有没有更好的方法在Python中连接连续的字符串元素？

Question

Problem Context

I am trying to create a chat log dataset from Whatsapp chats. Let me just provide the context of what problem I am trying to solve. Assume message to be M and response to be R . The natural way in which chats happen is not always alternate, for eg chats tend to happen like this

[ M, M, M, R, R, M, M, R, R, M ... and so on]

I am trying to concatenate continuously occurring strings of M's and R's. for the above example, I desire an output like this

Desired Output

[ "MMM", "RR", "MM" , "RR", "M ... and so on ]

An Example of Realistic Data:

 Input --> ["M: Hi", "M: How are you?", "R: Heyy", "R: Im cool", "R: Wbu?"] (length=5) Output --> ["M: Hi M: How are you?", "R: Heyy R: Im cool R: Wbu?"] (length = 2) 

Is there a fast and more efficient way of doing this? I have already read this Stackoverflow link to solve this problem. But, I didn't find a solution there.

So far, this is what I have tried .

final= []
temp = ''
change = 0
for i,ele in enumerate(chats):
    if i>0:
        prev = chats[i-1][0]
        current = ele[0]

        if current == prev:
            continuous_string += chats[i-1]  
            continue
        else:
            continuous_string += chats[i-1]
            final.append(temp)
            temp = ''

Explanation of my code: I have chats list in which the starting character of every message is 'M' and starting character of every response is 'R'. I keep track of prev value and current value in the list, and when there is a change (A transition from M -> R or R -> M), I append everything collected in the continuous_string to final list.

Again, my question is: Is there a shortcut in Python or a function to do the same thing effectively in less number of lines?

Answer 1

You can use the function groupby() :

from itertools import groupby

l = ['A', 'A', 'B', 'B']

[' '.join(g) for _, g in groupby(l)]
# ['A A', 'B B']

To group data from your example you need to add a key to the the groupby() function:

l = ["M: Hi", "M: How are you?", "R: Heyy", "R: Im cool", "R: Wbu?"]

[' '.join(g) for _, g in groupby(l, key=lambda x: x[0])]
# ['M: Hi M: How are you?', 'R: Heyy R: Im cool R: Wbu?']

As @TrebuchetMS mentioned in the comments the key lambda x: x.split(':')[0] might be more reliable. It depends on your data.

Answer 2

Algorithm

• Initialize a temporary item. This will help determine if the speaker has changed
• For each item
  • Extract the speaker
  • If it's the same, append to the text of the last item of the array
  • Else append a new item in the list containing the speaker and text

Implementation

def parse(x):
    parts = x.split(':')
    return parts[0], ' '.join(parts[1:]).strip()


def compress(l):
    ans = []
    prev = ''
    for x in l:
        curr, text = parse(x)
        if curr != prev:
            prev = curr
            ans.append(x)
        else:
            ans[len(ans) - 1] += f' {text}'
    return ans

Character names

IN:  ["M: Hi", "M: How are you?", "R: Heyy", "R: Im cool", "R: Wbu?"]
OUT: ['M: Hi How are you?', 'R: Heyy Im cool Wbu?']

String names

IN:  ["Mike: Hi", "Mike How are you?", "Mary: Heyy", "Mary: Im cool", "Mary: Wbu?"]
OUT: ['Mike: Hi How are you?', 'Mary: Heyy Im cool Wbu?']