Check if string is upper, lower, or mixed case in Python
PythonStringPython Problem Overview
I want to classify a list of string in Python depending on whether they are upper case, lower case, or mixed case
How can I do this?
Python Solutions
Solution 1 - Python
There are a number of "is methods" on strings. islower() and isupper() should meet your needs:
>>> 'hello'.islower()
True
>>> [m for m in dir(str) if m.startswith('is')]
['isalnum', 'isalpha', 'isdigit', 'islower', 'isspace', 'istitle', 'isupper']
Here's an example of how to use those methods to classify a list of strings:
>>> words = ['The', 'quick', 'BROWN', 'Fox', 'jumped', 'OVER', 'the', 'Lazy', 'DOG']
>>> [word for word in words if word.islower()]
['quick', 'jumped', 'the']
>>> [word for word in words if word.isupper()]
['BROWN', 'OVER', 'DOG']
>>> [word for word in words if not word.islower() and not word.isupper()]
['The', 'Fox', 'Lazy']
Solution 2 - Python
I want to give a shoutout for using re module for this. Specially in the case of case sensitivity.
We use the option re.IGNORECASE while compiling the regex for use of in production environments with large amounts of data.
>>> import re
>>> m = ['isalnum','isalpha', 'isdigit', 'islower', 'isspace', 'istitle', 'isupper', 'ISALNUM', 'ISALPHA', 'ISDIGIT', 'ISLOWER', 'ISSPACE', 'ISTITLE', 'ISUPPER']
>>>
>>>
>>> pattern = re.compile('is')
>>>
>>> [word for word in m if pattern.match(word)]
['isalnum', 'isalpha', 'isdigit', 'islower', 'isspace', 'istitle', 'isupper']
However try to always use the in operator for string comparison as detailed in this post
faster-operation-re-match-or-str
Also detailed in the one of the best books to start learning python with