convert string to regex pattern

You can puzzle this out:

• go over your strings characterwise
  • if the character is a text character add a 't' to a list
  • if the character is a number add a 'd' to a list
  • if the character is something else, add itself to the list

Use itertools.groupby to group consecutive identical letters into groups. Create a pattern from the group-key and the length of the group using some string literal formatting.

Code:

from itertools import groupby
from string import ascii_lowercase

lower_case = set(ascii_lowercase) # set for faster lookup

def find_regex(p):
    cum = []
    for c in p:
        if c.isdigit():
            cum.append("d")
        elif c in lower_case:
            cum.append("t")
        else:
            cum.append(c)

    grp = groupby(cum) 
    return ''.join(f'\\{what}{{{how_many}}}' 
                   if how_many>1 else f'\\{what}' 
                   for what,how_many in ( (g[0],len(list(g[1]))) for g in grp))

pattern = "1example4...whatit.ry2do"

print(find_regex(pattern))

Output:

\d\t{7}\d\.{3}\t{6}\.\t{2}\d\t{2}

The ternary in the formatting removes not needed {1} from the pattern.

See:

• str.isdigit()

If you now replace '\t'with '[a-z]' your regex should fit. You could also replace isdigit check using a regex r'\d' or a in set(string.digits) instead.

pattern = "1example4...whatit.ry2do"

pat = find_regex(pattern).replace(r"\t","[a-z]")
print(pat) # \d[a-z]{7}\d\.{3}[a-z]{6}\.[a-z]{2}\d[a-z]{2}

See

• string module for ascii_lowercase and digits