英文:
Is it possible to count all the items that satisfy predicate and the ones that don't in one iteration over the table?
问题
我有一个问题：
var x = from c in _context
[...]
group subResult by subResult.Id into g
select new
{
g.Key,
Count1 = g.Where(predicate).Count(),
Count2 = g.Where(!predicate).Count()
}
如果 g.Count()==Count1 + Count2，那么在 Count1 和 Count2 的谓词中是否有可能简化查询，以便不重复计算并且仍然在服务器端进行计数？
编辑：这是我得到的结果
英文:
I have a query:
var x = from c in _context
[...]
group subResult by subResult.Id into g
select new
{
g.Key,
Count1 = g.Where(predicate).Count(),
Count2 = g.Where(!predicate).Count()
}
Is it possible to simplify query if predicates used for Count1 and Count2 if g.Count()==Count1 + Count2 so counting isn't done twice and still on server-side?
EDIT: This is what I get
答案1
得分: 1
这是如何使用Aggregate：
var x =
from c in _context
[...]
group subResult by subResult.Id into g
let count = g.Aggregate((t: 0, f: 0), (a, x) => predicate(x) ? (a.t + 1, a.f) : (a.t, a.f + 1))
select new
{
g.Key,
Count1 = count.t,
Count2 = count.f
};
实际上，你想知道g的类型，如果它是一个IPartition<T>，那么Count是O(1)，然后Aggregate的性能会更差。
英文:
Here's how using Aggregate:
var x =
from c in _context
[...]
group subResult by subResult.Id into g
let count = g.Aggregate((t: 0, f : 0), (a, x) => predicate(x) ? (a.t + 1, a.f) : (a.t, a.f + 1))
select new
{
g.Key,
Count1 = count.t,
Count2 = count.f
};
You do actually want to know what type g is as if it is a IPartition<T> then Count is O(1) and then Aggregate performs more poorly.
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