python if list_item == re.match

Question

I’m trying to practice regex patterns with conditions in python (googlecollab), but stuck in (if… and…) by getting proper numbers from the list[000 to 999] – i need only numbers, ending with one digit ‘1’ (not 11, 111, 211 – I need only 001, 021, 031, 101), but it returns nothing with multiple condition… if I clear code starting with ‘and’ in condition – it returns all ones, elevens, hundred elevens…

list_ = []
list_.append('000')
for a in range(999):
    list_.append(str(a+1))

for i, el in  enumerate(list_):
    if len(el) == 1:
        list_[i] = '00'+el
    elif len(el) == 2:
        list_[i] = '0'+el

for item in list_:
    try:
        if item == re.match(r'dd1', item).group() 
        and item != re.match(r'd11', item).group():
            print(item) 
    except:
        pass

Asked By: haka

||

Source

Answer 1

To match only "numbers" which end with one(not more) digit 1 use the following regex pattern:

for i in list_:
    m = re.match(r'd(0|[2-9])1$', i)
    if m:
        print(i)

(0|[2-9]) – alternation group: to match either 0 or any in range 2-9

Answered By: RomanPerekhrest

python if list_item == re.match

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