How to find consecutive same items in a list

Question

I can't solve the following problem.

I have a list and items in it. I need to find 3 or more consecutive items and change them.

For example:

mylist = ["BLUE", "BLUE", "RED", "RED", "RED", "RED", "BLACK", "RED", "RED"]

Required output:

mylist = ["BLUE", "BLUE", "WHITE", "WHITE", "WHITE", "WHITE", "BLACK", "RED", "RED"]

There were 4 items found "RED" and changed to "WHITE". Less than 3 "RED" items are unchanged. There may be an elegant solution, but I still can't find it.

Will anyone help?

A for loop should be easy to use and help your problem even though it's not the most optimized solution — El Pandario, May 02 '22 at 13:49
This is quite useful: https://stackoverflow.com/questions/34443946/count-consecutive-characters — Douwe, May 02 '22 at 13:51

Andrej Kesely · Answer 1 · 2022-05-02T14:16:46.707

3

You can use itertools.groupby for the task:

from itertools import groupby

mylist = ["BLUE", "BLUE", "RED", "RED", "RED", "RED", "BLACK", "RED", "RED"]

out = []
for v, g in groupby(mylist):
    g = sum(1 for _ in g)
    out.extend(["WHITE" if g > 2 else v] * g)

print(out)

Prints:

['BLUE', 'BLUE', 'WHITE', 'WHITE', 'WHITE', 'WHITE', 'BLACK', 'RED', 'RED']

edited May 02 '22 at 14:16

answered May 02 '22 at 13:49

Andrej Kesely

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score 1 · Answer 2 · answered May 02 '22 at 14:13

I finally created this code. But there are not as many Pythonics as from Andrej Kesely. Thanks!

mylist = ["BLUE", "BLUE", "RED", "RED", "RED", "RED", "BLACK", "RED", "RED"]
counter = 0
for i in range(len(mylist)):
    if mylist[i] == "RED":
        counter += 1
        if counter >= 3:
            for k in reversed(range(counter)):
                mylist[i-k] = "WHITE"
    else:
        counter = 0
                
print(mylist)

How to find consecutive same items in a list

2 Answers2