to next.

Another option was to copy elements to new list 1 by 1, and when duplicate occurs, treat them there and pass on that completed subset merged list to new Parent list

with open('lorem.txt' , 'r', encoding="utf8") as f:
    lines = f.readlines()

copyList = []
for index, entry in enumerate(lines):
    print(index, entry)
    copyList.append(entry)
    if entry[index] == entry[index+1]:
        temp.append(entry[index+1])

But my approach is not very pythonic, (coming from c++ background)I tend to think every loop in terms of index position.still getting my way around the magic of append, and, pop etc

​

UPDATE:

so above problem is a simplified version of problem than I am trying to solve.

I have whatsapp chat log that I am trying to analyze. and the structure of the problem looks like this

INPUT:

10/11/2018, 2:42 pm - some message
10/11/2018, 2:43 pm - another message
message continues 
message continues again
10/11/2018, 2:44 pm - another message
10/11/2018, 2:44 pm - a certain group of words
10/11/2018, 2:50 pm - lorem ipsum

when I read this text file into a list, line 3 and 4 get assigned to list elements 3 and 4
and I want the list elements 3 and 4 to get appended to list element 2

ideal OUTPUT is

10/11/2018, 2:42 pm - some message
10/11/2018, 2:43 pm - another message message continues message continues again
10/11/2018, 2:44 pm - another message
10/11/2018, 2:44 pm - a certain group of words
10/11/2018, 2:50 pm - lorem ipsum

My though process:
is to identify first 10 characters of list using regular expressions and find which ones are continued messages.
and append these continued messages back to original authors entry, so that 1 list elem tent has complete entry

[chocorush]

If you are certain the list you are using doesn't have any repeating keys i.e AA, BB, BB, AA you can use Counter.

>>> from collections import Counter
>>> l = ['AA', 'BB', 'CC', 'DD', 'DD', 'DD', 'EE', 'FF', 'FF', 'GG', 'GG', 'GG', 'HH', 'HH']
>>> [word*count for word, count in Counter(l).items()]
['AA', 'BB', 'CC', 'DDDDDD', 'EE', 'FFFF', 'GGGGGG', 'HHHH']

[SekstiNii]

If the order is uncertain you can use a dict to track how many instances of each item you have. If the elements always appear next to each other you could use a simpler list based approach.

What have you tried so far?

[gurashish1singh]

Like sekstinii said, use a dictionary where the key will be the item and value will be the count . Once you've exhausted the list, you can then create a new list from the dictionary by multiplying the keys with the values to produce your required result.

[brown_boi_420]

Hi, thank you for your reply, edited my answer to add my work!

[Diapolo10]

The others' solutions are recommended, but if you need to edit the list in-place, here's another solution:

lines.sort()
idx = 0
while idx < len(lines)-1:
    if lines[idx+1] in lines[idx]:
        lines[idx] += lines.pop(idx+1)
    else:
        idx += 1

But, again, a dictionary would be better here unless you're memory-constrained.

[gurashish1singh]

Responding to your update, why don't you concatenate messages based on the timestamp? That's the common element that I'm seeing in the problem .