[英]RX getting a IObservable<List<T>> to an IObservable<T>
I'm pretty new to RX and cannot figure this out. 我是RX的新手,无法解决这个问题。
I have an IObservable<List<T>>
where List<T>
is guaranteed to have one element. 我有一个
IObservable<List<T>>
,其中List<T>
保证有一个元素。
How do I convert this to an IObservable<T>
. 如何将其转换为
IObservable<T>
。
I thought it would have something to do with Single
but that is listed as obsolete, and also doesn't return an IObservable<T>
anyway (as pretty sure it would return the Single
List<T>
element. 我认为它与
Single
但是它被列为过时的,并且也不会返回IObservable<T>
(因为很确定它会返回Single
List<T>
元素。
Is there some SelectMany
magic I can do here? 我可以在这里做一些
SelectMany
魔术吗?
You can just do this: 你可以这样做:
IObservable<List<T>> source;
var converted = source.Select(x => x[0]);
Or if you prefer LINQ query comprehension syntax then equivalent is: 或者,如果您更喜欢LINQ查询理解语法,那么等效的是:
var converted = from x in source select x[0];
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.