Python字典Update()函数及一次bug解决

一、用法
python字典中的update方法可用于字典的键值对更新、添加。示例：

dic = {"zhangsan": 89}
dic.update({"lisi":88})
dic
{'zhangsan': 89, 'lisi': 88}

二、bug描述
在构造数据的时候，如下更新字典bank_code，定义在for循环以外时，会导致所有的banl_code都是PSBC：

test_data = {
            "top": "2",
             "busId": "XXXX",
             "features":
                 {
                     "applist": "(38)嘻哈龟牛皮兔,10-10 极端令人惊异的网格块益智世界游戏"
                 },
            "multiFeatures": {}
}
s = 'ICBC CNCB PSBC'
dic = {"tag20": 0, "tag21": 0, "tag22": 1, "tag23": 1, "tag24": 0, "tag25": 1, "tag26": 0, "tag27": 0, "tag17": 0,
       "bank_code": "BCOM", "tag30": 0, "tag8": 0, "tag31": 1, "tag10": 0, "tag32": 0, "tag9": 0,
       }
for i in s.split(' '):
    dic['bank_code'] = i
    new_d = {}
    new_d[str(i)] = dic
    test_data['multiFeatures'].update(new_d)
print(test_data)

三、解决方法
将dic 定义在for循环以内，完美解决，不会导致所有的bank_code都是PSBC：

test_data = {
            "top": "2",
             "busId": "XXXX",
             "features":
                 {
                     "applist": "(38)嘻哈龟牛皮兔,10-10 极端令人惊异的网格块益智世界游戏"
                 },
            "multiFeatures": {}
}
s = 'ICBC CNCB PSBC'
for i in s.split(' '):
    dic = {"tag20": 0, "tag21": 0, "tag22": 1, "tag23": 1, "tag24": 0, "tag25": 1, "tag26": 0, "tag27": 0, "tag17": 0,
           "bank_code": "BCOM", "tag30": 0, "tag8": 0, "tag31": 1, "tag10": 0, "tag32": 0, "tag9": 0,
           }
    dic['bank_code'] = i
    new_d = {}
    new_d[str(i)] = dic
    test_data['multiFeatures'].update(new_d)
print(test_data)

最后

以上就是无情早晨为你收集整理的Python字典Update()函数及一次bug解决的全部内容，希望文章能够帮你解决Python字典Update()函数及一次bug解决所遇到的程序开发问题。

如果觉得靠谱客网站的内容还不错，欢迎将靠谱客网站推荐给程序员好友。