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我正在处理 ADX 上的一个大数据集，我需要在其中解开数据行并将它们转换为列。数据中的uniqueID由三个字段组合而成：组、时间戳和名称。

  datatable(Group:string, timestamp:datetime, channel_name:string , value:long)
[
    "A", datetime(2019-05-01 00:00:01), "channel_1", 12,
    "A", datetime(2019-05-01 00:00:02), "channel_1",  14,
    "A", datetime(2019-05-01 00:00:03),  "channel_1",  16,
    "A", datetime(2019-05-01 00:00:01), "channel_2", 12,
    "A", datetime(2019-05-01 00:00:02), "channel_2",  14,
    "A", datetime(2019-05-01 00:00:01),  "channel_3",  16,
    "B", datetime(2019-04-01 00:00:01),  "channel_1", 3,
    "B", datetime(2019-04-01 00:00:04),  "channel_1", 5,
    "B", datetime(2019-04-01 00:00:07),  "channel_2", 1,
    "B", datetime(2019-04-01 00:00:10),  "channel_3", 8,
]

一组的预期输出（结果可以有或没有组列——因为总是应用组过滤器）

group, timestamp, channel_1, channel_2, channel_3
    "A", datetime(2019-05-01 00:00:01), 12, 12, NULL,
    "A", datetime(2019-05-01 00:00:02), 14,14, NULL,
    "A", datetime(2019-05-01 00:00:03), 16, NULL,NULL,

尝试运行以下查询（基于），但这不会按预期取消堆叠列。这以与上述相同的格式返回数据。

| where timestamp > datetime(2019-04-01) and timestamp <datetime(2019-04-03) \\ filter1 Always applied
| where machine_name =='A' \\filter2 Always Applied

| where channel_name  in  ("channel1, channel2, channel3")
| summarize value=sum(value) by channel_name, timestamp
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1 回答 1

2

你可以试试这个：

datatable(group:string, timestamp:datetime, channel_name:string , value:long)
[
    "A", datetime(2019-05-01 00:00:01), "channel_1", 12,
    "A", datetime(2019-05-01 00:00:02), "channel_1", 14,
    "A", datetime(2019-05-01 00:00:03), "channel_1", 16,
    "A", datetime(2019-05-01 00:00:01), "channel_2", 12,
    "A", datetime(2019-05-01 00:00:02), "channel_2", 14,
    "A", datetime(2019-05-01 00:00:01), "channel_3", 16,
    "B", datetime(2019-04-01 00:00:01), "channel_1", 3,
    "B", datetime(2019-04-01 00:00:04), "channel_1", 5,
    "B", datetime(2019-04-01 00:00:07), "channel_2", 1,
    "B", datetime(2019-04-01 00:00:10), "channel_3", 8,
]
| where group == "A"
| summarize b = make_bag(pack(channel_name, value)) by timestamp
| project timestamp, channel_1 = tolong(b.channel_1), channel_2 = tolong(b.channel_2), channel_3 = tolong(b.channel_3)

或者这个（由于使用效率较低bag_unpack()）：

datatable(group:string, timestamp:datetime, channel_name:string , value:long)
[
    "A", datetime(2019-05-01 00:00:01), "channel_1", 12,
    "A", datetime(2019-05-01 00:00:02), "channel_1", 14,
    "A", datetime(2019-05-01 00:00:03), "channel_1", 16,
    "A", datetime(2019-05-01 00:00:01), "channel_2", 12,
    "A", datetime(2019-05-01 00:00:02), "channel_2", 14,
    "A", datetime(2019-05-01 00:00:01), "channel_3", 16,
    "B", datetime(2019-04-01 00:00:01), "channel_1", 3,
    "B", datetime(2019-04-01 00:00:04), "channel_1", 5,
    "B", datetime(2019-04-01 00:00:07), "channel_2", 1,
    "B", datetime(2019-04-01 00:00:10), "channel_3", 8,
]
| where group == "A"
| summarize b = make_bag(pack(channel_name, value)) by timestamp
| evaluate bag_unpack(b)

两者都输出此表：

| timestamp                   | channel_1 | channel_2 | channel_3 |
|-----------------------------|-----------|-----------|-----------|
| 2019-05-01 00:00:01.0000000 | 12        | 12        | 16        |
| 2019-05-01 00:00:02.0000000 | 14        | 14        |           |
| 2019-05-01 00:00:03.0000000 | 16        |           |           |
于 2021-01-20T21:51:07.227 回答