Change strings based on a condition in a list using list comprehension

Question

Let's say I have a list like so:

a = ['abc1', '2def', 'g3h']

And I am trying to make it like this using list comprehension:

['abc', 'def', 'gh']

What I've tried:

[''.join([x for y in a for x in y if x.isalpha()])]

Which produces:

# ['abcdefgh']

Is there a neat way of achieving ['abc', 'def', 'gh'] using list comprehension?

[re.sub('\d+', '', _) for _ in a] and import re

Sam Daniel
– Sam Daniel

2021-01-11 18:42:12 +00:00
Commented Jan 11, 2021 at 18:42 — Sam Daniel
– Sam Daniel, Commented Jan 11, 2021 at 18:42

Rakesh · Accepted Answer · 2021-01-11 18:38:40Z

2

Using str.join in a nested comprehension

Ex:

a = ['abc1', '2def', 'g3h']
print(["".join(j for j in i if j.isalpha()) for i in a])
# -> ['abc', 'def', 'gh']

answered Jan 11, 2021 at 18:38

Rakesh

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3 Comments

ShadowRanger Over a year ago

Or use not j.isdigit() or not j.isnumeric() if the goal is just to exclude numeric characters, not keep solely alphabetic characters.

Jonas Palačionis Over a year ago

@ShadowRanger, do you think using not isnumeric() would be faster than isalpha()? Would this have something to do with the differences in items that need to be checked? I mean there are more letters than numbers.

ShadowRanger Over a year ago

@JonasPalačionis: I wouldn't worry about the speed here; do what's correct (do you want to blacklist numeric stuff? Whitelist alphabetic characters? If there is a whitespace character, do you keep it or not?). Odds are, you won't notice any difference in speed either way; interpreter overhead is likely to swamp the tiny differences. And fastest to the wrong answer is still wrong.

Ignacio Alorre · Accepted Answer · 2021-01-11 18:43:37Z

1

If you can use regex, I would go for this:

import re
a = ['abc1', '2def', 'g3h']
[re.sub("\d+", "", x) for x in a]

answered Jan 11, 2021 at 18:43

Ignacio Alorre

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1 Comment

Jonas Palačionis Over a year ago

Thanks for the regex approach!

Vishesh Mangla · Accepted Answer · 2021-01-11 18:44:12Z

0

Use of map and filter

list(map(lambda x:"".join(filter(lambda y:y.isalpha(), x))  ,a))

answered Jan 11, 2021 at 18:44

Vishesh Mangla

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