python group by sum就是这样怎么做的

Question

描述：      我的数据来自mongodb这样的项目：

     { "_id" : "5620150715", "timeStamp" : NumberLong("1436889600000"), "userId" : 56, "WorldAmounts" : [ { "count" : NumberLong(10), "Address" : "aaa" }, { "count" : NumberLong(20), "Address" : "bbb" } ], "Amounts" : [ { "count" : NumberLong(10), "ipAddress" : "aaa" }, { "count" : NumberLong(20), "Address" : "bbb" } ], "date" : "20150715", "localAmounts" : [ ] }

问题：
      如何按所有项目分组，关键是＆＃34; userId＆＃34;和＃34; Address＆＃34;在WorldAmounts？

这样的测试数据：

[{'userId':1, 'localA': [{'address':'A','count':7},{'address':'B', 'count':10}], 'localB': [{'address':'X','count':5}], 'localALL':[{'address':'A','count':7},{'address':'B', 'count':10},{'address':'X','count':5}]},{'userId':2, 'localA': [{'address':'A','count':7},{'address':'B', 'count':10}], 'localB': [{'address':'X','count':5}], 'localALL':[{'address':'A','count':7},{'address':'B', 'count':10},{'address':'X','count':5}]},{'userId':1, 'localA': [{'address':'C','count':7}], 'localB': [{'address':'X','count':5}], 'localALL':[{'address':'A','count':7},{'address':'B', 'count':10},{'address':'X','count':5}]}]

结果如下：     如果＆＃39; userId＆＃39;如果不使用groupby，则localALL为LocalA + LocalB，如果相同＆＃39; address＆＃39;价值，＆＃39; count＆＃39;的总和否则附加本地

[{'userId':1, 'localA': [{'address':'C','count':7},{'address':'A','count':7},{'address':'B', 'count':10}], 'localB': [{'address':'X','count':10}], 'localALL':[{'address':'A','count':7},{'address':'B', 'count':10},{'address':'X','count':10}]},
{'userId':2, 'localA': [{'address':'A','count':7},{'address':'B', 'count':10}], 'localB': [{'address':'X','count':5}], 'localALL':[{'address':'A','count':7},{'address':'B', 'count':10},{'address':'X','count':5}]}
]

alxasin建议这样一个最简单的例子：

输入： [{＆＃39; abs＆＃39;：1，＆＃39; lad＆＃39;：[{＆＃39; count＆＃39;：2，＆＃39; desc＆＃39;：＆＃39; A＆＃ 39;}，{＆＃39;计数＆＃39;：3，＆＃39;降序＆＃39;：4}]}， {＆＃39; abs＆＃39;：1，＆＃39; lad＆＃39;：[{＆＃39; count＆＃39;：3，＆＃39; desc＆＃39;：＆＃39; A＆＃39; ;}，{＆＃39;计数＆＃39;：1，＆＃39;降序＆＃39;：5}]}]

输出：

[{＆＃39; abs＆＃39;：1，＆＃39; lst＆＃39;：[{＆＃39; count＆＃39;：5，＆＃39; desc＆＃39;：3}， {＆＃39; count＆＃39;：3，＆＃39; desc＆＃39;：4}，{＆＃39; count＆＃39;：1，＆＃39; desc＆＃39;：5}]}]