find the lengths of all sublists containing common repeated element-CodePudding

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'

CodePudding user response：

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

CodePudding user response：

You can create 2 variable to keep count of the repeated letter. Traverse the array and when you found t increase t, when you find a f check the tcount first if it is bigger than 1 it means there is a repeat print the count of the repetition.

tcount = 0;
fcount = 0;
for e in g:
   if e=="T":
      tcount  
      if fcount>1
         print(fcount)
      fcount=0
  //do same operation for F