Question

我有一个变量输出，它可以取值＃1，＃2，我正在努力获得＆＃34;原因＆＃34;如果它存在，正如您从＃1中看到的那样它总是不存在，有人可以建议如何做到这一点吗？

output =
#1: {"took":42,"timed_out":false,"_shards":{"total":5,"successful":5,"failed":0},"hits":{"total":0,"max_score":null,"hits":[]}}
#2: {"took":88,"timed_out":false,"_shards":{"total":5,"successful":5,"failed":0},"hits":{"total":1,"max_score":1.0,"hits":[{"_index":"dispatcher","_type":"lookaheadDV","_id":"832238","_score":1.0, "_source" : {"reasons": ["12345 is associated with data in an invalid state"]}}]}}

输出： -

12345 is associated with data in an invalid state

Answer 1

好吧，从逻辑开始：

hits内的hits可能始终存在...此外，hits看起来是一个列表，因此您可以测试hits（hits内部}）存在，如果它不是空的（例如，你可以测试它的长度）。

一旦你知道hits在那里，就必须进入那个对象并检查你要找的值是否存在。

如果在Python中有字典，则可以使用以下语法检索值：

new_value = some_dictionary.get('some_key', None)

如果没有与该键相关联的值，那么最后的值“None”是Python给我的值。你可以在那里放置你想要的任何值，然后在以后检查：

new_value = some_dictionary.get('some_key', 'BETTER ASK STACKOVERFLOW')

你必须尝试一下。使用REPL。

此外，以下内容不是我在Python中见过的任何内容。是什么给了你这个错误信息：

12345 is associated with data in an invalid state

Answer 2

这应该这样做：

adict = {"took":42,"timed_out":False,"_shards":{"total":5,"successful":5,"failed":0},"hits":{"total":0,"max_score":"null","hits":[]}}
bdict = {"took":88,"timed_out":False,"_shards":{"total":5,"successful":5,"failed":0},"hits":{"total":1,"max_score":1.0,"hits":[{"_index":"dispatcher","_type":"lookaheadDV","_id":"832238","_score":1.0, "_source" : {"reasons": ["12345 is associated with data in an invalid state"]}}]}}

def get_reasons(some_dict):
    output = ""
    try:
        output = some_dict.get("hits").get("hits")[0].get("_source").get("reasons")
    except:
        pass
    return output

print get_reasons(adict)
print get_reasons(bdict)

不会为第一个字典打印任何内容并打印

12345 is associated with data in an invalid state

为第二本词典。

PS：我在您的词典中将false更改为False并将null更改为"null"。

Answer 3

在必要时忽略KeyError。可能有多种原因，因此以下代码将它们全部收集到列表中，然后使用', '作为分隔符连接在一起。

value = { "took":88,"timed_out":False,"_shards":{"total":5,"successful":5,"failed":0},"hits":{"total":1,"max_score":1.0,"hits":[{"_index":"dispatcher","_type":"lookaheadDV","_id":"832238","_score":1.0, "_source" : {"reasons": ["12345 is associated with data in an invalid state"]}}]}}

reasons = []

for i in value['hits']['hits']:
    try:
        reasons.extend(i['_source']['reasons'])
    except KeyError:
        pass

reason = ', '.join(reasons)

如何在“理由”领域获得价值？

3 个答案: