python re.compile() and re.findall()

Question

So I try to print only the month, and when I use :

regex = r'([a-z]+) \d+'
re.findall(regex, 'june 15')

And it prints : june But when I try to do the same for a list like this :

regex = re.compile(r'([a-z]+) \d+')
l = ['june 15', 'march 10', 'july 4']
filter(regex.findall, l)

it prints the same list like they didn't take in count the fact that I don't want the number.

filter keeps the whole thing if bool(condition)=True all items in your list match and are thus kept — Kevin
– Kevin, Commented Dec 10, 2018 at 0:10
if each element is only one date use [re.sub(regex,'\\1',x) for x in l] — Kevin
– Kevin, Commented Dec 10, 2018 at 0:18

Chiheb Nexus · Accepted Answer · 2018-12-10 00:16:56Z

5

Use map instead of filter like this example:

import re

a = ['june 15', 'march 10', 'july 4']
regex = re.compile(r'([a-z]+) \d+')
# Or with a list comprehension
# output = [regex.findall(k) for k in a]
output = list(map(lambda x: regex.findall(x), a))
print(output)

Output:

[['june'], ['march'], ['july']]

Bonus:

In order to flatten the list of lists you can do:

output = [elm for k in a for elm in regex.findall(k)]
# Or:
# output = list(elm for k in map(lambda x: regex.findall(x), a) for elm in k)

print(output)

Output:

['june', 'march', 'july']

edited Dec 10, 2018 at 0:16

answered Dec 10, 2018 at 0:10

Chiheb Nexus

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2 Comments

larsks Over a year ago

Or just a list comprehension: output = [regex.findall(x) for x in a]

Senseikaii Over a year ago

Great !! But now I have all the month in a list inside a list. How can I deal with it and put them in a simple list instead ?

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