Run Length Encoding in Python with List Comprehension

Question

I have a more basic Run Length Encoding question compared to many of the questions about this topic that have already been answered. Essentially, I\'m trying to take the str

Answer 1

You can use itertools.groupby():

from itertools import groupby

grouped = [list(g) for k, g in groupby(string)]

This will produce your per-letter groups as a list of lists.

You can turn that into a RLE in one step:

rle = ''.join(['{}{}'.format(k, sum(1 for _ in g)) for k, g in groupby(string)])

Each k is the letter being grouped, each g an iterator producing N times the same letter; the sum(1 for _ in g) expression counts those in the most efficient way possible.

Demo:

>>> from itertools import groupby
>>> string = 'aabccccaaa'
>>> [list(g) for k, g in groupby(string)]
[['a', 'a'], ['b'], ['c', 'c', 'c', 'c'], ['a', 'a', 'a']]
>>> ''.join(['{}{}'.format(k, sum(1 for _ in g)) for k, g in groupby(string)])
'a2b1c4a3'

Answer 2

Consider using the more_itertools.run_length tool.

Demo

import more_itertools as mit


iterable = "aabccccaaa"
list(mit.run_length.encode(iterable))
# [('a', 2), ('b', 1), ('c', 4), ('a', 3)]

Code

"".join(f"{x[0]}{x[1]}" for x in mit.run_length.encode(iterable))  # python 3.6
# 'a2b1c4a3'

"".join(x[0] + str(x[1]) for x in mit.run_length.encode(iterable))
# 'a2b1c4a3'

Alternative itertools/functional style:

"".join(map(str, it.chain.from_iterable(x for x in mit.run_length.encode(iterable))))
# 'a2b1c4a3'

Note: more_itertools is a third-party library that installable via pip install more_itertools.