Validate string

Question

I am trying to solve the following problem:

Given a string str, the task is to check if the string is a valid identifier or not. In order to qualify as a valid identifier, the string must satisfy the following conditions:

It must start with either underscore(_) or any of the characters from the ranges [‘a’, ‘z’] and [‘A’, ‘Z’]. There must not be any white space in the string. And, all the subsequent characters after the first character must not consist of any special characters like ,ドル #, % etc.

Examples:

Input: str= "_code123"

Output: Valid

Input: str = "123code_"

Output: Invalid

My attempted solution, which works:

def isValid(str1, n):
 # If first character is invalid
 if (((ord(str1[0]) >= ord('a') and
 ord(str1[0]) <= ord('z')) or
 (ord(str1[0]) >= ord('A') and
 ord(str1[0]) <= ord('Z')) or
 ord(str1[0]) == ord('_')) == False):
 return False
 # Traverse the string for the rest of the characters
 for i in range(1, len(str1)):
 if (((ord(str1[i]) >= ord('a') and
 ord(str1[i]) <= ord('z')) or
 (ord(str1[i]) >= ord('A') and
 ord(str1[i]) <= ord('Z')) or
 (ord(str1[i]) >= ord('0') and
 ord(str1[i]) <= ord('9')) or
 ord(str1[i]) == ord('_')) == False):
 return False
 # is a valid identifier
 return True
# Driver code
str1 = "_code123"
n = len(str1)
if (isValid(str1, n)):
 print("Valid")
else:
 print("Invalid")

How can I make the above code follow OOP principles, without so many nested ifs?

• GoodDeeds: You could use a regular expression.
• DevScheffer: Show some input examples and expected output
• nerdyGuy: @DevScheffer Check the driver code. It has an example
• DevScheffer: I mean more than just one example, some valid and invalid cases
• nerdyGuy: @DevScheffer Added the example, but your solution doesnt seem to work if _ is at the end of the expression.

Answer 1

Have some read on regex it will help a lot for this kind of problem.

https://regexr.com/

This site is great for practice

import re
def isValid(str1):
 res=re.match(r'^[_A-Za-z]\w+$',str1)
 return res
 
str1 = "_code123"
n = len(str1)
if (isValid(str1)):
 print("Valid")
else:
 print("Invalid")

• nerdyGuy: Thank you! Can you please provide a bit of an explanation regarding what each of the reg ex expressions do? :)
• DevScheffer: Go to the site and put my expression the site will explain automatically

Answer 2

I think you can use that code.

if re.match('^[a-zA-Z_]+', line) is not None and re.match('^[\w-]+$', line) is not None:
 return True

[a-zA-Z_]+ Your string may start with a-z, A-Z and underscore

Answer 3

There is nothing to say that nested ifs break OOP principles.

On the other hand, Python has certain features to help you out. First, you don't need to use ord so much because comparing two one-character strings does the same thing:

ord('a') < ord('b')

is equivalent to

'a' < 'b'

Second, Python allows you to chain math comparison operators, so instead of

'a' < 'b' and 'b' < 'c''

you can just type

'a' < 'b' < 'c'

But there is still an easier way: using the isalpha and isalphanumeric methods, you can avoid using the comparison operators, converting your code to:

Also, Python allows you to work directly on each character while looping by using the for c in str construct, saving you much indexing.

A more natural implementation of your code would be:

def is_valid(str1, n):
 if not (str1[0] == '_' or str1[0].isalpha()):
 return False
 for c in str1[1:]:
 if not (c == '_' or c.isalnum()):
 return False
 return True

You could even use a list comprehension for the second part, by doing:

if not all(c == '_' or c.isalnum() for c in str1[1:]):
 return False