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'

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

  
import re

ab = re.compile("^([A-Z]{1}[0-9]{1})+$")
ab.match(string)
  

I believe that should work for an uppercase, number pattern.

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())

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

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

If you need the full string to exactly match the regex, see @Ali Sajjad’s answer using re.fullmatch

Credit: @LondonRob and @conradkleinespel in the comments.

Careful! (Maybe you want to check if FULL string matches)

The re.match(...) will not work if you want to match the full string.

For example;

• re.match("[a-z]+", "abcdef") ✅ will give a match
• But! re.match("[a-z]+", "abcdef 12345") ✅ will also give a match because there is a part in string which matches (maybe you don’t want that when you’re checking if the entire string is valid or not)

Solution

Use re.fullmatch(...). This will only match if the

if re.fullmatch("[a-z]+", my_string):
    print("Yes")

Example

• re.fullmatch("[a-z]+", "abcdef") ✅ Yes
• re.fullmatch("[a-z]+", "abcdef 12345") ❌ No

One liner: bool(re.fullmatch("[a-z]+", my_string))

Ali Sajjad’s answer should be the default, i.e. fullmatch to avoid false positives.

However, it’s also important to know that you’re always checking not None for "yes, it’s a match":

The two possibilities are therefore:

if re.fullmatch("[a-z]+", my_string)!=None:

or, as in Ali’s answer:

if bool(re.fullmatch("[a-z]+", my_string)):

To my way of thinking both of these are really quite horribly unreadable. So a simple utility function is needed for readability:

def is_match(pattern, string, flags=re.IGNORECASE | re.DOTALL): # or "is_full_match", as desired
    return re.fullmatch(pattern, string, flags)!=None

Those 2 flags are (usually) the most helpful default flags settings in my experience, rather than "0".

In practice, of course, you may need to examine the Match object delivered by re.fullmatch. But for cases where you just need to find whether there’s a match…