当某事物的id值发生变化时,如何及时写下记录?当ture-> false和false->为某些列表时为每条记录的id?
例如表
id value
1 0
2 0
2 0
2 0
1 0
2 1 --> the only changes here
2 1
1 0
2 0 --> and here (node with id 2 changed 1 -> 0 )
1 1 --> node with id 1 changed 0 -> 1
结果表
2 1
2 0
1 1
我的想法不起作用,有点奇怪,我正在考虑功能性或linq制作相同的方法。
let oop = ref (filteredsq
|> Seq.distinctBy(fun (node,v,k) -> k)
|> Seq.map(fun (node,v,k) -> k, false )
|> Array.ofSeq )
[for (node,value,key) in filteredsq do
let i = ref 0
for (k,v) in !oop do
if key = k && value <> v then
(!oop).[!i] <- (k,value)
yield node
i := !i + 1 ]
谢谢
答案 0 :(得分:3)
我认为如果您定义如下函数:
let getChanges f items =
items
|> Seq.map (fun x -> f x, x)
|> Seq.pairwise
|> Seq.choose (fun ((a, _), (b, x)) -> if a <> b then Some x else None)
然后你可以这样做:
filteredsq
|> Seq.groupBy (fun (_, _, k) -> k)
|> Seq.collect (fun (_, items) ->
items
|> getChanges (fun (_, value, _) -> value)
|> Seq.map (fun (node, _, _) -> node))
|> Seq.toList
答案 1 :(得分:1)
我会做像
这样的事情List
|> List.toSeq
|> Seq.pairwise
|> Seq.pick (fun ((fa,fb),(sa,sb)) -> if fb <> sb then Some(sa,sb) else None)
答案 2 :(得分:1)
<强>更新强>
open System.Collections.Generic
let filter (acc:'a) (f:('a -> 'b -> bool * 'a)) (s:'b seq) =
let rec iter (acc:'a) (e:IEnumerator<'b>) =
match e.MoveNext() with
| false -> Seq.empty
| true -> match f acc e.Current with
| (true,newAcc) -> seq { yield e.Current; yield! iter newAcc e}
| (false,newAcc) -> seq { yield! iter newAcc e}
iter acc (s.GetEnumerator())
let skipUntilChange (f : 'a -> 'b) (s : 'a seq) =
s |> Seq.skip 1
|> filter (s |> Seq.head |> f)
(fun a b -> if a = f b then false,f b else true,f b)
[(1,0);(2,0);(2,0);(2,0);(1,0);(2,1);(2,1);(1,0);(2,0);]
|> Seq.mapi (fun c (i,v) -> (i,v,c))
|> Seq.groupBy (fun (i,v,c) -> i)
|> Seq.map (snd >> skipUntilChange (fun (_,v,_) -> v))
|> Seq.concat |> Seq.sortBy (fun (i,v,c) -> c)
|> Seq.map (fun (i,v,c) -> (i,v))
|> printfn "%A"
答案 3 :(得分:1)
我不确定我是否完全理解您的问题,但以下根据您的样本给出正确的输出。我们的想法是首先过滤掉没有正确密钥的值,然后使用Seq.pairwaise
(如jpalmer的解决方案)来查找值发生变化的位置:
let input = [ (1, 0); (2, 0); (2, 0); (2, 0); (1, 0); (2, 1); (2, 1); (1, 0); (2, 0) ]
let findValueChanges key input =
input
|> Seq.filter (fun (k, v) -> k = key) // Get values with the right key
|> Seq.pairwise // Make tuples with previous & next value
|> Seq.filter (fun ((_, prev), (_, next)) -> prev <> next) // Find changing points
|> Seq.map snd // Return the new key-value pair (after change)
如果您想查找所有不同密钥的更改,则可以使用Seq.groupBy
查找所有可能的密钥(然后您不需要findValueChanges
中的第一行):
input
|> Seq.groupBy fst
|> Seq.map (fun (k, vals) -> findValueChanges k vals)
(对于您的输入,键1的值没有变化,因为值始终为1, 0
)
答案 4 :(得分:1)
我只是使用内部可变字典来跟踪每个键的最后看到的值,并且当任何值与该键的最后一个值不同时,得到(键,值):
let filterChanges (s:('a*'b) seq) =
let dict = new System.Collections.Generic.Dictionary<'a,'b>()
seq {
for (key,value) in s do
match dict.TryGetValue(key) with
| false,_ -> dict.[key] <- value
| true,lastValue ->
if lastValue <> value then
yield (key,value)
dict.[key] <- value
}
测试:
> filterChanges [(1,0);(2,0);(2,0);(2,0);(1,0);(2,1);(2,1);(1,0);(2,0);(1,1)];;
val it : seq<int * int> = seq [(2, 1); (2, 0); (1, 1)]