[ {'_id': ObjectId('5abb4f9ca7e2c54c757b3e48'), 'amount': 8400, 'buyerEmail': 'otzYzwMh24edWk8NxSJOqCSZREe0', 'from': 'weixin', 'orderDate': datetime.datetime(2018, 3, 28, 8, 17, 28, 940000), 'orderid': '2018032816173212079', 'real': 8400, 'rechargeDate': datetime.datetime(2018, 3, 28, 8, 17, 41, 877000), 'status': 1, 'tradeNo': '4200000099201803287230332578', 'uid': ObjectId('5abb36051a62067bf7e30178') }, {'_id': ObjectId('5b6699f6df03ec3294d7c0a4'), 'amount': 100, 'buyerEmail': 'otzYzwC3YwRdu7QrWLXqS3VRJybI', 'from': 'weixin', 'orderDate': datetime.datetime(2018, 8, 5, 6, 32, 21, 442000), 'orderid': '2018080514322245193', 'real': 100, 'rechargeDate': datetime.datetime(2018, 8, 5, 6, 33, 0, 381000), 'status': 1, 'tradeNo': '4200000148201808052403940202', 'uid': ObjectId('5a5738411a62061972e128cb') }, ..... ] 如何把同一個uid的整合一起，并只顯示uid real from rechargeDate的元素?類似mongodb聚合的$group理想結果： [ {'uid':xxxxxxxxxxxxxxxx: [ { 'real':xxx, 'from':xxx, 'recharge':xxx, }, { 'real':xxx, 'from':xxx, 'recharge':xxx, }, { 'real':xxx, 'from':xxx, 'recharge':xxx, }, ] }, {'uid':xxxxxxxxxxxxxxxx: [ { 'real':xxx, 'from':xxx, 'recharge':xxx, }, { 'real':xxx, 'from':xxx, 'recharge':xxx, }, { 'real':xxx, 'from':xxx, 'recharge':xxx, }, ] }, ... ]