Java 如何使用聚合合并包含timeseries数据数组的两个或多个文档?
我希望合并两个或多个文档,以便将这些选定文档中的timeseries数据合并到一个文档中,例如,如果数据在该时间戳不存在,则该数据应为null 以下是文档(可以有多个文档):- 文件1:-Java 如何使用聚合合并包含timeseries数据数组的两个或多个文档?,java,mongodb,Java,Mongodb,我希望合并两个或多个文档,以便将这些选定文档中的timeseries数据合并到一个文档中,例如,如果数据在该时间戳不存在,则该数据应为null 以下是文档(可以有多个文档):- 文件1:- { "unit_id":38, "mac_id":"abcdefghijkl", "data":{ "ch1":[21,22,31,31,22,11], "ch2":[55,55,56,57,88,90] }, "updatetime":["2019
{
"unit_id":38,
"mac_id":"abcdefghijkl",
"data":{
"ch1":[21,22,31,31,22,11],
"ch2":[55,55,56,57,88,90]
},
"updatetime":["2019-07-15 17:20:16",
"2019-07-15 17:20:23",
"2019-07-15 17:21:23",
"2019-07-15 17:22:23",
"2019-07-15 17:23:25",
"2019-07-15 17:24:25"]
}
文件2:-
{
"unit_id":39,
"mac_id":"abcdefgavdca",
"data":{
"ch1":[43.21,55.12,44.21,66.21,11.21,22.10]
},
"updatetime":["2019-07-15 17:25:00",
"2019-07-15 17:26:23",
"2019-07-15 17:27:23",
"2019-07-15 17:28:23",
"2019-07-15 17:29:25",
"2019-07-15 17:30:25"]
}
文件3:-
{
"unit_id":40,
"mac_id":"abcdefgzxcvs",
"data":{
"ch1":[21,22,31,31,22,11],
},
"updatetime":["2019-07-15 17:35:16",
"2019-07-15 17:36:23",
"2019-07-15 17:37:23",
"2019-07-15 17:38:23",
"2019-07-15 17:39:25",
"2019-07-15 17:40:25"]
}
结果应该是这样的:-
updatetime ch1[d1] ch2[d1] ch1[d2] ch1[d3]
2019-07-15 17:20:16 21 55 nan nan
2019-07-15 17:20:23 22 55 nan nan
2019-07-15 17:21:23 31 56 nan nan
2019-07-15 17:22:23 31 57 nan nan
2019-07-15 17:23:25 22 88 nan nan
2019-07-15 17:24:25 11 90 nan nan
2019-07-15 17:25:00 nan nan 43.21 nan
2019-07-15 17:26:23 nan nan 55.12 nan
2019-07-15 17:27:23 nan nan 44.21 nan
2019-07-15 17:28:23 nan nan 66.21 nan
2019-07-15 17:29:25 nan nan 11.21 nan
2019-07-15 17:30:25 nan nan 22.1 nan
2019-07-15 17:35:16 nan nan nan 21
2019-07-15 17:36:23 nan nan nan 22
2019-07-15 17:37:23 nan nan nan 31
2019-07-15 17:38:23 nan nan nan 31
2019-07-15 17:39:25 nan nan nan 22
2019-07-15 17:40:25 nan nan nan 11
如何实现上述结果,我认为我应该使用聚合查询,但我没有任何使用它的经验。不完全如此,但您可以随时使用json。 它将为您提供实现所需的所有数据
db.collection.aggregate([
{
$project: {
unit_id: true,
updatetime: true,
data: { $objectToArray: "$data" },
}
},
{
$project: {
mydata: {
$map: {
input: "$updatetime",
as: "input",
in: {
unit_id: "$unit_id",
date: "$$input",
data: {
$arrayToObject: {
$map: {
input: "$data",
as: "newInput",
in: {
k: "$$newInput.k",
v: { $ifNull: [{ $arrayElemAt: ["$$newInput.v", { $indexOfArray: ["$updatetime", "$$input"] }] }, "nan"] }
}
}
}
}
}
}
}
}
},
{
$unwind: "$mydata"
},
{
$group: {
_id: "result",
data: {
$push: "$mydata"
}
}
}
])
响应
不完全是这样,但您可以随时使用json。 它将为您提供实现所需的所有数据
db.collection.aggregate([
{
$project: {
unit_id: true,
updatetime: true,
data: { $objectToArray: "$data" },
}
},
{
$project: {
mydata: {
$map: {
input: "$updatetime",
as: "input",
in: {
unit_id: "$unit_id",
date: "$$input",
data: {
$arrayToObject: {
$map: {
input: "$data",
as: "newInput",
in: {
k: "$$newInput.k",
v: { $ifNull: [{ $arrayElemAt: ["$$newInput.v", { $indexOfArray: ["$updatetime", "$$input"] }] }, "nan"] }
}
}
}
}
}
}
}
}
},
{
$unwind: "$mydata"
},
{
$group: {
_id: "result",
data: {
$push: "$mydata"
}
}
}
])
响应
可能有多少个
通道
,如ch1、ch2?1.从所有文件中提取所有时间戳并将其列出2。检查每一份文件并填写频道的可用时间戳channel@ShivamMishra我猜不出来。像ch1、ch2这样的频道可能有多少?1.从所有文件中提取所有时间戳并将其列出2。检查每一份文件并填写频道的可用时间戳channel@ShivamMishra我猜不出来。有唯一id的列表,请参阅所有三个文档“unit_id”是唯一id,我添加了unit_id
,并更新了答案。请先检查project
pipeline,然后检查第二个project
pipeline中的map
函数。这没问题,我知道了,但假设我想在操作符[38,39,40]中使用$in。这样我只得到了三个单位id的聚合数据。是否需要按unit\u id
查询或过滤数据?实际上两者都是,仅筛选选定的数据,即[38,39,40]存在唯一id列表,请参阅所有三个文档“unit_id”是唯一id,我添加了unit_id
,并更新了答案。请先检查project
pipeline,然后检查第二个project
pipeline中的map
函数。这没问题,我知道了,但假设我想在操作符[38,39,40]中使用$in。这样我只得到了三个单位id的聚合数据。是否需要按unit\u id
查询或过滤数据?实际上两者都是,仅对选定数据进行过滤,即[38,39,40]