[FIXED] find the lengths of all sublists containing common repeated element

Issue

I need to find all the sublists from a list where the element is 'F' and that must come one after other

g= ['T','F','F,'F','F','T','T','T','F,'F','F','T]

so, here in this case there are two sublists present in this list which contains element 'F' in repeat

i.e; ['F','F,'F','F'] in index 1,2,3,4 which is in repeat ,so answer is 4
and
['F','F,'F'] in index 8,9,10 which is again in continuous index,so answer is 3

Note: The list contains only two elements 'T' and 'F' and every time we are doing these operations for element 'F'

Solution

You can get the lengths of consecutive sequences with itertools.groupby:

from itertools import groupby

data = ['T','F','F','F','F','T','T','T','F','F','F','T']

# Consecutive sequences of "F".
#   "groupby(data)" produces an iterator that calculates on-the-fly.
#   The iterator returns consecutive keys and groups from the iterable "data".
seqs = [list(g) for k, g in groupby(data) if k == 'F']
print(seqs)
# [['F', 'F', 'F', 'F'], ['F', 'F', 'F']]
seq_lens = [len(k) for k in seqs]
print(seq_lens)
# [4, 3]

Also cool is max length of such consecutive sequences:

max_len_seq = len(max(seqs, key=len))
print(max_len_seq)
# 4

See itertools.groupby for more info:

class groupby:
    # [k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B
    # [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D
    ...
    etc

Answered By - Aeronautix