count how many ones are in first consecutive repetition in list

Question

For example my list is l=[1113213211] and I want the program to print how many "characters" are in first consecutive repetition of ones, I say ones because they are the the first but it can be any number. For example if list is l=[1113213211] I want my program to print: there are 3 ones then 1 three then 1 two then 1 one then 1 three then 1 two then 2 ones. How can I do that in Python3?

P.S. That list I mentioned before can be different. It can be l=[12325228961112333] or something else.

Did you mean `l = [1, 1, 1, 3, 2, 1, 3, 2, 1, 1]` or a string of digits? — sshashank124, Jan 03 '20 at 14:38
https://stackoverflow.com/questions/773/how-do-i-use-itertools-groupby check this question — André Cascais, Jan 03 '20 at 14:52

score 2 · Answer 1 · answered Jan 03 '20 at 14:48

You could use itertools.groupby like,

>>> x = [1113213211]
>>> import itertools
>>> g = itertools.groupby(''.join(str(v) for v in x))
>>> for k,grp in g:
...   print(f'{k} is present {len(list(grp))} times consequitively')
... 
1 is present 3 times consequitively
3 is present 1 times consequitively
2 is present 1 times consequitively
1 is present 1 times consequitively
3 is present 1 times consequitively
2 is present 1 times consequitively
1 is present 2 times consequitively

Vasco Lopes · Answer 2 · 2020-01-03T14:59:34.500

What you want is to iterate over the number and check if it is the same as the last one and do something accordingly. The following code should do it:

number = 1113213211 
number = [int(d) for d in str(number)] # split number into digits

list = [] # to store lists that represent (number, times seen)
lastSeen = None
for i in number: # iterate over all digits
    if lastSeen == None: # initial case
        lastSeen = [i,1]
    else:
        if i == lastSeen[0]: # if the same: add 1
            lastSeen[1] +=1
        else: # if not the same, add it to the list 
            list.append(lastSeen)
            lastSeen = [i,1]
print (list)
# [[1, 3], [3, 1], [2, 1], [1, 1], [3, 1], [2, 1]]

It is not homework, I am first grade of High school and this is one of three parts of one task on national competition — Ante Matković, Jan 03 '20 at 14:57

score 0 · Answer 3 · answered Jan 03 '20 at 14:51

0

itertools groupby was made for this job:

from itertools import groupby

x = [1,1,1,1,2,2,2,3,3,2,2,1,1]
[(k,len(list(g))) for k,g in groupby(x)]

[(1, 4), (2, 3), (3, 2), (2, 2), (1, 2)]

answered Jan 03 '20 at 14:51

score 0 · Answer 4 · answered Jan 03 '20 at 14:59

Was this the sort of thing you were looking for?

l = '12325228961112333'


def count_characters(s):
    answer = "There are "
    count = 0
    character = s[0]
    for ch in s:
        if(ch == character):
            count += 1
        else:
            answer += ("{} {}s ").format(count, character)
            character = ch
            count = 1
    answer += ("{} {}s ").format(count, character)
    return answer


print(count_characters(l))

count how many ones are in first consecutive repetition in list

4 Answers4