不使用内置函数或导入库的Python文本文件单词计数作业难题

Hey there! Let's break down what's going wrong with your code and fix it step by step.

What's Causing the Issue? #

Your current code has a critical problem in the else block: when you encounter a word that's already in compare_list, you're looping through the entire master_list every single time to increment the count. This means if a word appears 5 times, you'll end up adding 5 to its count multiple times (once for each occurrence), leading to wildly inflated and incorrect numbers.

The Fix: Targeted Count Increment #

Instead of re-scanning the entire list every time you find a duplicate word, just find the index of the existing word in compare_list and increment the corresponding position in count_list directly. Here's the revised code:

# Most common word counter
fh = open("romeo.txt")
master_list = fh.read().split()
fh.close()  # Don't forget to close the file!

compare_list = []
count_list = []

for word in master_list:
    if word not in compare_list:
        compare_list.append(word)
        count_list.append(1)
    else:
        # Find the index of the existing word and increment its count
        word_index = compare_list.index(word)
        count_list[word_index] += 1

# Now find the most common word
max_count = max(count_list)
most_common_index = count_list.index(max_count)
most_common_word = compare_list[most_common_index]

print(f"Total words: {len(master_list)}")
print(f"Unique words: {len(compare_list)}")
print(f"Counts per word: {list(zip(compare_list, count_list))}")
print(f"The most common word is '{most_common_word}' with {max_count} occurrences.")

Key Improvements: #

• Fixed Counting Logic: We use compare_list.index(word) to directly find where the duplicate word lives, then increment its count by 1 (no more re-counting the entire list).
• Proper File Handling: Added fh.close() to properly close the text file after reading it—this is good practice to avoid wasting system resources.
• Final Result Calculation: Added code to actually find and print the most common word, which was missing from your original code.
• Readable Output: Used zip() to pair words with their counts for clearer, easier-to-interpret printing.

Optional: Case Insensitivity #

If your assignment considers "Romeo" and "romeo" as the same word, you can normalize all words to lowercase (or uppercase) before processing to avoid skewed results:

# Modify the loop to use lowercase words
for word in master_list:
    normalized_word = word.lower()
    if normalized_word not in compare_list:
        compare_list.append(normalized_word)
        count_list.append(1)
    else:
        word_index = compare_list.index(normalized_word)
        count_list[word_index] += 1

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