情况是这样：
json文件中存在一个值为
"headers":{
"connection":["close"],
"content_language":["en"],
"content_length":["3137"],
"content_type":["text/html"],
"server":["squid/3.1.23"],
"unknown":[
{"key":"mime_version","value":["1.0"]},
{"key":"date","value":["Sat, 25 Mar 2017 06:11:38 GMT"]},
{"key":"x_squid_error","value":["ERR_INVALID_URL 0"]},
{"key":"x_cache","value":["MISS from unknown"]},
{"key":"x_cache_lookup","value":["NONE from unknown:8080"]}
]
}由于之前的脚本的处理过于简单粗暴。现实要将"unknown"给替换成字典中的值。
以下是我处理的一段Test code ，在Ipython中：
import json
f = open('file.json','r')
test_line = f.readline()
jsonstr = json.loads(test_line)
he = jsonstr['headers']
# 输出正常的
for (k,v) in he.items():
print k,':',v[0]
输出的是：
"connection":"close",
"content_language":"en",
"content_length":"3137",
"content_type":"text/html",
"server":"squid/3.1.23",
"unknown":[
{"value":["1.0"],"key":"mime_version"}问题：
1， 怎么处理“unknown”中的list，用for的话，怎么输出？
2， 怎么处理“unknown”使其能输出如下的结果：
"connection":"close",
"content_language":"en",
"content_length":"3137",
"content_type":"text/html",
"server":"squid/3.1.23",
"mime_version"："1.0",
"date":"Sat, 25 Mar 2017 06:11:38 GMT",
"x_squid_error":"ERR_INVALID_URL 0",
"x_cache":"MISS from unknown",
"x_cache_lookup":"NONE from unknown:8080"谢谢！～
大家讲道理2017-04-18 10:35:02
# 无非就是list套dict，一层一层往下写就是了
# 输出正常的
for (k,v) in he.items():
if k != 'unknown':
print k,':',v[0]
else:
# unknown对应的值是list
for it in v:
# it是dict
print it.get('key'), ':', it.get('value')[0]
ringa_lee2017-04-18 10:35:02
要学会优雅的处理数据，只要把unknown 取出来再合并进去就行了。
unknown = headers['unknown']
headers.pop('unknown')
set(map(lambda x: (x['key'], x['value'][0])))
headers = dict(headers.items() + unknown)
阿神2017-04-18 10:35:02
headers.update({_['key']: _['value'][0] for _ in headers['unknown']})
headers.pop('unknown')
print headers