simple_salesforce python 中的父子关系查询,从有序字典中提取
parent-child relationship query in simple_salesforce python, extracting from ordered dicts
我正在尝试使用 python 中的 simple_salesforce
包从 salesforce 查询信息。
问题是它将作为父子关系一部分的字段嵌套到有序字典中的有序字典中
我想..从机会对象中找到与该记录关联的 ID 和帐户 ID。
SOQL 查询可能类似于..
query = "select id, account.id from opportunity where closedate = last_n_days:5"
在 SOQL(Salesforce 对象查询语言)中,点表示数据库中的父子关系。所以我试图从机会对象中获取 id,然后从该记录上的帐户对象中获取相关的 id。
出于某种原因,Id 正常,但 account.id 嵌套在有序字典中的有序字典中:
q = sf.query_all(query)
这会拉回一个有序的字典..
OrderedDict([('totalSize', 455),
('done', True),
('records',
[OrderedDict([('attributes',
OrderedDict([('type', 'Opportunity'),
('url',
我会拉 ordereddict
的 records
部分来创建 df
df = pd.DataFrame(q['records'])
这给了我 3 列,一个名为 'attributes'
、Id
的有序字典和另一个名为 'Account'
的有序字典。我正在寻找一种方法来从嵌套的有序字典 'Account'
中提取 ('BillingCountry', 'United States')
片段
[OrderedDict([('attributes',
OrderedDict([('type', 'Opportunity'),
('url',
'/services/data/v34.0/sobjects/Opportunity/0061B003451RhZgiHHF')])),
('Id', '0061B003451RhZgiHHF'),
('Account',
OrderedDict([('attributes',
OrderedDict([('type', 'Account'),
('url',
'/services/data/v34.0/sobjects/Account/001304300MviPPF3Z')])),
('BillingCountry', 'United States')]))])
编辑:阐明我在寻找什么。
我想以一个数据框结束,每个查询字段都有一列。
当我使用 df = pd.DataFrame(sf.query_all(query)['records'])
将 'records'
片段放入 DataFrame 时,它给我:
attributes Id Account
OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B003451RhZgiHHF')]) 0061B003451RhZgiHHF OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0013000000MvkRQQAZ')])), ('BillingCountry', 'United States')])
OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B00001Pa52QQAR')]) 0061B00001Pa52QQAR OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0011300001vQPxqAAG')])), ('BillingCountry', 'United States')])
OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B00001TRu5mQAD')]) 0061B00001TRu5mQAD OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0011300001rfRTrAAE')])), ('BillingCountry', 'United States')])
删除 'attributes'
列后,我希望输出为
Id BillingCountry
0061B003451RhZgiHHF 'United States'
0061B00001Pa52QQAR 'United States'
0061B00001TRu5mQAD 'United States'
Pandas 是一个了不起的表格数据工具。但是虽然它可以包含 Python 个对象,但这并不是它的最佳选择。我建议您在将数据插入 pandas.Dataframe
:
之前从查询中提取数据
提取记录:
提取所需字段作为字典列表非常简单:
records = [dict(id=rec['Id'], country=rec['Account']['BillingCountry'])
for rec in data['records']]
将记录插入数据框:
有了字典列表,数据框就像:
df = pd.DataFrame(records)
测试代码:
import pandas as pd
from collections import OrderedDict
data = OrderedDict([
('totalSize', 455),
('done', True),
('records', [
OrderedDict([
('attributes', OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B003451RhZgiHHF')])),
('Id', '0061B003451RhZgiHHF'),
('Account', OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0013000000MvkRQQAZ')])),
('BillingCountry', 'United States')])),
]),
OrderedDict([
('attributes', OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B00001Pa52QQAR')])),
('Id', '0061B00001Pa52QQAR'),
('Account', OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0011300001vQPxqAAG')])),
('BillingCountry', 'United States')])),
]),
OrderedDict([
('attributes', OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B00001TRu5mQAD')])),
('Id', '0061B00001TRu5mQAD'),
('Account', OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0011300001rfRTrAAE')])),
('BillingCountry', 'United States')])),
]),
])
])
records = [dict(id=rec['Id'], country=rec['Account']['BillingCountry'])
for rec in data['records']]
for r in records:
print(r)
print(pd.DataFrame(records))
测试结果:
{'country': 'United States', 'id': '0061B003451RhZgiHHF'}
{'country': 'United States', 'id': '0061B00001Pa52QQAR'}
{'country': 'United States', 'id': '0061B00001TRu5mQAD'}
country id
0 United States 0061B003451RhZgiHHF
1 United States 0061B00001Pa52QQAR
2 United States 0061B00001TRu5mQAD
Pandas 可以阅读有序的字典。
import pandas as pd
from simple_salesforce import Salesforce
sf = Salesforce(username='your_username',
password='your_password',
security_token='your_token')
query = "select id, account.id from opportunity where closedate = last_n_days:5"
df = pd.DataFrame(sf.query_all(query)['records']).drop(columns='attributes')
我正在尝试使用 python 中的 simple_salesforce
包从 salesforce 查询信息。
问题是它将作为父子关系一部分的字段嵌套到有序字典中的有序字典中
我想..从机会对象中找到与该记录关联的 ID 和帐户 ID。
SOQL 查询可能类似于..
query = "select id, account.id from opportunity where closedate = last_n_days:5"
在 SOQL(Salesforce 对象查询语言)中,点表示数据库中的父子关系。所以我试图从机会对象中获取 id,然后从该记录上的帐户对象中获取相关的 id。
出于某种原因,Id 正常,但 account.id 嵌套在有序字典中的有序字典中:
q = sf.query_all(query)
这会拉回一个有序的字典..
