CODEWITHSUNDEEP
pythonstringlist-comprehension
user268254
user268254

Reputation: 43

Splitting a string into consecutive counts?

For example, if the given string is this:

"aaabbbbccdaeeee"

I want to say something like:

3 a, 4 b, 2 c, 1 d, 1 a, 4 e

It is easy enough to do in Python with a brute force loop, but I am wondering if there is a more Pythonic / cleaner one-liner type of approach.

My brute force:

while source!="":
    leading = source[0]
    c=0
    while source!="" and source[0]==leading:
        c+=1
        source=source[1:]
    print(c, leading)

Upvotes: 4

Views: 58

Answers (2)

Grant Williams
Grant Williams

Reputation: 1537

There are a number of different ways to solve the problem. @dawg has already posted the best solution, but if for some reason you aren't allowed to use Counter() (maybe a job interview or school assignment) then you can actually solve the problem in a few ways.

from collections import Counter, defaultdict

def counter_counts(s):
    """ Preferred method using Counter()


    Arguments:
        s {string} -- [string to have each character counted]

    Returns:
        [dict] -- [dictionary of counts of each char]
    """

    return Counter(s)

def default_counts(s):
    """ Alternative solution using defaultdict


    Arguments:
        s {string} -- [string to have each character counted]

    Returns:
        [dict] -- [dictionary of counts of each char]
    """

    counts = defaultdict(int)  # each key is initalized to 0
    for char in s:
        counts[char] += 1  # increment the count of each character by 1

    return counts

def vanilla_counts_1(s):
    """ Alternative solution using a vanilla dicitonary


    Arguments:
        s {string} -- [string to have each character counted]

    Returns:
        [dict] -- [dictionary of counts of each char]
    """

    counts = {}
    for char in s:
        # we have to manually check that each value is in the dictionary before attempting to increment it
        if char in counts:
            counts[char] += 1
        else:
            counts[char] = 1

    return counts

def vanilla_counts_2(s):
    """ Alternative solution using a vanilla dicitonary
    This version uses the .get() method to increment instead of checking if a key already exists


    Arguments:
        s {string} -- [string to have each character counted]

    Returns:
        [dict] -- [dictionary of counts of each char]
    """

    counts = {}
    for char in s:
         # the second argument in .get() is the default value if we dont find the key
        counts[char] = counts.get(char, 0) + 1 

    return counts

And just for fun lets take a look at how each method performs.

For s = "aaabbbbccdaeeee" and 10,000 runs:

Counter: 0.0330204963684082s
defaultdict: 0.01565241813659668s
vanilla 1: 0.01562952995300293s
vanilla 2: 0.015581130981445312s

(actually rather surprising results)

Now let's test what happens if we set our string to the entire plaintext version of the book of Genesis and 1,000 runs:

Counter: 8.500739336013794s
defaultdict: 14.721554040908813s
vanilla 1: 18.089043855667114s
vanilla 2: 27.01840090751648s

Looks like the overhead of creating the Counter() object becomes much less important!

(These weren't very scientific tests, but it was a bit of fun).

Upvotes: 0

dawg
dawg

Reputation: 104014

Use a Counter for a count of each distinct letter in the string regardless of position:

>>> s="aaabbbbccdaeeee"
>>> from collections import Counter
>>> Counter(s)
Counter({'a': 4, 'b': 4, 'e': 4, 'c': 2, 'd': 1})

You can use groupby if the position in the string has meaning:

from itertools import groupby
li=[]
for k, l in groupby(s):
    li.append((k, len(list(l))))

print li

Prints:

[('a', 3), ('b', 4), ('c', 2), ('d', 1), ('a', 1), ('e', 4)]

Which can be reduce to a list comprehension:

[(k,len(list(l))) for k, l in groupby(s)]

You can even use a regex:

>>> [(m.group(0)[0], len(m.group(0))) for m in re.finditer(r'((\w)\2*)', s)] 
[('a', 3), ('b', 4), ('c', 2), ('d', 1), ('a', 1), ('e', 4)]

Upvotes: 9

Related Questions