Extract Keywords from a List of Strings – Python

Last Updated : 10 Feb, 2025

We are given a list of strings and our task is to extract all words that are valid Python keywords. [Python keywords are reserved words that define the language’s syntax (e.g., is, True, global, try).] For example: a = [“Gfg is True”, “Its a global win”, “try Gfg”] then the output will be: [‘is’, ‘True’, ‘global’, ‘try’].

Using a Set

In this method we iterate over each string and split it into words, for each word we use keyword.iskeyword() to check if it is a valid Python keyword then we use a set to keep track of the keywords we have already added so that each keyword appears only once preserving the order of first occurrence.

Python

import keyword

a = ["Gfg is True", "Its a global win", "try Gfg"]

seen = set()         # To store keywords that have been added
res = []             # List to store unique keywords in order

for s in a:
    for w in s.split():
        if keyword.iskeyword(w) and w not in seen:  # Check if w is a keyword and not already added
            seen.add(w)
            res.append(w)

print(res)

Output

['is', 'True', 'global', 'try']

Explanation: Loop through each string and then each word in the string to check if the word is a Python keyword and ensure it hasn’t been added before (using the set seen).

Using List Comprehension

In this method we first create a flattened list of words from all strings while filtering out only the valid Python keywords using keyword.iskeyword(). We then use dict.fromkeys() to remove duplicates while preserving the order of appearance.

Python

import keyword

a = ["Gfg is True", "Its a global win", "try Gfg"]

words = [w for s in a for w in s.split() if keyword.iskeyword(w)]  # Extract keywords from each string
res = list(dict.fromkeys(words))  # Remove duplicates while preserving order

print(res)

Output

['is', 'True', 'global', 'try']

Explanation:

List comprehension flattens the list of strings into individual words and filters for keywords.
dict.fromkeys(words) removes duplicate keywords while keeping the original order (since dictionaries preserve insertion order in Python 3.7+).

Using filter with a Lambda Function

This method is similar to Method 2 but instead of a list comprehension we use filter() with keyword.iskeyword to extract the keywords from the flattened list and then remove duplicates using dict.fromkeys().

Python

import keyword

a = ["Gfg is True", "Its a global win", "try Gfg"]

words = list(filter(keyword.iskeyword, [w for s in a for w in s.split()]))  # Filter keywords
res = list(dict.fromkeys(words))  # Remove duplicates while preserving order

print(res)

Output

['is', 'True', 'global', 'try']

Explanation:

Inner list comprehension gathers all words and filter(keyword.iskeyword, …) retains only the Python keywords.
We remove duplicates and preserve order with dict.fromkeys().

Using Nested Loops

In this method we again iterate over each string and word, check if the word is a Python keyword and add it to the result list only if it is not already present. However, using if w not in res for duplicate checking is less efficient for larger datasets due to O(n) membership tests on a list.

Python

import keyword

a = ["Gfg is True", "Its a global win", "try Gfg"]

res = []  # List to store unique keywords
for s in a:
    for w in s.split():
        if keyword.iskeyword(w) and w not in res:  # Check membership in list (less efficient)
            res.append(w)

print(res)

Output

['is', 'True', 'global', 'try']

Explanation:

We use nested loops to process each string and each word within it.
The condition w not in res ensures that duplicates are not added but this check is O(n) per lookup, making it less efficient for larger input sizes.