Count consecutive characters

Totals (without sub-groupings)

#!/usr/bin/python3 -B

charseq = 'abbcccdddd'
distros = { c:1 for c in charseq  }

for c in range(len(charseq)-1):
    if charseq[c] == charseq[c+1]:
        distros[charseq[c]] += 1

print(distros)

I’ll provide a brief explanation for the interesting lines.

distros = { c:1 for c in charseq  }

The line above is a dictionary comprehension, and it basically iterates over the characters in charseq and creates a key/value pair for a dictionary where the key is the character and the value is the number of times it has been encountered so far.

Then comes the loop:

for c in range(len(charseq)-1):

We go from 0 to length - 1 to avoid going out of bounds with the c+1 indexing in the loop’s body.

if charseq[c] == charseq[c+1]:
    distros[charseq[c]] += 1

At this point, every match we encounter we know is consecutive, so we simply add 1 to the character key. For example, if we take a snapshot of one iteration, the code could look like this (using direct values instead of variables, for illustrative purposes):

# replacing vars for their values
if charseq[1] == charseq[1+1]:
    distros[charseq[1]] += 1

# this is a snapshot of a single comparison here and what happens later
if 'b' == 'b':
    distros['b'] += 1

You can see the program output below with the correct counts:

➜  /tmp  ./counter.py
{'b': 2, 'a': 1, 'c': 3, 'd': 4}

You only need to change len(word) to len(word) - 1. That said, you could also use the fact that False‘s value is 0 and True‘s value is 1 with sum:

sum(word[i] == word[i+1] for i in range(len(word)-1))

This produces the sum of (False, True, True, False) where False is 0 and True is 1 – which is what you’re after.

If you want this to be safe you need to guard empty words (index -1 access):

sum(word[i] == word[i+1] for i in range(max(0, len(word)-1)))

And this can be improved with zip:

sum(c1 == c2 for c1, c2 in zip(word[:-1], word[1:]))

Consecutive counts:

You can use itertools.groupby:

s = "111000222334455555"

from itertools import groupby

groups = groupby(s)
result = [(label, sum(1 for _ in group)) for label, group in groups]

After which, result looks like:

[("1": 3), ("0", 3), ("2", 3), ("3", 2), ("4", 2), ("5", 5)]

And you could format with something like:

", ".join("{}x{}".format(label, count) for label, count in result)
# "1x3, 0x3, 2x3, 3x2, 4x2, 5x5"

Total counts:

Someone in the comments is concerned that you want a total count of numbers so "11100111" -> {"1":6, "0":2}. In that case you want to use a collections.Counter:

from collections import Counter

s = "11100111"
result = Counter(s)
# {"1":6, "0":2}

Your method:

As many have pointed out, your method fails because you’re looping through range(len(s)) but addressing s[i+1]. This leads to an off-by-one error when i is pointing at the last index of s, so i+1 raises an IndexError. One way to fix this would be to loop through range(len(s)-1), but it’s more pythonic to generate something to iterate over.

For string that’s not absolutely huge, zip(s, s[1:]) isn’t a a performance issue, so you could do:

counts = []
count = 1
for a, b in zip(s, s[1:]):
    if a==b:
        count += 1
    else:
        counts.append((a, count))
        count = 1

The only problem being that you’ll have to special-case the last character if it’s unique. That can be fixed with itertools.zip_longest

import itertools

counts = []
count = 1
for a, b in itertools.zip_longest(s, s[1:], fillvalue=None):
    if a==b:
        count += 1
    else:
        counts.append((a, count))
        count = 1

If you do have a truly huge string and can’t stand to hold two of them in memory at a time, you can use the itertools recipe pairwise.

def pairwise(iterable):
    """iterates pairwise without holding an extra copy of iterable in memory"""
    a, b = itertools.tee(iterable)
    next(b, None)
    return itertools.zip_longest(a, b, fillvalue=None)

counts = []
count = 1
for a, b in pairwise(s):
    ...

This is my simple code for finding maximum number of consecutive 1’s in binaray string in python 3:

count= 0
maxcount = 0
for i in str(bin(13)):
    if i == '1':
        count +=1
    elif count > maxcount:
        maxcount = count;
        count = 0
    else:
        count = 0
if count > maxcount: maxcount = count        
maxcount

If we want to count consecutive characters without looping, we can make use of pandas:

In [1]: import pandas as pd

In [2]: sample = 'abbcccddddaaaaffaaa'
In [3]: d = pd.Series(list(sample))

In [4]: [(cat[1], grp.shape[0]) for cat, grp in d.groupby([d.ne(d.shift()).cumsum(), d])]
Out[4]: [('a', 1), ('b', 2), ('c', 3), ('d', 4), ('a', 4), ('f', 2), ('a', 3)]

The key is to find the first elements that are different from their previous values and then make proper groupings in pandas:

In [5]: sample = 'abba'
In [6]: d = pd.Series(list(sample))

In [7]: d.ne(d.shift())
Out[7]:
0     True
1     True
2    False
3     True
dtype: bool

In [8]: d.ne(d.shift()).cumsum()
Out[8]:
0    1
1    2
2    2
3    3
dtype: int32

There is no need to count or groupby. Just note the indices where a change occurs and subtract consecutive indicies.

