我'm working with some data that has ever-changing keys for each record. I'收集了每个记录的主密钥列表并将它们存储在列表中 . 我想迭代遍历列表以使用 try catch
拉出适当的值,但由于列表项中的项为 str
且字典为 dict
,我收到错误 .
举个例子,注意嵌套键的可能性 . 否则我的方法会简单得多:
records = [{
"id": 2017215,
"name": "foo bar"
"campaign": {
"id": 161,
"name": "Historical Data Campaign"
}
},
{
"id": 2017215,
"name": "foo bar",
"last_updated": "2018-01-01",
"campaign": {
"id": 161,
"name": "Historical Data Campaign"
}
}]
keys = [ ['id'], ['name'], ['campaign'], ['campaign']['id'], ['campaign']['name'], ['last_updated'] ]
for record in records:
for key in keys:
print (record + key) # assumption was this would generate "record['id']" but errors due to mismatched types.
我是否比实际需要更难?最大的问题是密钥可以从记录更改为记录,在某些情况下,数据中存在嵌套密钥 .
1 回答
首先,如果你做了一些事情,你可以使上面的代码片段工作
您的代码段中的一个问题是
['campaign']['id']
等元素的出现 . Python假设['campaign']
部分是文字列表定义,然后['id']
正在索引该列表(但't because it'不是整数) .更一般地说,我建议查看字典的.items()方法 . 您可以将上面的代码段转换为
这将处理任何键,但如果您有两个以上级别的嵌套字典,您可能希望将其重新实现为递归函数 .