Check if string matches pattern

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
^

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)

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.