Python - Elegant way to transform list to list of all combinations

Question

I have a list of elements like:

x = ['A, B', '2', '3', 'Jan, Feb']
#      ^                   ^
#    comma separated single string

I want to transform this list into a list of lists with all the combinations of each element that is comma separated so that it looks like:

[['A', '2', '3', 'Jan'], ['A', '2', '3', 'Feb']
 ['B', '2', '3', 'Jan'], ['B', '2', '3', 'Feb']]

What is the elegant way to achieve this using Python's in-built functions/libraries?

Bob, you've done a good job describing what you want to do. But [SO] isn't based on descriptions, but rather questions and answers. Do you have a specific question to ask? — Robᵩ, Jan 26 '17 at 19:41

Moinuddin Quadri · Accepted Answer · 2017-01-27T15:28:33.397

5

You may firstly split the elements of list based on comma , and then use itertools.product to get the desired list as:

from itertools import product
x = ['A, B', '2', '3', 'Jan, Feb']

new_list = list(product(*[i.split(', ') for i in x]))

where content of new_list will be:

[('A', '2', '3', 'Jan'), 
 ('A', '2', '3', 'Feb'), 
 ('B', '2', '3', 'Jan'), 
 ('B', '2', '3', 'Feb')]

Based on Dan D's comment, below is the timeit comparison of intermediate List V/s intermediate generator comparison on Python 2. Intermediate list expression is slightly faster:

mquadri$ python -m timeit "from itertools import product; list(product(*(i.split(', ') for i in ['A, B', '2', '3', 'Jan, Feb'])))"
100000 loops, best of 3: 4.83 usec per loop

mquadri$ python -m timeit "from itertools import product; list(product(*[i.split(', ') for i in ['A, B', '2', '3', 'Jan, Feb']]))"
100000 loops, best of 3: 3.78 usec per loop

edited Jan 27 '17 at 15:28

answered Jan 26 '17 at 19:42

Moinuddin Quadri

46,825
13
96
126

You could avoid making a useless list by passing the generator expression to the splat: `list(product(*(i.split(', ') for i in x)))`. – Dan D. Jan 26 '17 at 19:46
@DanD. Intermediate list will be comparatively faster in this case. Reason being that the generator expression is useless here because in order to be used with *product*, it has to be completely evaluated – Moinuddin Quadri Jan 26 '17 at 19:49
Editted answer with `timeit` comparison of both – Moinuddin Quadri Jan 26 '17 at 19:54
Thanks a lot... Exactly what I was looking for – Bob Ezuba Jan 26 '17 at 19:55

Python - Elegant way to transform list to list of all combinations

1 Answers1