Searching for similar values within a regex string

Question

I'm trying to do a search with regex within two lists that have similar strings, but not the same, how to fix the fault below?

Script:

import re

list1 = [
'juice',
'potato']

list2 = [
'juice;44',
'potato;55',
'apple;66']

correlation = []
for a in list1:
    r = re.compile(r'\b{}\b'.format(a), re.I)
    for b in list2:
        if r.search(b):
            pass
        else:
            correlation.append(b)

print(correlation)

Output:

['potato;55', 'apple;66', 'juice;44', 'apple;66']

Desired Output:

['apple;66']

Regex:

You search each item of list1 in each item of list2 and if e.g. 'juice' isn't in 'potato;55' it is added to correlation. — Michael Butscher
– Michael Butscher, Commented Sep 11, 2020 at 0:23
Set a flag to False before the inner for-loop, set it in the inner loop to True if you found a match, after the loop add to correlation if flag is False yet. — Michael Butscher
– Michael Butscher, Commented Sep 11, 2020 at 0:26
I understand the logic, but could you please provide a snippet of code as an example? Please — Luis Henrique
– Luis Henrique, Commented Sep 11, 2020 at 0:28
The fastest gun in the West won again. You may want to look at the other, arguably better answers. — wp78de
– wp78de, Commented Sep 11, 2020 at 0:57

Wiktor Stribiżew · Accepted Answer · 2020-09-11 00:48:42Z

2

You can create a single regex pattern to match terms from list1 as whole words, and then use filter:

import re

list1 = ['juice', 'potato']
list2 = ['juice;44', 'potato;55', 'apple;66']

rx = re.compile(r'\b(?:{})\b'.format("|".join(list1)))
print( list(filter(lambda x: not rx.search(x), list2)) )
# => ['apple;66']

See the Python demo.

The regex is \b(?:juice|potato)\b, see its online demo. The \b is a word boundary, the regex matches juice or potato as whole words. filter(lambda x: not rx.search(x), list2) removes all items from list2 that match the regex.

answered Sep 11, 2020 at 0:48

Wiktor Stribiżew

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Comments

Michael Butscher · Accepted Answer · 2020-09-11 00:41:54Z

1

First, inner and outer for-loop must be swapped to make this work.

Then you can set a flag to False before the inner for-loop, set it in the inner loop to True if you found a match, after the loop add to correlation if flag is False yet.

This finally looks like:

import re

list1 = [
'juice',
'potato']

list2 = [
'juice;44',
'potato;55',
'apple;66']

correlation = []
for b in list2:
    found = False

    for a in list1:
        r = re.compile(r'\b{}\b'.format(a), re.I)
        if r.search(b):
            found = True

    if not found:
        correlation.append(b)

print(correlation)

answered Sep 11, 2020 at 0:41

Michael Butscher

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Comments

Barmar · Accepted Answer · 2020-09-11 00:50:36Z

1

Convert list1 into a single regexp that matches all the words. Then append the element of list2 if it doesn't match the regexp.

regex = re.compile(r'\b(?:' + '|'.join(re.escape(word) for word in ROE) + r')\b')
correlation = [a for a in list2 if not regex.search(a)]

answered Sep 11, 2020 at 0:50

Barmar

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3 Comments

Wiktor Stribiżew Over a year ago

If you use re.escape, you cannot rely on \b as word boundaries. In this case, you should use (?<!\w) and (?!\w) as word boundaries since you assume that "words" may start/end with special characters.

Barmar Over a year ago

Good point. I just use re.escape() habitually whenever the input data is supposed to be matched literally, to prevent errors from embedded punctuation.

Barmar Over a year ago

If the input words begin or end with punctuation, I'm not even sure what the expected result is supposed to be.

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