python regular expression does not return the value as expected

Question

I only need to print out the non-match item which "333-33". When I run the following code, it prints all the items such as "111-11", "222-22", "333-33"...

"333-33" should return the None type...

Any suggestions? Thank you very much in advance.

ALLOW_LIST = ["(4[0-4]{2})-\d{2}", "111-11", "222-22"]

item=[]

codes = ["111-11", "222-22", "333-33"]
for code in codes:
    for i in ALLOW_LIST:
        if re.search(i, code) is None:
            item.append(code)
print(item)

Answer 1

You can use the all function with a generator expression that iterates through ALLOW_LIST to ensure that none of the patterns matches the current code before outputting it:

[code for code in codes if all(not re.search(i, code) for i in ALLOW_LIST)]

Answer 2

From what I understand you want to include only items matched by allow list.

Your if statement will append items that are not matched : if re.search(i, code) is None:

You should edit out your if statement to include items that are matched:

ALLOW_LIST = ["(4[0-4]{2})-\d{2}", "111-11", "222-22"]

item=[]

codes = ["111-11", "222-22", "333-33"]
for code in codes:
    for i in ALLOW_LIST:
        if re.search(i, code):
            item.append(code)
print(item)

Answer 3

You could form a single regex alternation here:

ALLOW_LIST = ["4[0-4]{2}-\d{2}", "111-11", "222-22"]
regex = r'^(?:' + '|'.join(ALLOW_LIST) + r')$'

items = []
codes = ["111-11", "222-22", "333-33"]
for code in codes:
    if not re.search(regex, code):
        items.append(code)

print(items)  # ['333-33']