5 Best Ways to Check if a String is a Valid Keyword in Python

💡 Problem Formulation: In Python development, knowing whether a string is a reserved keyword can be pivotal, especially when generating code dynamically or implementing a syntax parser. For instance, you might need to check if the string "for" is a valid keyword, which should return True as 'for' is a reserved keyword in Python.

Method 1: Using the keyword module

Python’s standard library provides a module called keyword that can be used to check if a string is a Python keyword. The keyword.iskeyword(string) function checks if a string is one of the language’s reserved keywords, which are crucial tokens with a specific meaning defined by the language grammar.

Here’s an example:

import keyword

def is_keyword(string):
    return keyword.iskeyword(string)

print(is_keyword("try"))

Output: True

This code snippet defines a function is_keyword that uses the keyword.iskeyword() function to determine if the passed string is a reserved keyword. In the example, the string "try" is correctly identified as a keyword.

Method 2: Checking against the keyword list

The keyword module also provides a list of all keywords under keyword.kwlist. We can check if a string is in this list to determine if it is a keyword. Note that this is less dynamic since it involves a manual check against a predefined list of keywords.

Here’s an example:

import keyword

def is_keyword(string):
    return string in keyword.kwlist

print(is_keyword("elif"))

Output: True

This code snippet demonstrates how to use the keyword.kwlist to check if a string "elif" is part of the keywords in Python. It’s a straightforward method that leverages Python’s built-in knowledge of its syntax.

Method 3: Using the tokenize module

The tokenize module provides lexical scanning for Python source code. By tokenizing a string, we can check if it corresponds to a token type that identifies keywords (called NAME tokens).

Here’s an example:

import tokenize
import io

def is_keyword(string):
    tokens = list(tokenize.tokenize(io.BytesIO(string.encode('utf-8')).readline))
    return (tokens[1].type == tokenize.NAME and 
            tokens[1].string == string)

print(is_keyword("while"))

Output: True

This example uses the tokenize module to tokenize a byte stream of our string. It checks whether the first token (other than the initial ENCODING token) is of type NAME and matches our string, indicating it’s a keyword.

Method 4: Exception Handling

This method tries to use the string as a variable name. If Python throws a SyntaxError, it indicates that the string is a keyword. However, be cautious with this approach, as it depends on executing code, which can have side effects.

Here’s an example:

def is_keyword(string):
    try:
        exec(f"{string} = 42")
        return False
    except SyntaxError:
        return True

print(is_keyword("global"))

Output: True

This code is attempting to execute a string assignment which, if successful, would mean the string is not a keyword. A SyntaxError would denote that Python recognizes the string as a reserved keyword, as seen with "global".

Bonus One-Liner Method 5: Using a Lambda Function

For succinctness, you can create a one-liner function using lambda that invokes keyword.iskeyword to check for a keyword. It’s essentially a compressed version of Method 1.

Here’s an example:

import keyword
is_keyword = lambda s: keyword.iskeyword(s)

print(is_keyword("def"))

Output: True

The lambda function here serves as an inline, anonymous function that checks if a given string is a keyword. It’s short and functional, especially for quick checks within your code.

Summary/Discussion

• Method 1: Using the keyword Module. Simple and reliable. Best for when you need a quick check against Python’s current list of keywords.
• Method 2: Checking against the keyword list. Direct and easy to understand. However, it requires loading the keyword list into memory and isn’t as elegant as using keyword.iskeyword().
• Method 3: Using the tokenize Module. Good for complex parsing tasks and ensures the string is exactly a keyword, not a part of one. However, it’s an overkill for simple keyword identification.
• Method 4: Exception Handling. Interesting use of Python’s dynamic nature but can be risky if not sandboxed properly, as it involves executing code.
• Bonus Method 5: Lambda Function. Quick and concise for inline usage. The downside is the lambda’s lack of explicitness for future code review purposes.