Arrays 数组与x27之间的和；s元素和用Cypher语言创建新数组

我需要将数组中的一个元素与前面的元素求和，然后用这些元素创建一个新数组。 现在我以手动方式使用此代码：

match (a:User) 
with collect(a.capital) as cap 
with cap as cap, length(cap) as len
return cap[0], 
       cap[0]+cap[1], 
       cap[0]+cap[1]+cap[2],
       cap[0]+cap[1]+cap[2]+cap[3], 
       cap[0]+cap[1]+cap[2]+cap[3]+cap[4], 
       cap[0]+cap[1]+cap[2]+cap[3]+cap[4]+cap[5], 
       cap[0]+cap[1]+cap[2]+cap[3]+cap[4]+cap[5]+cap[6], 
       len

但我需要使用一个查询密码对不同长度的数组执行此操作。

Cypher具有以下功能：

WITH [12,28,74,45] AS cap
RETURN reduce(x = 0, v IN cap | x + v)
// 159

您还可以利用：

更新

WITH [12,28,74,45] AS cap
UNWIND range(1, size(cap)) AS i
WITH apoc.coll.sum(cap[0..i-1])
     + apoc.coll.sum(cap[0..i]) AS sum
RETURN collect(sum)

结果:

╒═══════════════════════╕
│"collect(sum)"         │
╞═══════════════════════╡
│[12.0,52.0,154.0,273.0]│
└───────────────────────┘

---

谢谢，我用这种方式在我的数据库上进行了尝试，它正在工作：

   WITH [12,28,74,45] AS cap
   UNWIND range(1, size(cap)) AS i
   WITH apoc.coll.sum(cap[0..i]) AS sum
   RETURN collect(sum)

---

按照你的例子，我需要一个新的数组[12,12+28,12+28+74,12+28+74+45]

   WITH [12,28,74,45] AS cap
   UNWIND range(1, size(cap)) AS i
   WITH apoc.coll.sum(cap[0..i]) AS sum
   RETURN collect(sum)