响应式扩展观察者的问题
本文关键字：问题 观察者 扩展 响应 | 更新日期: 2023-09-27 18:08:26
我正在开发一个使用响应式扩展的应用程序，遇到了以下问题:
假设我有两个观察者p和Q，我想建立第三个观察者R，如果两个p值没有Q, R输出0。如果P之后是Q，则R输出传递这些值的方法的结果，例如:
P0 Q0 -> R0 = f(P0,Q0)
P1 -> R1 = 0
P2 Q1 -> R2 = f(P2,Q1)
P3 -> R3 = 0
P4 -> R4 = 0
P5 Q2 -> R5 = f(P5,Q2)
(...)
和值按以下顺序进入观察器:
P0 Q0 P1 P2 Q1 P3 P4 P5 Q2
谢谢你的帮助。
我想我有一个解决办法。
如果我假设你有以下定义:
IObservable<int> ps = ...;
IObservable<int> qs = ...;
Func<int, int, int> f = ...;
首先，我创建一个函数字典来计算最终值:
var fs = new Dictionary<string, Func<int, int, int?>>()
{
{ "pp", (x, y) => 0 },
{ "pq", (x, y) => f(x, y) },
{ "qp", (x, y) => null },
{ "qq", (x, y) => null },
};
"p" &"q"在那里
然后你可以创建一个合并的观察对象，像这样:
var pqs =
(from p in ps select new { k = "p", v = p })
.Merge(from q in qs select new { k = "q", v = q });
我现在知道哪个序列产生了哪个值。
接下来，我发布合并列表，因为我不知道源可观察对象是热的还是冷的——所以发布它们会使它们成为热的——然后我将发布的可观察对象压缩到自己，分别跳过1和0。然后我知道每一对值和它们来自的原始可观测值。这样就很容易应用字典函数(过滤掉任何空值)。
在这里:
var rs =
from kvv in pqs.Publish(_pqs =>
_pqs.Skip(1).Zip(_pqs, (pq1, pq0) => new
{
k = pq0.k + pq1.k,
v1 = pq1.v,
v0 = pq0.v
}))
let r = fs[kvv.k](kvv.v0, kvv.v1)
where r.HasValue
select r.Value;
这对你有用吗?
总体思路很简单:您合并p和Q，使用BufferWithCount(2)获得值对，然后根据您的逻辑处理对:
<>之前
P.Merge(Q).BufferWithCount(2).Select(values =>
{
var first = values[0];
var second = values[1];
if (first is P && second is P ||
first is Q && second is Q)
{
return 0;
}
if (first is P)
{
return selector(first, second);
}
else // suppose Q, P is a valid sequence as well.
{
return selector(second, first);
}
});
之前现在困难的部分是合并p和Q，如果它们是不同的类型，然后在Select中区分它们。如果它们是相同的类型，你可以使用一些简单的方法，比如Enigmativity提出的方法，即
<>之前
var pqs =
(from p in ps select new { k = "p", v = p })
.Merge(from q in qs select new { k = "q", v = q });
之前现在困难的部分是，如果它们是不同的类型，要合并它们，我们需要一些通用的包装类型，例如，Data。从Haskell:
<>之前
public abstract class Either<TLeft, TRight>
{
private Either()
{
}
public static Either<TLeft, TRight> Create(TLeft value)
{
return new Left(value);
}
public static Either<TLeft, TRight> Create(TRight value)
{
return new Right(value);
}
public abstract TResult Match<TResult>(
Func<TLeft, TResult> onLeft,
Func<TRight, TResult> onRight);
public sealed class Left : Either<TLeft, TRight>
{
public Left(TLeft value)
{
this.Value = value;
}
public TLeft Value
{
get;
private set;
}
public override TResult Match<TResult>(
Func<TLeft, TResult> onLeft,
Func<TRight, TResult> onRight)
{
return onLeft(this.Value);
}
}
public sealed class Right : Either<TLeft, TRight>
{
public Right(TRight value)
{
this.Value = value;
}
public TRight Value
{
get;
private set;
}
public override TResult Match<TResult>(
Func<TLeft, TResult> onLeft,
Func<TRight, TResult> onRight)
{
return onRight(this.Value);
}
}
}
之前有趣的是，在system . react .dll中已经有类似的类了，不幸的是它是内部的，所以我们需要自己的实现。现在我们可以将P和Q都放入任意一个，并继续求解(我已经将其一般化了一点，因此您可以返回任何结果，而不是只返回int):<>之前
public static IObservable<TResult> SmartZip<TLeft, TRight, TResult>(
IObservable<TLeft> leftSource,
IObservable<TRight> rightSource,
Func<TLeft, TRight, TResult> selector)
{
return Observable
.Merge(
leftSource.Select(Either<TLeft, TRight>.Create),
rightSource.Select(Either<TLeft, TRight>.Create))
.BufferWithCount(2)
.Select(values =>
{
// this case was not covered in your question,
// but I've added it for the sake of completeness.
