如何在Python中检测字符串是否包含字典内的指定词汇？

Got it, let's break this down. You have a dictionary where one key maps to a list of keywords, and you want to check if an input string (like "I love simplicity") includes any of those keywords. Here are two reliable ways to implement this:

Method 1: Clean and Split the Input String #

This approach works well for basic cases, where you want to match exact words after cleaning up punctuation and case differences:

import string

# Your dictionary data
review_keywords = {"value for money": ["rescheduled", "cost", "low", "high", "simplicity", "booking", "price-performance", "satisfied", "satisfaction", "pricing", "prices"]}

# Extract the list of target keywords
target_words = review_keywords["value for money"]

def contains_target_keyword(input_str):
    # Step 1: Clean the input string - lowercase and remove punctuation
    cleaned_str = input_str.lower().translate(str.maketrans('', '', string.punctuation))
    # Step 2: Split into individual words
    input_words = cleaned_str.split()
    # Step 3: Check if any target word exists in the input words
    return any(word in input_words for word in target_words)

# Test cases
print(contains_target_keyword("I love simplicity"))  # Output: True
print(contains_target_keyword("The price is way too high!"))  # Output: True
print(contains_target_keyword("This service is amazing"))  # Output: False

Key Notes for Method 1: #

• We convert the input to lowercase to avoid missing matches like "Simplicity" vs "simplicity".
• We strip punctuation so words like "simplicity!" are treated the same as "simplicity".
• any() stops checking as soon as it finds a match, making it efficient.

Method 2: Regular Expression (More Robust) #

If you want to ensure whole-word matches (so "lower" doesn't trigger a match for "low") and handle edge cases like words with special characters (e.g., "price-performance"), regex is a better choice:

import re

# Your dictionary data
review_keywords = {"value for money": ["rescheduled", "cost", "low", "high", "simplicity", "booking", "price-performance", "satisfied", "satisfaction", "pricing", "prices"]}

target_words = review_keywords["value for money"]

# Build a regex pattern: match any target word as a whole word, case-insensitive
# re.escape() handles special characters in keywords (like the hyphen in price-performance)
pattern = re.compile(
    r'\b(' + '|'.join(re.escape(word) for word in target_words) + r')\b',
    re.IGNORECASE
)

def contains_target_keyword(input_str):
    # Check if the pattern exists anywhere in the input string
    return bool(pattern.search(input_str))

# Test cases
print(contains_target_keyword("I love Simplicity!"))  # Output: True
print(contains_target_keyword("The lower price is great"))  # Output: False (doesn't match "low")
print(contains_target_keyword("Great price-performance on this booking"))  # Output: True

Key Notes for Method 2: #

• \b ensures we only match whole words (not substrings).
• re.IGNORECASE makes the match case-insensitive.
• re.escape() prevents special characters in keywords from breaking the regex pattern.

Either method will work for your use case—pick the one that fits your specific needs!

内容的提问来源于stack exchange，提问作者user3310469