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

Question

Hi

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

Answer 1

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

Answer 2

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

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

Answer 3

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"

Answer 4

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

Answer 5

You can use this code:

def is_valid(string):
    words = string.split('_')
    for word in words:
        if not word.istitle():
            return False, word
    return True, words
x="Alpha_beta_Gamma"
assert is_valid(x)==(False,'beta')
x="Alpha_Beta_Gamma"
assert is_valid(x)==(True,['Alpha', 'Beta', 'Gamma'])

This way you know if is valid and what word is wrong