Python关键字计数器统计异常求助：重复输入关键字未正确计数

Let's break down what's going wrong with your keyword counter and fix it up step by step.

The Root of the Problem #

Looking at your loop logic, there are two critical issues throwing off the count:

• You’re using a counter variable that’s unnecessary and actively breaking the count. Every time you match a keyword, you first set dict1[x] = counter (which is always 0), then increment it to 1. Even if the keyword appears multiple times, this overwrites the existing count back to 0 before adding 1—so you’ll never get a number higher than 1.
• Your conditional check after setting dict1[x] = counter is redundant because you just added the key to the dictionary, so the else block will never run.

Fixed Code #

Here’s the corrected version of your script, with clear explanations of the changes:

phrase = input('Enter Python source code:').split(' ')
keywords = {"and", "del", "from", "not", "while", "as", "elif", "global", "or", "with", "assert", "else", "if", "pass", "yield", "break", "except", "import", "print", "class", "exec", "in", "raise", "continue", "finally", "is", "return", "def", "for", "lambda", "try"}
dict1 = {}

for x in phrase:
    if x in keywords:
        # Use dict.get() to safely get the current count (0 if not present) and increment by 1
        dict1[x] = dict1.get(x, 0) + 1

# Sort the dictionary alphabetically by keyword
sorted_dict = dict(sorted(dict1.items()))

# Clean up the print loop by unpacking key-value pairs
for keyword, count in sorted_dict.items():
    print(f"{keyword}: {count}")

Key Improvements #

• Removed the useless counter variable—we don’t need it because we’re tracking counts directly in the dictionary.
• Used dict.get(x, 0) to handle both cases (keyword seen before or not) in one line: if the keyword is already in dict1, we get its current count; if not, we default to 0. Then we add 1 to that value.
• Simplified the final print loop by unpacking key-value pairs directly, making the code cleaner and easier to read.

Testing this with input while while will now correctly output while: 2 as expected.

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