Python遍历字典集合重复输出修复及结果存储咨询

错误原因 #

你原代码的问题在于循环逻辑冗余：duck_collection共有3个元素，for循环会执行3次，你在每次循环内部都重复计算、打印了3个账号的差值，最终得到3*3=9行输出，且全程没有用到遍历得到的循环变量duck。

修改后代码 #

duck_1 = {
    "first_name": "Davey",
    "followers": 12865,
    "following": 120,
    "weapons": ["wit", "steely stare", "devilish good looks"]
}
duck_2 = {
    "first_name": "Jim",
    "followers": 123,
    "following": 5000,
    "weapons": ["squeak"]
}
duck_3 = {
    "first_name": "Celest",
    "followers": 40189,
    "following": 1, 
    "weapons": ["politics", "dance moves", "chess grandmaster", "immortality"]
}

duck_collection = [ duck_1, duck_2, duck_3 ]
# 初始化存储差值的列表（Python变量名不支持空格，因此调整为驼峰命名）
TrendyDucks = []

for duck in duck_collection:
    # 计算当前遍历账号的差值
    diff = duck['followers'] - duck['following']
    print(diff)
    TrendyDucks.append(diff)

运行效果 #

• 打印输出共3行，完全符合要求：

12745
-4877
40188

• 最终TrendyDucks列表的值为[12745, -4877, 40188]，满足存储需求。

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