OrderedDict([('totalSize', 455),
('done', True),
('records',
[OrderedDict([('attributes',
OrderedDict([('type', 'Opportunity'),
('url',
我会拉 ordereddict
的 records
部分来创建 df
df = pd.DataFrame(q['records'])
这给了我 3 列,一个名为 'attributes'
、Id
的有序字典和另一个名为 'Account'
的有序字典。我正在寻找一种方法来从嵌套的有序字典 'Account'
('BillingCountry', 'United States')
片段
[OrderedDict([('attributes',
OrderedDict([('type', 'Opportunity'),
('url',
'/services/data/v34.0/sobjects/Opportunity/0061B003451RhZgiHHF')])),
('Id', '0061B003451RhZgiHHF'),
('Account',
OrderedDict([('attributes',
OrderedDict([('type', 'Account'),
('url',
'/services/data/v34.0/sobjects/Account/001304300MviPPF3Z')])),
('BillingCountry', 'United States')]))])
编辑:阐明我在寻找什么。
我想以一个数据框结束,每个查询字段都有一列。
当我使用 df = pd.DataFrame(sf.query_all(query)['records'])
将 'records'
片段放入 DataFrame 时,它给我:
attributes Id Account
OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B003451RhZgiHHF')]) 0061B003451RhZgiHHF OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0013000000MvkRQQAZ')])), ('BillingCountry', 'United States')])
OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B00001Pa52QQAR')]) 0061B00001Pa52QQAR OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0011300001vQPxqAAG')])), ('BillingCountry', 'United States')])
OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B00001TRu5mQAD')]) 0061B00001TRu5mQAD OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0011300001rfRTrAAE')])), ('BillingCountry', 'United States')])
删除 'attributes'
列后,我希望输出为
Id BillingCountry
0061B003451RhZgiHHF 'United States'
0061B00001Pa52QQAR 'United States'
0061B00001TRu5mQAD 'United States'
Pandas 是一个了不起的表格数据工具。但是虽然它可以包含 Python 个对象,但这并不是它的最佳选择。我建议您在将数据插入 pandas.Dataframe
:
提取记录:
提取所需字段作为字典列表非常简单:
records = [dict(id=rec['Id'], country=rec['Account']['BillingCountry'])
for rec in data['records']]
将记录插入数据框:
有了字典列表,数据框就像:
df = pd.DataFrame(records)
测试代码:
import pandas as pd
from collections import OrderedDict
data = OrderedDict([
('totalSize', 455),
('done', True),
('records', [
OrderedDict([
('attributes', OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B003451RhZgiHHF')])),
('Id', '0061B003451RhZgiHHF'),
('Account', OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0013000000MvkRQQAZ')])),
('BillingCountry', 'United States')])),
]),
OrderedDict([
('attributes', OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B00001Pa52QQAR')])),
('Id', '0061B00001Pa52QQAR'),
('Account', OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0011300001vQPxqAAG')])),
('BillingCountry', 'United States')])),
]),
OrderedDict([
('attributes', OrderedDict([('type', 'Opportunity'), ('url', '/services/data/v34.0/sobjects/Opportunity/0061B00001TRu5mQAD')])),
('Id', '0061B00001TRu5mQAD'),
('Account', OrderedDict([('attributes', OrderedDict([('type', 'Account'), ('url', '/services/data/v34.0/sobjects/Account/0011300001rfRTrAAE')])),
('BillingCountry', 'United States')])),
]),
])
])
records = [dict(id=rec['Id'], country=rec['Account']['BillingCountry'])
for rec in data['records']]
for r in records:
print(r)
print(pd.DataFrame(records))
测试结果:
{'country': 'United States', 'id': '0061B003451RhZgiHHF'}
{'country': 'United States', 'id': '0061B00001Pa52QQAR'}
{'country': 'United States', 'id': '0061B00001TRu5mQAD'}
country id
0 United States 0061B003451RhZgiHHF
1 United States 0061B00001Pa52QQAR
2 United States 0061B00001TRu5mQAD
Pandas 可以阅读有序的字典。
import pandas as pd
from simple_salesforce import Salesforce
sf = Salesforce(username='your_username',
password='your_password',
security_token='your_token')
query = "select id, account.id from opportunity where closedate = last_n_days:5"
df = pd.DataFrame(sf.query_all(query)['records']).drop(columns='attributes')