使用組合組合先前的值

Question

如何使用Combine框架重寫ReactiveSwift/ReactiveCocoa代碼？ 我附上了屏幕截圖，文檔中的combinePrevious是什么意思。

let producer = SignalProducer<Int, Never>([1, 2, 3]).combinePrevious(0)
producer.startWithValues { value in
    print(value) // print: (0, 1), (1, 2), (2, 3)
}

Answer 1

我對ReactiveSwift/ReactiveCocoa並不完全熟悉，但根據您的描述，您可以使用.scan ，這似乎是一個比combinePrevious更通用的函數。

它需要一個初始結果——你可以把它變成一個元組——以及一個帶有存儲值和當前值的閉包，並返回一個新的存儲值——在你的情況下，一個帶有(previous, current)的元組：

let producer = [1,2,3].publisher
                      .scan((0,0)) { ($0.1, $1) }

producer.sink { 
   print($0) 
}

Answer 2

這些是我想出的自定義運算符（稱為withPrevious ）。 有兩種重載，一種是初始值nil ，另一種是提供初始值，這樣您就不必處理可選項。

extension Publisher {

    /// Includes the current element as well as the previous element from the upstream publisher in a tuple where the previous element is optional.
    /// The first time the upstream publisher emits an element, the previous element will be `nil`.
    ///
    ///     let range = (1...5)
    ///     cancellable = range.publisher
    ///         .withPrevious()
    ///         .sink { print ("(\($0.previous), \($0.current))", terminator: " ") }
    ///      // Prints: "(nil, 1) (Optional(1), 2) (Optional(2), 3) (Optional(3), 4) (Optional(4), 5) ".
    ///
    /// - Returns: A publisher of a tuple of the previous and current elements from the upstream publisher.
    func withPrevious() -> AnyPublisher<(previous: Output?, current: Output), Failure> {
        scan(Optional<(Output?, Output)>.none) { ($0?.1, $1) }
            .compactMap { $0 }
            .eraseToAnyPublisher()
    }

    /// Includes the current element as well as the previous element from the upstream publisher in a tuple where the previous element is not optional.
    /// The first time the upstream publisher emits an element, the previous element will be the `initialPreviousValue`.
    ///
    ///     let range = (1...5)
    ///     cancellable = range.publisher
    ///         .withPrevious(0)
    ///         .sink { print ("(\($0.previous), \($0.current))", terminator: " ") }
    ///      // Prints: "(0, 1) (1, 2) (2, 3) (3, 4) (4, 5) ".
    ///
    /// - Parameter initialPreviousValue: The initial value to use as the "previous" value when the upstream publisher emits for the first time.
    /// - Returns: A publisher of a tuple of the previous and current elements from the upstream publisher.
    func withPrevious(_ initialPreviousValue: Output) -> AnyPublisher<(previous: Output, current: Output), Failure> {
        scan((initialPreviousValue, initialPreviousValue)) { ($0.1, $1) }.eraseToAnyPublisher()
    }
}

Answer 3

Cocoacasts 有一個很好的例子：

https://cocoacasts.com/combine-essentials-combining-publishers-with-combine-zip-operator

zip 運算符可用於創建發布者，該發布者發出前一個元素和發布者發出的當前元素。 我們將同一個發布者傳遞給 Publishers.Zip 結構的初始化程序兩次，但將 dropFirst 運算符應用於第二個發布者。 這僅僅意味着第二個發布者不會發出原始發布者的第一個元素。

import Combine

let numbers = [1, 2, 3, 4, 5].publisher

Publishers.Zip(numbers, numbers.dropFirst(1))

用法：

import Combine

let numbers = [1, 2, 3, 4, 5].publisher

Publishers.Zip(numbers, numbers.dropFirst(1))
    .sink(receiveValue: { values in
        print(values)
    })
    
// (1, 2)
// (2, 3)
// (3, 4)
// (4, 5)