我有像这样的弹性搜索聚合查询。
{
"aggs": {
"customer": {
"aggs": {
"Total_Sale": {
"sum": {
"field": "amount"
}
}
},
"terms": {
"field": "org",
"size": 50000
}
}
}
}
它会导致桶聚合如下
{
"aggregations": {
"customer": {
"buckets": [
{
"Total_Sale": { "value": 9999 },
"doc_count": 8,
"key": "cats"
},
{
"Total_Sale": { "value": 8888 },
"doc_count": 6,
"key": "tigers"
},
{
"Total_Sale": { "value": 444},
"doc_count": 5,
"key": "lions"
},
{
"Total_Sale": { "value": 555 },
"doc_count": 2,
"key": "wolves"
}
]
}
}
}
我想要另一个基于doc_count的范围桶聚合。因此,最终结果是
{
"buckets": [
{
"Sum_of_Total_Sale": 555, // If I can form bucket, I can get this using sum_bucket. So, getting bucket is important.
"Sum_of_doc_count": 2,
"doc_count": 1,
"key": "*-3",
"to": 3.0
},
{
"Sum_of_Total_Sale": 9332,
"Sum_of_doc_count": 11,
"doc_count": 2,
"from": 4.0,
"key": "4-6",
"to": 6.0
},
{
"Sum_of_Total_Sale": 9999,
"Sum_of_doc_count": 8,
"doc_count": 1,
"from": 7.0,
"key": "7-*"
}
]
}
答案 0 :(得分:0)
据我所知,没有一种汇总可以让您一口气做到这一点。但是,我不时使用一种技术来克服此限制。想法是重复相同的terms/sum
聚合,然后对您感兴趣的每个范围使用bucket_selector
管道聚合。
POST index/_search
{
"size": 0,
"aggs": {
"*-3": {
"terms": {
"field": "org",
"size": 1000
},
"aggs": {
"Total_Sale": {
"sum": {
"field": "amount"
}
},
"*-3": {
"bucket_selector": {
"buckets_path": {
"docCount": "_count"
},
"script": "params.docCount <= 3"
}
}
}
},
"*-3_Total_Sales": {
"sum_bucket": {
"buckets_path": "*-3>Total_Sale"
}
},
"*-3_Total_Docs": {
"sum_bucket": {
"buckets_path": "*-3>_count"
}
},
"4-6": {
"terms": {
"field": "org",
"size": 1000
},
"aggs": {
"Total_Sale": {
"sum": {
"field": "amount"
}
},
"4-6": {
"bucket_selector": {
"buckets_path": {
"docCount": "_count"
},
"script": "params.docCount >= 4 && params.docCount <= 6"
}
}
}
},
"4-6_Total_Sales": {
"sum_bucket": {
"buckets_path": "4-6>Total_Sale"
}
},
"4-6_Total_Docs": {
"sum_bucket": {
"buckets_path": "4-6>_count"
}
},
"7-*": {
"terms": {
"field": "org",
"size": 1000
},
"aggs": {
"Total_Sale": {
"sum": {
"field": "amount"
}
},
"7-*": {
"bucket_selector": {
"buckets_path": {
"docCount": "_count"
},
"script": "params.docCount >= 7"
}
}
}
},
"7-*_Total_Sales": {
"sum_bucket": {
"buckets_path": "7-*>Total_Sale"
}
},
"7_*_Total_Docs": {
"sum_bucket": {
"buckets_path": "7-*>_count"
}
}
}
}
您将获得一个类似以下的答案,其中包含您在xyz_Total_Sales
和xyz_Total_Docs
结果中寻找的数字:
"aggregations": {
"*-3": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "wolves",
"doc_count": 2,
"Total_Sale": {
"value": 555
}
}
]
},
"7-*": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "cats",
"doc_count": 8,
"Total_Sale": {
"value": 9999
}
}
]
},
"4-6": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "tigers",
"doc_count": 6,
"Total_Sale": {
"value": 8888
}
},
{
"key": "lions",
"doc_count": 5,
"Total_Sale": {
"value": 444
}
}
]
},
"*-3_Total_Sales": {
"value": 555
},
"*-3_Total_Docs": {
"value": 2
},
"4-6_Total_Sales": {
"value": 9332
},
"4-6_Total_Docs": {
"value": 11
},
"7-*_Total_Sales": {
"value": 9999
},
"7_*_Total_Docs": {
"value": 8
}
}