w = "111000222334455555"
iw = [0] + [i+1 for i in range(len(w)-1) if w[i] != w[i+1]] + [len(w)]
dw = [w[i] for i in range(len(w)-1) if w[i] != w[i+1]] + [w[-1]]
cw = [ iw[j] - iw[j-1] for j in range(1, len(iw) ) ]

print(dw)  # digits
['1', '0', '2', '3', '4']
print(cw)  # counts
[3, 3, 3, 2, 2, 5]

w = 'XXYXYYYXYXXzzzzzYYY'
iw = [0] + [i+1 for i in range(len(w)-1) if w[i] != w[i+1]] + [len(w)]
dw = [w[i] for i in range(len(w)-1) if w[i] != w[i+1]] + [w[-1]]
cw = [ iw[j] - iw[j-1] for j in range(1, len(iw) ) ]
print(dw)  # characters
print(cw)  # digits

['X', 'Y', 'X', 'Y', 'X', 'Y', 'X', 'z', 'Y']
[2, 1, 1, 3, 1, 1, 2, 5, 3]

A one liner that returns the amount of consecutive characters with no imports:

def f(x):s=x+" ";t=[x[1] for x in zip(s[0:],s[1:],s[2:]) if (x[1]==x[0])or(x[1]==x[2])];return {h: t.count(h) for h in set(t)}

That returns the amount of times any repeated character in a list is in a consecutive run of characters.

alternatively, this accomplishes the same thing, albeit much slower:

def A(m):t=[thing for x,thing in enumerate(m) if thing in [(m[x+1] if x+1<len(m) else None),(m[x-1] if x-1>0 else None)]];return {h: t.count(h) for h in set(t)}

In terms of performance, I ran them with

site = 'https://web.njit.edu/~cm395/theBeeMovieScript/'
s = urllib.request.urlopen(site).read(100_000)
s = str(copy.deepcopy(s))
print(timeit.timeit('A(s)',globals=locals(),number=100))
print(timeit.timeit('f(s)',globals=locals(),number=100))

which resulted in:

12.528256356999918
5.351301653001428

This method can definitely be improved, but without using any external libraries, this was the best I could come up with.

In python

your_string = "wwwwweaaaawwbbbbn"
current = ''
count = 0
for index, loop in enumerate(your_string):
    current = loop
    count = count + 1
    if index == len(your_string)-1:
        print(f"{count}{current}", end ='')
        break

    if your_string[index+1] != current:
        print(f"{count}{current}",end ='')
        count = 0
        continue

This will output

5w1e4a2w4b1n

#I wrote the code using simple loops and if statement
s='feeekksssh' #len(s) =11
count=1  #f:0, e:3, j:2, s:3 h:1
l=[]
for i in range(1,len(s)): #range(1,10)
    if s[i-1]==s[i]:
        count = count+1
    else:
        l.append(count)
        count=1
    if i == len(s)-1: #To check the last character sequence we need loop reverse order
        reverse_count=1
        for i in range(-1,-(len(s)),-1): #Lopping only for last character
            if s[i] == s[i-1]:
                reverse_count = reverse_count+1
            else:
                l.append(reverse_count)
                break
print(l)

Today I had an interview and was asked the same question. I was struggling with the original solution in mind:

s = 'abbcccda'

old = ''
cnt = 0
res = ''
for c in s:
    cnt += 1
    if old != c:
        res += f'{old}{cnt}'
        old = c
        cnt = 0  # default 0 or 1 neither work
print(res)
#  1a1b2c3d1

Sadly this solution always got unexpected edge cases result(is there anyone to fix the code? maybe i need post another question), and finally timeout the interview.

After the interview I calmed down and soon got a stable solution I think(though I like the groupby best).

s = 'abbcccda'

olds = []
for c in s:
    if olds and c in olds[-1]:
        olds[-1].append(c)
    else:
        olds.append([c])
print(olds)
res = ''.join([f'{lst[0]}{len(lst)}' for lst in olds])
print(res)

#  [['a'], ['b', 'b'], ['c', 'c', 'c'], ['d'], ['a']]
#  a1b2c3d1a1

Here is my simple solution:

def count_chars(s):
    size = len(s)
    count = 1
    op = ''
    for i in range(1, size):
        if s[i] == s[i-1]:
            count += 1
        else:
            op += "{}{}".format(count, s[i-1])
            count = 1
    if size:
        op += "{}{}".format(count, s[size-1])

    return op

data_input = 'aabaaaabbaaaaax'
start = 0
end = 0
temp_dict = dict()
while start < len(data_input):
  if data_input[start] == data_input[end]:
     end = end + 1
  if end == len(data_input):
     value = data_input[start:end]
     temp_dict[value] = len(value)
     break
  if data_input[start] != data_input[end]:
     value = data_input[start:end]
     temp_dict[value] = len(value)
     start = end
print(temp_dict)

PROBLEM: we need to count consecutive characters and return characters with their count.

def countWithString(input_string:str)-> str:
    count = 1
    output = ''
 
    for i in range(1,len(input_string)):
        if input_string[i]==input_string[i-1]:
            count +=1
        else:
            output += f"{count}{input_string[i-1]}"
            count = 1
    # Used to add last string count (at last else condition will not run and data will not be inserted to ouput string)
    output += f"{count}{input_string[-1]}"
    return output

countWithString(input)

input:'aaabbbaabbcc'
output:'3a3b2a2b2c'

Time Complexity: O(n)
Space Complexity: O(1)

[enter image description here][1]

temp_str = "aaaajjbbbeeeeewwjjj"
def consecutive_charcounter(input_str):
    counter = 0
    temp_list = []
    for i in range(len(input_str)):
        if i==0:
            counter+=1
        elif input_str[i]== input_str[i-1]:
            counter+=1
            if i == len(input_str)-1:
                temp_list.extend([input_str[i - 1], str(counter)])
        else:
            temp_list.extend([input_str[i-1],str(counter)])
            counter = 1
    print("".join(temp_list))

consecutive_charcounter(temp_str)