Question

我正在尝试压缩看起来像这样的JSON文件：

{
"teams": [
  {
    "teamname": "1",
    "members": [
      {
        "firstname": "John", 
        "lastname": "Doe",
        "orgname": "Anon",
        "phone": "916-555-1234",
        "mobile": "",
        "email": "john.doe@wildlife.net"
      },
      {
        "firstname": "Jane",
        "lastname": "Doe",
        "orgname": "Anon",
        "phone": "916-555-4321",
        "mobile": "916-555-7890",
        "email": "jane.doe@wildlife.net"
      }
    ]
  },
  {
    "teamname": "2",
    "members": [
      {
        "firstname": "Mickey",
        "lastname": "Moose",
        "orgname": "Moosers",
        "phone": "916-555-0000",
        "mobile": "916-555-1111",
        "email": "mickey.moose@wildlife.net"
      },
      {
        "firstname": "Minny",
        "lastname": "Moose",
        "orgname": "Moosers",
        "phone": "916-555-2222",
        "mobile": "",
        "email": "minny.moose@wildlife.net"
      }
    ]
  }       
]

}

我希望将其导出到excel表。我目前的代码是：

from pandas.io.json import json_normalize
import json
import pandas as pd

inputFile = 'E:\\teams.json'
outputFile = 'E:\\teams.xlsx'

f = open(inputFile)
data = json.load(f)
f.close()

df = pd.DataFrame(data)

result1 = json_normalize(data, 'teams' )
print result1

导致此输出：

members                                              teamname
0  [{u'firstname': u'John', u'phone': u'916-555-...        1
1  [{u'firstname': u'Mickey', u'phone': u'916-555-...      2

每行中嵌套有2个成员的数据。我想有一个输出表，显示所有4个成员的数据及其相关的团队名称。

Answer 1

这是一种方法。应该给你一些想法。

df = pd.concat(
    [
        pd.concat([pd.Series(m) for m in t['members']], axis=1) for t in data['teams']
    ], keys=[t['teamname'] for t in data['teams']]
)

                                     0                         1
1 email          john.doe@wildlife.net     jane.doe@wildlife.net
  firstname                       John                      Jane
  lastname                         Doe                       Doe
  mobile                                            916-555-7890
  orgname                         Anon                      Anon
  phone                   916-555-1234              916-555-4321
2 email      mickey.moose@wildlife.net  minny.moose@wildlife.net
  firstname                     Mickey                     Minny
  lastname                       Moose                     Moose
  mobile                  916-555-1111                          
  orgname                      Moosers                   Moosers
  phone                   916-555-0000              916-555-2222

要获得一个包含团队名称和成员作为行的好表，列中的所有属性：

df.index.levels[0].name = 'teamname'
df.columns.name = 'member'

df.T.stack(0).swaplevel(0, 1).sort_index()

要将团队名称和成员作为实际列，只需重置索引。

df.index.levels[0].name = 'teamname'
df.columns.name = 'member'

df.T.stack(0).swaplevel(0, 1).sort_index().reset_index()

整件事

import json
import pandas as pd

json_text = """{
"teams": [
  {
    "teamname": "1",
    "members": [
      {
        "firstname": "John", 
        "lastname": "Doe",
        "orgname": "Anon",
        "phone": "916-555-1234",
        "mobile": "",
        "email": "john.doe@wildlife.net"
      },
      {
        "firstname": "Jane",
        "lastname": "Doe",
        "orgname": "Anon",
        "phone": "916-555-4321",
        "mobile": "916-555-7890",
        "email": "jane.doe@wildlife.net"
      }
    ]
  },
  {
    "teamname": "2",
    "members": [
      {
        "firstname": "Mickey",
        "lastname": "Moose",
        "orgname": "Moosers",
        "phone": "916-555-0000",
        "mobile": "916-555-1111",
        "email": "mickey.moose@wildlife.net"
      },
      {
        "firstname": "Minny",
        "lastname": "Moose",
        "orgname": "Moosers",
        "phone": "916-555-2222",
        "mobile": "",
        "email": "minny.moose@wildlife.net"
      }
    ]
  }       
]
}"""


data = json.loads(json_text)

df = pd.concat(
    [
        pd.concat([pd.Series(m) for m in t['members']], axis=1) for t in data['teams']
    ], keys=[t['teamname'] for t in data['teams']]
)

df.index.levels[0].name = 'teamname'
df.columns.name = 'member'

df.T.stack(0).swaplevel(0, 1).sort_index().reset_index()

Answer 2

这应该是你追求的目标。

json_normalize(data,record_path=['teams','members'],meta=[['teams','teamname']])

output:
         email                firstname lastname mobile      orgname    phone       teams.teamname
0   john.doe@wildlife.net       John    Doe                   Anon      916-555-1234    1
1   jane.doe@wildlife.net       Jane    Doe     916-555-7890  Anon      916-555-4321    1
2   mickey.moose@wildlife.net   Mickey  Moose   916-555-1111  Moosers   916-555-0000    2
3   minny.moose@wildlife.net    Minny   Moose                 Moosers   916-555-2222    2

的说明

from pandas.io.json import json_normalize import pandas as pd

我最近才学会了如何使用json_normalize函数，所以我的解释可能不对。

从我正在呼叫的第0层＆＃39;
开始
json_normalize(data) output: teams 0 [{'teamname': '1', 'members': [{'firstname': '...

有1列和1行。一切都在＆＃39;团队内部。列。

调查我正在呼叫的第1层＆＃39;通过使用record_path =

json_normalize(data,record_path='teams') output: members teamname 0 [{'firstname': 'John', 'lastname': 'Doe', 'org... 1 1 [{'firstname': 'Mickey', 'lastname': 'Moose', ... 2

在第1层，我们已经扁平化了＃teamname＆＃39;但是会员内部会有更多人。

使用record_path =查看第2层。该符号最初是不直观的。我现在通过[＆＃39; layer＆＃39;，＆＃39; deeplayer＆＃39;]来记住它，其结果是layer.deeperlayer。

json_normalize(data,record_path=['teams','members']) output: email firstname lastname mobile orgname phone 0 john.doe@wildlife.net John Doe Anon 916-555-1234 1 jane.doe@wildlife.net Jane Doe 916-555-7890 Anon 916-555-4321 2 mickey.moose@wildlife.net Mickey Moose 916-555-1111 Moosers 916-555-0000 3 minny.moose@wildlife.net Minny Moose Moosers 916-555-2222

请原谅我的输出，我不知道如何在回复中制作表格。

最后，我们使用meta =
添加第1层列
json_normalize(data,record_path=['teams','members'],meta=[['teams','teamname']]) output: email firstname lastname mobile orgname phone teams.teamname 0 john.doe@wildlife.net John Doe Anon 916-555-1234 1 1 jane.doe@wildlife.net Jane Doe 916-555-7890 Anon 916-555-4321 1 2 mickey.moose@wildlife.net Mickey Moose 916-555-1111 Moosers 916-555-0000 2 3 minny.moose@wildlife.net Minny Moose Moosers 916-555-2222 2

注意我们如何需要meta = [[]]的列表列表来引用第1层。如果我们想要第0层和第1层的列，我们可以这样做：

json_normalize(data,record_path=['layer1','layer2'],meta=['layer0',['layer0','layer1']])

json_normalize的结果是pandas数据帧。

展平双嵌套JSON

2 个答案:

整件事