Is it possible to calculate the number of occurence of distinct values in a list field.
For example, let the following data:
[
{
"page":1,
"colors":[
{
"color": red
},
{
"color": white
},
{
"color": red
}
]
},
{
"page":2,
"colors":[
{
"color": yellow
},
{
"color": yellow
}
]
}
]
Is it possible to get a result as the follwing:
{
"page":1,
"colors_count":[
{
"Key": red,
"Count": 2
},
{
"Key": white,
"Count": 1
},
]
},
{
"page":2,
"colors_count":[
{
"Key": yellow,
"Count": 2
}
]
}
I tried using term aggregation but I got the number of distinct values, so for page:1 i got red:1 and white:1.
CodePudding user response:
Yes, you can do it. you will have to use nested_field type and nested_Agg
Mapping:
PUT colors
{
"mappings": {
"properties": {
"page" : { "type": "keyword" },
"colors": {
"type": "nested",
"properties": {
"color": {
"type": "keyword"
}
}
}
}
}
}
Insert Documents:
PUT colors/_doc/1
{
"page": 1,
"colors": [
{
"color": "red"
},
{
"color": "white"
},
{
"color": "red"
}
]
}
PUT colors/_doc/2
{
"page": 2,
"colors": [
{
"color": "yellow"
},
{
"color": "yellow"
}
]
}
Query:
GET colors/_search
{
"size" :0,
"aggs": {
"groupByPage": {
"terms": {
"field": "page"
},
"aggs": {
"colors": {
"nested": {
"path": "colors"
},
"aggs": {
"genres": {
"terms": {
"field": "colors.color"
}
}
}
}
}
}
}
}
Output:
{
"took": 3,
"timed_out": false,
"_shards": {
"total": 1,
"successful": 1,
"skipped": 0,
"failed": 0
},
"hits": {
"total": {
"value": 2,
"relation": "eq"
},
"max_score": null,
"hits": []
},
"aggregations": {
"groupByPage": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "1", // page field value
"doc_count": 1,
"colors": {
"doc_count": 3,
"genres": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "red",
"doc_count": 2
},
{
"key": "white",
"doc_count": 1
}
]
}
}
},
{
"key": "2", // page field value
"doc_count": 1,
"colors": {
"doc_count": 2,
"genres": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "yellow",
"doc_count": 2
}
]
}
}
}
]
}
}
}