ElasticSearch在另一个聚合结果上使用聚合
有一个对话列表，每个对话都有一个消息列表。每个消息都有一个不同的字段和一个action字段。我们需要考虑的是，在对话的第一条消息中使用了动作A，在几条消息中使用了动作A.1之后，过了一会儿A.1.1，依此类推（有一个聊天机器人意图列表）。
将对话的消息动作分组将类似于： A > A > A > A.1 > A > A.1 > A.1.1 ...
我需要使用ElasticSearch创建一个报告，该报告将返回actions group每次会话的；接下来，我需要对类似的东西进行分组并actions
groups添加一个计数；最终将导致Map<actionsGroup, count>as 'A > A.1 > A > A.1 > A.1.1',
3。
构建actions groupI需要消除每组重复项；而不是A > A > A > A.1 > A > A.1 > A.1.1我需要拥有A >
A.1 > A > A.1 > A.1.1。
：
{ "collapse":{
"field":"context.conversationId",
"inner_hits":{
"name":"logs",
"size": 10000,
"sort":[
{
"@timestamp":"asc"
}
]
}
},
"aggs":{
},
}
- 我需要将崩溃的结果映射到单个结果中
A > A.1 > A > A.1 > A.1.1。我已经看到，在这种情况下，或者aggr有可能在结果中使用脚本，并且有可能创建一个我需要aggr执行的动作列表，但它会对所有消息进行操作，而不仅仅是对我组合的消息进行操作陷入崩溃。是否可以使用aggr内部塌陷或类似的解决方案？ - 我需要对
A > A.1 > A > A.1 > A.1.1所有折叠的结果values（）进行分组，添加一个计数并得出Map<actionsGroup, count>。
conversationId使用字段将对话消息分组aggr（我不知道该怎么做）- 使用脚本来迭代所有值并
actions group为每个对话创建。（不确定是否可行） aggr在所有值上使用另一个，并将重复项分组，返回Map<actionsGroup, count>。
添加一些其他详细信息
对应：
"mappings":{ "properties":{
"@timestamp":{
"type":"date",
"format": "epoch_millis"
}
"context":{
"properties":{
"action":{
"type":"keyword"
},
"conversationId":{
"type":"keyword"
}
}
}
}
}
对话的样本文件：
Conversation 1.{
"@timestamp": 1579632745000,
"context": {
"action": "A",
"conversationId": "conv_id1",
}
},
{
"@timestamp": 1579632745001,
"context": {
"action": "A.1",
"conversationId": "conv_id1",
}
},
{
"@timestamp": 1579632745002,
"context": {
"action": "A.1.1",
"conversationId": "conv_id1",
}
}
Conversation 2.
{
"@timestamp": 1579632745000,
"context": {
"action": "A",
"conversationId": "conv_id2",
}
},
{
"@timestamp": 1579632745001,
"context": {
"action": "A.1",
"conversationId": "conv_id2",
}
},
{
"@timestamp": 1579632745002,
"context": {
"action": "A.1.1",
"conversationId": "conv_id2",
}
}
Conversation 3.
{
"@timestamp": 1579632745000,
"context": {
"action": "B",
"conversationId": "conv_id3",
}
},
{
"@timestamp": 1579632745001,
"context": {
"action": "B.1",
"conversationId": "conv_id3",
}
}
预期结果：
{ "A -> A.1 -> A.1.1": 2,
"B -> B.1": 1
}
Something similar, having this or any other format.
回答：
我用scripted_metric弹性的解决了。而且，的index状态已从初始状态更改。
剧本：
{ "size": 0,
"aggs": {
"intentPathsCountAgg": {
"scripted_metric": {
"init_script": "state.messagesList = new ArrayList();",
"map_script": "long currentMessageTime = doc['messageReceivedEvent.context.timestamp'].value.millis; Map currentMessage = ['conversationId': doc['messageReceivedEvent.context.conversationId.keyword'], 'time': currentMessageTime, 'intentsPath': doc['brainQueryRequestEvent.brainQueryRequest.user_data.intentsHistoryPath.keyword'].value]; state.messagesList.add(currentMessage);",
"combine_script": "return state",
"reduce_script": "List messages = new ArrayList(); Map conversationsMap = new HashMap(); Map intentsMap = new HashMap(); String[] ifElseWorkaround = new String[1]; for (state in states) { messages.addAll(state.messagesList);} messages.stream().forEach((message) -> { Map existingMessage = conversationsMap.get(message.conversationId); if(existingMessage == null || message.time > existingMessage.time) { conversationsMap.put(message.conversationId, ['time': message.time, 'intentsPath': message.intentsPath]); } else { ifElseWorkaround[0] = ''; } }); conversationsMap.entrySet().forEach(conversation -> { if (intentsMap.containsKey(conversation.getValue().intentsPath)) { long intentsCount = intentsMap.get(conversation.getValue().intentsPath) + 1; intentsMap.put(conversation.getValue().intentsPath, intentsCount); } else {intentsMap.put(conversation.getValue().intentsPath, 1L);} }); return intentsMap.entrySet().stream().map(intentPath -> [intentPath.getKey().toString(): intentPath.getValue()]).collect(Collectors.toSet()) "
}
}
}
}
格式化脚本（为了提高可读性-使用.ts）：
scripted_metric: { init_script: 'state.messagesList = new ArrayList();',
map_script: `
long currentMessageTime = doc['messageReceivedEvent.context.timestamp'].value.millis;
Map currentMessage = [
'conversationId': doc['messageReceivedEvent.context.conversationId.keyword'],
'time': currentMessageTime,
'intentsPath': doc['brainQueryRequestEvent.brainQueryRequest.user_data.intentsHistoryPath.keyword'].value
];
state.messagesList.add(currentMessage);`,
combine_script: 'return state',
reduce_script: `
List messages = new ArrayList();
Map conversationsMap = new HashMap();
Map intentsMap = new HashMap();
boolean[] ifElseWorkaround = new boolean[1];
for (state in states) {
messages.addAll(state.messagesList);
}
messages.stream().forEach(message -> {
Map existingMessage = conversationsMap.get(message.conversationId);
if(existingMessage == null || message.time > existingMessage.time) {
conversationsMap.put(message.conversationId, ['time': message.time, 'intentsPath': message.intentsPath]);
} else {
ifElseWorkaround[0] = true;
}
});
conversationsMap.entrySet().forEach(conversation -> {
if (intentsMap.containsKey(conversation.getValue().intentsPath)) {
long intentsCount = intentsMap.get(conversation.getValue().intentsPath) + 1;
intentsMap.put(conversation.getValue().intentsPath, intentsCount);
} else {
intentsMap.put(conversation.getValue().intentsPath, 1L);
}
});
return intentsMap.entrySet().stream().map(intentPath -> [
'path': intentPath.getKey().toString(),
'count': intentPath.getValue()
]).collect(Collectors.toSet())`
答案：
{ "took": 2,
"timed_out": false,
"_shards": {
"total": 5,
"successful": 5,
"skipped": 0,
"failed": 0
},
"hits": {
"total": {
"value": 11,
"relation": "eq"
},
"max_score": null,
"hits": []
},
"aggregations": {
"intentPathsCountAgg": {
"value": [
{
"smallTalk.greet -> smallTalk.greet2 -> smallTalk.greet3": 2
},
{
"smallTalk.greet -> smallTalk.greet2 -> smallTalk.greet3 -> smallTalk.greet4": 1
},
{
"smallTalk.greet -> smallTalk.greet2": 1
}
]
}
}
}
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