Regex Pattern Comprising From the Elements in List

Question

I am trying to make a script that reads the list element sequentially and concatenate them each into regex pattern.

for example,

I have a list like this:

lst = ['B.', 'Article', 'III']

And want to comprise regex something like this:

re.search(r lst[0]\s+lst[1]\s+lst[2]).group()

so that it can match below regardless of white_spaces between each elements from the list:

candidate_1 = 'B.      Article III'
candidate_2 = 'B.        Article III'

@Robᵩ I don't know how to concatenate r with str. if I do r lst[0]\s+lst[1]\s+lst[2] it reutrns error, and if I do rlst[0]\s+lst[1]\s+lst[2] it also definitely returns error — snapper
– snapper, Commented Feb 21, 2018 at 7:02

Robᵩ · Accepted Answer · 2018-02-21 07:00:58Z

3

Try str.join(), like so:

r'\s+'.join(lst)

Here is a complete program:

import re

def list2pattern(l):
    return r'\s+'.join(l)

lst = ['B.', 'Article', 'III']
assert re.search(list2pattern(lst), 'B. Article III')
assert re.search(list2pattern(lst), 'B.      Article III')
assert not re.search(list2pattern(lst), 'B.Article III')
assert not re.search(list2pattern(lst), 'George')

answered Feb 21, 2018 at 7:00

Robᵩ

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4 Comments

snapper Over a year ago

I have more fundamental question. I have tried type(list2pattern(lst)) and it returns str, where my question is then, how the machine distinguish between str and r str if they are all memorized as str?

CristiFati Over a year ago

@delinco: What are "str" and "r str"? Ah, probably you mean r"": docs.python.org/3/reference/….

snapper Over a year ago

the r that always added before expressing regex pattern

Robᵩ Over a year ago

@delinco - See this question: stackoverflow.com/questions/2081640/…

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