How to check if a character is upper-case in Python?

I have a string like this

>>> x="Alpha_beta_Gamma"
>>> words = [y for y in x.split('_')]
>>> words
['Alpha', 'beta', 'Gamma']

I want output saying X is non conformant as the the second element of the list words starts with a lower case and if the string x = "Alpha_Beta_Gamma" then it should print string is conformant

To test that all words start with an upper case use this:

print all(word[0].isupper() for word in words)

Maybe you want str.istitle

>>> help(str.istitle)
Help on method_descriptor:

istitle(...)
    S.istitle() -> bool

    Return True if S is a titlecased string and there is at least one
    character in S, i.e. uppercase characters may only follow uncased
    characters and lowercase characters only cased ones. Return False
    otherwise.

>>> "Alpha_beta_Gamma".istitle()
False
>>> "Alpha_Beta_Gamma".istitle()
True
>>> "Alpha_Beta_GAmma".istitle()
False

x="Alpha_beta_Gamma"
is_uppercase_letter = True in map(lambda l: l.isupper(), x)
print is_uppercase_letter
>>>>True

So you can write it in 1 string

words = x.split("_")
for word in words:
    if word[0] == word[0].upper() and word[1:] == word[1:].lower():
        print word, "is conformant"
    else:
        print word, "is non conformant"

You can use this regex:

^[A-Z][a-z]*(?:_[A-Z][a-z]*)*$

Sample code:

import re

strings = ["Alpha_beta_Gamma", "Alpha_Beta_Gamma"]
pattern = r'^[A-Z][a-z]*(?:_[A-Z][a-z]*)*$'

for s in strings:
    if re.match(pattern, s):
        print s + " conforms"
    else:
        print s + " doesn't conform"

As seen on codepad