How do I check if a string matches this pattern?
Uppercase letter, number(s), uppercase letter, number(s)...
Example, These would match:
A1B2 B10L1 C1N200J1 These wouldn't ('^' points to problem)
a1B2 ^ A10B ^ AB400 ^ 7 Answers
import re pattern = re.compile("^([A-Z][0-9]+)+$") pattern.match(string) One-liner: re.match(r"pattern", string) # No need to compile
import re >>> if re.match(r"hello[0-9]+", 'hello1'): ... print('Yes') ... Yes You can evalute it as bool if needed
>>> bool(re.match(r"hello[0-9]+", 'hello1')) True Please try the following:
import re name = ["A1B1", "djdd", "B2C4", "C2H2", "jdoi","1A4V"] # Match names. for element in name: m = re.match("(^[A-Z]\d[A-Z]\d)", element) if m: print(m.groups()) import re import sys prog = re.compile('([A-Z]\d+)+') while True: line = sys.stdin.readline() if not line: break if prog.match(line): print 'matched' else: print 'not matched' import re ab = re.compile("^([A-Z]{1}[0-9]{1})+$") ab.match(string)
I believe that should work for an uppercase, number pattern.
regular expressions make this easy ...
[A-Z] will match exactly one character between A and Z
\d+ will match one or more digits
() group things (and also return things... but for now just think of them grouping)
+ selects 1 or more
As stated in the comments, all these answers using re.match implicitly matches on the start of the string. re.search is needed if you want to generalize to the whole string.
import re pattern = re.compile("([A-Z][0-9]+)+") # finds match anywhere in string bool(re.search(pattern, 'aA1A1')) # True # matches on start of string, even though pattern does not have ^ constraint bool(re.match(pattern, 'aA1A1')) # False Credit: @LondonRob and @conradkleinespel in the comments.
