我的操作相对昂贵,因此我愿意执行一次该操作并从中创建2个Observables
。
这是它的外观:
let outputObservable1: Observable<Bool>
let outputObservable2: Observable<Bool>
(outputObservable1, outputObservable2) = inputObservable1.zip(inputObservable2).map { booleanCondition1, booleanCondition2 in
// different condition combinations create different outputObservables
}
我猜这里map
不是正确的运算符,因为它只会产生一个可观察的值。如何混合和匹配条件并立即返回2 Observables
?
答案 0 :(得分:1)
根据我的理解,您只需要使用map
let inputs = Observable.zip(inputObservable1, inputObservable2)
.share() // you only need this if one of your inputs is making a network request.
let outputObservable1 = inputs
.map { first, second in
return true // or false depending on the values of first & second.
}
let outputObservable2 = inputs
.map { first, second in
return true // or false depending on the values of first & second.
}