elasticsearch 我想将Elasticsearch中的重复值合并为一个,并使用不同的过滤器查看结果
我通过弹性搜索收集日志。日志收集如下elasticsearch 我想将Elasticsearch中的重复值合并为一个,并使用不同的过滤器查看结果,elasticsearch,filter,distinct,elasticsearch,Filter,Distinct,我通过弹性搜索收集日志。日志收集如下 ex. { "name" : "John" "team" : "IT" "startTime" : "21:00" "result" : "pass" }, { "name" : "James" "team" : "HR"
ex.
{
"name" : "John"
"team" : "IT"
"startTime" : "21:00"
"result" : "pass"
},
{
"name" : "James"
"team" : "HR"
"startTime" : "21:04"
"result" : "pass"
},
{
"name" : "Paul"
"team" : "IT"
"startTime" : "21:05"
"result" : "pass"
},
{
"name" : "Jackson"
"team" : "Marketing"
"startTime" : "21:30"
"result" : "fail"
},
{
"name" : "John"
"team" : "IT"
"startTime" : "21:41"
"result" : "pass"
},
.....and so on
如果对这些收集的日志运行下面的查询
GET logData/_search
{
"size": 0,
"aggs": {
"Documents_per_team": {
"terms": {
"field": "team"
}
}
}
}
将公开以下结果
"aggregations" : {
"Documents_per_team" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "IT",
"doc_count" : 70
},
{
"key" : "Marketing",
"doc_count" : 55
},
{
"key" : "HR",
"doc_count" : 11
}
]
}
}
}
我想要的是,如果文档的名称在此结果中重复,则消除重复
[现状]
- 如上图所示,IT团队数量为70
- 如果John表演了50次,Kate表演了10次,Paul表演了10次,那么IT团队的计数3就暴露了。(因为有三名IT团队成员)
谢谢你有两个选择:
名称
s在每个团队级别都是唯一的。如果他们没有,你就需要。此外,假设名称
映射为关键字
类型,就像团队
一样。如果没有,您需要将它们替换为您的_字段。关键字
1.基数
2.脚本度量
注意:如果您还希望看到基础部门与名称的细分,您可以修改
组合脚本
以返回整个状态,即返回状态
,您有两个选项:
名称
s在每个团队级别都是唯一的。如果他们没有,你就需要。此外,假设名称
映射为关键字
类型,就像团队
一样。如果没有,您需要将它们替换为您的_字段。关键字
1.基数
2.脚本度量
注意:如果您还希望看到基础部门与名称的细分,您可以修改合并脚本
以返回整个状态,即返回状态
{
"size": 0,
"aggs": {
"Documents_per_team": {
"terms": {
"field": "team"
},
"aggs": {
"unique_names_per_team": {
"cardinality": {
"field": "name"
}
}
}
}
}
}
{
"size": 0,
"aggs": {
"Documents_per_team": {
"scripted_metric": {
"init_script": "state.by_department = [:]; state.dept_vs_name = [:];",
"map_script": """
def dept = doc['team'].value;
def name = doc['name'].value;
def name_already_considered = state.by_department.containsKey(dept) && state.dept_vs_name[dept].containsKey(name);
if (name_already_considered) {
return;
}
if (state.by_department.containsKey(dept)) {
state.by_department[dept] += 1;
} else {
state.by_department[dept] = 1
}
if (!state.dept_vs_name.containsKey(dept)) {
// init new map & set is first member
state.dept_vs_name[dept] = [name:true];
} else if (!state.dept_vs_name[dept].containsKey(name)) {
state.dept_vs_name[dept][name] = true;
}
""",
"combine_script": "return state.by_department",
"reduce_script": "return states"
}
}
}
}