Creating a filtered list by comparing one list to another other list - Python

Question

I have a list and I need to get certain elements out of it. The way I did it was to compare the second list to the first list and then append to a third list. Here is my code:

l1 = ["0 = 0","1 = 1","2 = 2", "And", 
      "Something", "Wicked", "This", "Way", "Comes",
      f"zero.0 = 0", f"one.0 = 1", f"two.0 = 2",
      f"zero.1 = 0", f"one.1 = 1", f"two.1 = 2",
      f"zero.2 = 0", f"one.2 = 1", f"two.2 = 2",
      "purple", "monkey", "dishwasher"]
l3 = []
for element in l1:
   for i in range(3):
       l2 = [f"zero.{i} = 0", f"one.{i} = 1", f"two.{i} = 2"]
       for j in l2:
           if j in element:
               l3.append(element)

The results of this nested for loop is:

['zero.0 = 0', 'one.0 = 1', 'two.0 = 2', 'zero.1 = 0', 'one.1 = 1', 'two.1 = 2', 'zero.2 = 0', 'one.2 = 1', 'two.2 = 2']

The result is exactly what I want but I'm looking for a more Pythonic way of writing this. Any help would be greatly appreciated.

@Linny Yeah, I figured an edit would clear it up a bit. Sorry about that. — Hellyeah
– Hellyeah, Commented Sep 23, 2020 at 18:17

Ben A · Accepted Answer · 2020-09-23 18:35:40Z

Unnecessary format strings

Unless you intend to format a string right then and there, you don't need an f before it.

Regex

This can be reduced to one line using regex:

import re

l3 = [item for item in l1 if re.match(r"\w{3,5}[.]\d\s[=]\s\d{1,}", item)]

Explanation

\w{3,5} matches any word character (equal to [a-zA-Z0-9_])
- {3,5} Quantifier — Matches between 3 and 5 times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below [.]
- . matches the character . literally (case sensitive)
- \d matches a digit (equal to [0-9])
- \s matches any whitespace character (equal to [\r\n\t\f\v ])
Match a single character present in the list below [=]
- = matches the character = literally (case sensitive)
- \s matches any whitespace character (equal to [\r\n\t\f\v ])
\d{1,} matches a digit (equal to [0-9])
- {1,} Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)

Very nice @Linny. I wanted to use l3 = [x for x in l1 if any(y in x for y in l2)] originally, but that only yielded ['zero.2 = 0', 'one.2 = 1', 'two.2 = 2'] for l3. — Hellyeah
– Hellyeah, Commented Sep 23, 2020 at 18:55

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Creating a filtered list by comparing one list to another other list - Python

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Unnecessary format strings

Regex

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Creating a filtered list by comparing one list to another other list - Python

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Unnecessary format strings

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