多句子按指定关键词分类报错：‘in’要求左操作数为字符串而非列表

Let's break down what's going wrong and fix your code step by step.

The Root Cause of the Error #

Your original code uses if y in i: where y is a list (like ["web", "online"]) and i is a single string sentence. Python's in operator expects the left operand to be a single element (like a string) when checking membership in another string—you can't ask if an entire list is a substring of a string, which is why you get the in requires string as left operand, not list error.

Revised Code to Handle Keyword Lists #

First, let's adjust the function to properly check for any keyword in the sentence, assign the corresponding category, and avoid global variable issues (your original b list is global, which can cause unexpected behavior on repeated calls):

# Your input data
data = ["my web portal is not working","online is better than offline", "i like going to pharmacy shop for medicines"]
words = ["web", "online"]

def classify_sentences(sentences, keywords):
    classified_results = []
    for sentence in sentences:
        matched_category = None
        # Check each keyword to see if it exists in the sentence
        for keyword in keywords:
            if keyword in sentence:
                matched_category = keyword  # Use the keyword as the category (customize this if needed)
                break  # Stop checking once we find a match
        # Append the matched category or "other" if no keywords were found
        classified_results.append(matched_category if matched_category else "other")
    return classified_results

# Test the function
print(classify_sentences(data, words))
# Output: ['web', 'online', 'other']

Key Improvements #

• No global variables: We create the result list inside the function, so each call returns a fresh set of classifications without leftover data.
• Proper keyword checking: Instead of comparing a list to a string, we iterate through each keyword and check if it's present in the sentence.
• Clear category mapping: The code assigns the matching keyword as the category (you can easily change this to custom category names, like "web_service" instead of "web" if needed).

Simplified Version (If You Don't Need Specific Keyword Categories) #

If you just need to mark sentences as "relevant" (containing any keyword) or "other", you can use a list comprehension with any() for a more concise solution:

def classify_sentences(sentences, keywords):
    return [
        "relevant" if any(key in sentence for key in keywords)
        else "other"
        for sentence in sentences
    ]

This will return ['relevant', 'relevant', 'other'] for your input data.

内容的提问来源于stack exchange，提问作者anant