reduce 函数¶
reduce()将表达式逐个应用于列表中的元素，然后和累加器中的当前结果累加，最后返回完整结果。该函数将遍历给定列表中的每个元素e，在e上运行表达式并和累加器的当前结果累加，将新的结果存储在累加器中。这个函数类似于函数式语言（如 Lisp 和 Scala）中的 fold 或 reduce 方法。
openCypher 兼容性¶
在 openCypher 中，reduce()函数没有定义。nGQL 使用了 Cypher 方式实现reduce()函数。
语法¶
ngql
reduce(<accumulator> = <initial>, <variable> IN <list> | <expression>)
| 参数 | 说明 |
|---|---|
| accumulator | 在遍历列表时保存累加结果。 |
| initial | 为accumulator提供初始值的表达式或值。 |
| variable | 为列表引入一个变量，决定使用列表中的哪个元素。 |
| list | 列表或列表表达式。 |
| expression | 该表达式将对列表中的每个元素运行一次，并将结果累加至accumulator。 |
Note
返回值的类型取决于提供的参数，以及表达式的语义。
示例¶
```ngql nebula> RETURN reduce(totalNum = 10, n IN range(1, 3) | totalNum + n) AS r; +----+ | r | +----+ | 16 | +----+
nebula> RETURN reduce(totalNum = -4 * 5, n IN [1, 2] | totalNum + n * 2) AS r; +-----+ | r | +-----+ | -14 | +-----+
nebula> MATCH p = (n:player{name:"LeBron James"})<-[:follow]-(m) \ RETURN nodes(p)[0].age AS src1, nodes(p)[1].age AS dst2, \ reduce(totalAge = 100, n IN nodes(p) | totalAge + n.age) AS sum; +------+------+-----+ | src1 | dst2 | sum | +------+------+-----+ | 34 | 31 | 165 | | 34 | 29 | 163 | | 34 | 33 | 167 | | 34 | 26 | 160 | | 34 | 34 | 168 | | 34 | 37 | 171 | +------+------+-----+
nebula> LOOKUP ON player WHERE player.name == "Tony Parker" YIELD id(vertex) AS VertexID \ | GO FROM \(-.VertexID over follow \ WHERE properties(edge).degree != reduce(totalNum = 5, n IN range(1, 3) | properties(\)\().age + totalNum + n) \ YIELD properties(\)\().name AS id, properties(\)$).age AS age, properties(edge).degree AS degree; +---------------------+-----+--------+ | id | age | degree | +---------------------+-----+--------+ | "Tim Duncan" | 42 | 95 | | "LaMarcus Aldridge" | 33 | 90 | | "Manu Ginobili" | 41 | 95 | +---------------------+-----+--------+ ```