在1个列表中检查部分匹配，而在另一列表中找到部分匹配-列表理解可能吗？

Question

somewhat of a python/programming newbie here.

I have written code that does what I need it to:

import re
syns = ['professionals|experts|specialists|pros', 'repayed|payed back', 'ridiculous|absurd|preposterous', 'salient|prominent|significant' ]
new_syns = ['repayed|payed back', 'ridiculous|crazy|stupid', 'salient|prominent|significant', 'winter-time|winter|winter season', 'professionals|pros']

def pipe1(syn):
    # Find first word/phrase in list element up to and including the 1st pipe
    r = r'.*?\|'
    m = re.match(r, syn)
    m = m.group()
    return m

def find_non_match():
    # Compare 'new_syns' with 'syns' and create new list from non-matches in 'new_syns'
    p = '@#&'   # Place holder created
    joined = p.join(syns)
    joined = p + joined   # Adds place holder to beginning of string too
    non_match = []
    for syn in new_syns:
        m = pipe1(syn)
        m = p + m
        if m not in joined:
            non_match.append(syn)
    return non_match

print find_non_match()

Printed output:

['winter-time|winter|winter season']

The code checks if the word/phrase up to and including the first pipe for each element in new_syns is a match for the same partial match in syns list. The purpose of the code is to actually find the non-matches and then append them to a new list called non_match , which it does.

However, I am wonder if it is possible to achieve the same purpose, but in much fewer lines using list comprehension. I have tried, but I am not getting what I want exactly. This is what I have come up with so far:

import re
syns = ['professionals|experts|specialists|pros', 'repayed|payed back', 'ridiculous|absurd|preposterous', 'salient|prominent|significant' ]
new_syns = ['repayed|payed back', 'ridiculous|crazy|stupid', 'salient|prominent|significant', 'winter-time|winter|winter season', 'professionals|pros']

def pipe1(syn):
    # Find first word/phrase in list element up to and including the 1st pipe
    r = r'.*?\|'
    m = re.match(r, syn)
    m = '@#&' + m.group() # Add unusual symbol combo to creatte match for beginning of element
    return m

non_match = [i for i in new_syns if pipe1(i) not in '@#&'.join(syns)]
print non_match

Printed output:

['winter-time|winter|winter season', 'professionals|pros'] # I don't want 'professionals|pros' in the list

The caveat in the list comprehension is that when joining syns with @#& , I don't have the @#& at the beginning of the now joined string, whereas in my original code above that does not use the list comprehension I add @#& to the beginning of the joined string. The result is that 'professionals|pros' have slipped through the net. But I don't know how to pull that off within the list comprehension.

So my question is "Is this possible with the list comprehension?".

Answer 1

I think you want something like:

non_match = [i for i in new_syns if not any(any(w == s.split("|")[0] 
                                                for w in i.split("|")) 
                                            for s in syns)]

This doesn't use regular expressions, but does give the result

non_match == ['winter-time|winter|winter season']

The list includes any items from new_syns where none ( not any ) of the '|' -separated words w are in any of the first word ( split("|")[0] ) of each synonym group s from syns