python2.7：迭代字典列表并评估所需的值[关闭]

Question

代码：

import json
import urllib
URL = 'www.xyz.com'
data = json.load(urllib.urlopen(url)
print data   
[[{"Region":"Europe", "Details":[{"gender":"male", "name":"john", "age":"24", "status":"Ok"}, {"gender":"female", "name":"Rebecca", "age":"22", "status":"None"}], "country":"Germany"}], [{"Region":"Asia", "Details":[{"gender":"male", "name":"kim", "age":"27", "status":"None"}, {"gender":"male", "name":"jen", "age":"22", "status":"None"}], "country":"China"}]]
# here is what I have tried
for i in data:
     for j in i:
          for key in j.keys():
               dicx = j[key]
               for k in dicx:
                 if isinstance(k, dict) and k['status']=='None':
                 print (i['region'], k['name'])) #I wnant to store this value in a variable rather than printing.
#this is giving me the following output.  

Europe, Rebecca
Asia, kim
Asia, jen

场景：根据上面的数据，我想检查所有“区域”（键）是否“状态”（键）是“Ok”（值）或“None”（值），如果“状态”（ key) 是“None”(value) 那么它应该返回我那个特定“Region”(key) 的详细信息(value) 以及“name”(key)。

例如：- 期望的输出（参考上述数据）

Europe, Rebecca

Asia, (kim, jen)

任何帮助将不胜感激。

Answer 1

这很丑陋，但可能确实需要你问：

matches = []
for i in data:  #
                # no changes here
                if isinstance(k, dict) and k['status']=='None':
                    # print (i[0]['Region'], k['name'])
                    matches.append(([i[0]['Region'], k['name']]))
from collections import defaultdict
res = defaultdict(list)

for k, v in matches:
    res[k].append(v)

for k, v in res.iteritems():
    if len(v) == 1:
        print'{}, {}'.format(k, v[0])
    else:
        print '{}, ({})'.format(k, ', '.join(map(str, v)))

编辑#1：

...
for k, v in matches:
    result[k].append(v)

mystr = ''

for k, v in result.iteritems():
    if len(v) == 1:
        mystr += '{}, {}'.format(k, v[0])
    else:
        mystr += ' {}, ({})'.format(k, ', '.join(map(str, v)))

print type(mystr)
print mystr

输出：

<type 'str'>
Europe, Rebecca Asia, (kim, jen)

具有 3 个区域的示例：

<type 'str'>
Europe, Rebecca America, (David, Ricardo) Asia, (kim, jen)