- Published on
Most Common Word
- Authors
- Name
- Alex Noh
Problem
You are given a string and a list of banned words. Find the most frequent word in the list except for the words that are in the banned list.
Things you should know
1. List Comprehension
List Comprehension is arguably one of the most powerful features of the Python language.
# finding a list of jobs with names consisting of more than one word.
jobs = ['fire fighter', 'police officer', 'software engineer', 'lawyer', 'teacher']
filtered = []
for job in jobs:
if len(job.split(' ')) > 1:
filtered.append(job)
# you can also use 'filter'. make sure to do the type conversion(to list).
filtered = list(filter(lambda x: len(x.split(' ')) > 1, jobs))
# done easier and more compact with the list comprehension
filtered = [job for job in jobs if len(job.split(' ')) > 1]
2. Counter object
Utilize the collections.Counter function instead of reinventing the wheel.
This function conveniently stores the frequency of each element, eliminating the need for custom implementations.
The most popular and useful method is counter.most_common(). You can specify the number of items you need by using the n parameter
from collections import Counter
counter = Counter(['a', 'b', 'a', 'c', 'a', 'b'])
print(counter.most_common()) # [('a', 3), ('b', 2), ('c', 1)]
print(counter.most_common(n=1)) # [('a', 3)]
3. Python typings
Python is famous for its dynamic typing, allowing flexibility in variable assignment without explicit type declarations.
However, starting from Python 3.7, developers can leverage the benefits of a gradually-improving type system, thanks to the introduction of optional type hints.
This enhancement gives you better readability, improved code hint in IDEs.
Here are some of the basic examples using typing
from typing import List, Dict
def greet(name: str) -> str:
return f"Hello, {name}!"
print(result) # Output: Hello, Alice!
names: List[str] = ["Alice", "Bob", "Charlie"]
scores: Dict[str, int] = {"Alice": 85, "Bob": 90, "Charlie": 80}
Solutions
def most_common_word(paragraph: str, banned: List[str]):
chunks = re.sub(r'[^A-Za-z]+', ' ', paragraph).lower().split(' ')
counter = Counter([word for word in chunks if word not in banned and word != ''])
return counter.most_common(1)[0][0]