if (values.Count < 2)
{
return default(TResult);
}
var first = values[0];
var second = values[1];
// pattern-matching in C# is really ugly.
return first.Match(
left => second.Match(
_ => default(TResult),
right => selector(left, right)),
right => second.Match(
left => selector(left, right),
_ => default(TResult)));
});
}
之前这里是所有这些可怕的丑陋的东西的一个小演示。
<>之前
private static void Main(string[] args)
{
var psource = Observable
.Generate(1, i => i < 100, i => i, i => i + 1)
.Zip(Observable.Interval(TimeSpan.FromMilliseconds(10.0)), (i, _) => i);
var qsource = Observable
.Generate(1, i => i < 100, i => (double)i * i, i => i + 1)
.Zip(Observable.Interval(TimeSpan.FromMilliseconds(30.0)), (i, _) => i);
var result = SmartZip(
psource,
qsource,
(p, q) => q / p).ToEnumerable();
foreach (var item in result)
{
Console.WriteLine(item);
}
}
如果我正确理解了你的问题，那么下面是一个可以处理这种情况的通用函数:
public static IObservable<T> MyCombiner<T>(IObservable<T> P, IObservable<T> Q, T defaultValue,Func<T,T,T> fun)
{
var c = P.Select(p => new { Type = 'P', Value = p })
.Merge(Q.Select(p => new { Type = 'Q', Value = p }));
return c.Zip(c.Skip(1), (a, b) =>
{
if (a.Type == 'P' && b.Type == 'P')
return new { Ok = true, Value = defaultValue };
if (a.Type == 'P' && b.Type == 'Q')
return new { Ok = true, Value = fun(a.Value, b.Value) };
else
return new { Ok = false, Value = default(T) };
}).Where(b => b.Ok).Select(b => b.Value);
}
假设我们有两个方法
- Before，每当第一个可观察对象在第二个可观察对象之前产生元素时，通过使用选择器函数将两个可观察序列合并为一个可观察序列。
- 不包含，每当第一个可观察对象出现两个项目时，将一个可观察序列合并到另一个可观察序列中，而第二个可观察对象没有任何项目。
用这种方法问题几乎解决了。
IObservable<TP> P = // observer P
IObservable<TQ> Q = // observer Q
var PP = P.Without((prev, next) => 0, Q);
var PQ = P.Before(Q, (p,q) => f(p,q)); // apply the function
var ResultSecuence = PP.Merge(PQ);
这里有两个方法
public static class Observer
{
/// <summary>
/// Merges two observable sequences into one observable sequence by using the selector function
/// whenever the first observable produces an element rigth before the second one.
/// </summary>
/// <param name="first"> First observable source.</param>
/// <param name="second">Second observable source.</param>
/// <param name="resultSelector">Function to invoke whenever the first observable produces an element rigth before the second one.</param>
/// <returns>
/// An observable sequence containing the result of combining elements of both sources
/// using the specified result selector function.
/// </returns>
public static IObservable<TResult> Before<TLeft, TRight, TResult>(this IObservable<TLeft> first, IObservable<TRight> second, Func<TLeft, TRight, TResult> resultSelector)
{
var result = new Subject<TResult>();
bool firstCame = false;
TLeft lastLeft = default(TLeft);
first.Subscribe(item =>
{
firstCame = true;
lastLeft = item;
});
second.Subscribe(item =>
{
if (firstCame)
result.OnNext(resultSelector(lastLeft, item));
firstCame = false;
});
return result;
}
/// <summary>
/// Merges an observable sequence into one observable sequence by using the selector function
/// every time two items came from <paramref name="first"/> without any item of any observable
/// in <paramref name="second"/>
/// </summary>
/// <param name="first"> Observable source to merge.</param>
/// <param name="second"> Observable list to ignore.</param>
/// <param name="resultSelector">Function to invoke whenever the first observable produces two elements without any of the observables in the secuence produces any element</param>
/// <returns>
/// An observable sequence containing the result of combining elements
/// using the specified result selector function.
/// </returns>
public static IObservable<TResult> Without<TLeft, TResult>(this IObservable<TLeft> first, Func<TLeft, TLeft, TResult> resultSelector,params IObservable<object>[] second)
{
var result = new Subject<TResult>();
bool firstCame = false;
TLeft lastLeft = default(TLeft);
first.Subscribe(item =>
{
if (firstCame)
result.OnNext(resultSelector(lastLeft, item));
firstCame = true;
lastLeft = item;
});
foreach (var observable in second)
observable.Subscribe(item => firstCame = false);
return result;
}
}