修复Python字典统计字符串数字频率功能的代码问题

功能要求 #

• 通过函数参数接收字符串text变量
• 初始化一个新字典
• 遍历字符串的每个字符，判断是否为数字
• 统计字符串中数字的频率，忽略空格、字母及标点符号
• 以数字为键（确保键唯一）、对应频率为值填充字典
• 返回该字典

错误代码示例 #

def count_numbers(text):
    # Initialize a new dictionary.
    dictionary = {}
    # Complete the for loop to iterate through each "text" character.
    for i in text:
        # Complete the if-statement using a string method to check if the
        # character is a number.
        if i == int:

            # Complete the if-statement using a logical operator to check if
            # the number is not already in the dictionary.
            if i in dictionary:

                # Use a dictionary operation to add the number as a key
                # and set the initial count value to zero.
                    dictionary.update({text, i})
            # Use a dictionary operation to increment the number count value
            # for the existing key.
            i += i
    return dictionary

问题分析 #

1. 数字判断逻辑错误：用i == int判断字符是否为数字完全无效，正确做法是使用字符串方法char.isdigit()
2. 条件逻辑颠倒：原代码中判断数字已在字典时执行新增操作，实际应该是数字不在字典时初始化计数，存在时累加
3. 字典操作错误：dictionary.update({text, i})语法错误且逻辑错误，初始化键值应该用num_counts[char] = 0
4. 计数累加错误：i += i是对字符本身进行拼接，而非对字典中的计数进行累加，应操作字典对应键的值

修复后的代码 #

def count_numbers(text):
    num_counts = {}
    for char in text:
        # 判断当前字符是否为数字
        if char.isdigit():
            # 若数字未在字典中，初始化计数为0
            if char not in num_counts:
                num_counts[char] = 0
            # 对应数字计数加1
            num_counts[char] += 1
    return num_counts

也可以用更简洁的写法（利用字典get方法简化逻辑）：

def count_numbers(text):
    num_counts = {}
    for char in text:
        if char.isdigit():
            num_counts[char] = num_counts.get(char, 0) + 1
    return num_counts

测试验证 #

运行以下测试用例，输出与预期一致：

print(count_numbers("1001000111101"))
# 预期输出: {'1': 7, '0': 6}

print(count_numbers("Math is fun! 2+2=4"))
# 预期输出: {'2': 2, '4': 1}

print(count_numbers("This is a sentence."))
# 预期输出: {}

print(count_numbers("55 North Center Drive"))
# 预期输出: {'5': 2}

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