反应式编程 – 如何使用ReactiveCocoa对输入信号进行两级排序？

我一直在使用Reactive Cocoa,我遇到了一个有趣的问题.我可以设想任何数量的丑陋,有状态的解决方案,但我非常有信心有一种优雅,功能性的方式,无论出于何种原因,我的脑子里都没有实现.也许你可以帮忙！

这里的输入信号是两部分字符串,例如“< letter>,< number>”.所需的排序规则是对于给定的字母,输入值应该以< number>的顺序出现在输出处. (即A,2应该永远不会出现在A,1之前)和所有字母之间的字母< letter>输出不应违反alpha顺序. (即,在出现至少一个以A开头的字符串之前,不应出现以B开头的字符串.)除了这些规则所规定的情况之外,期望事物将按照它们提交给输入的顺序到达输出.

请考虑以下代码：

RACSubject* input = [RACSubject subject];

RACSignal* output = [input <SOME CHAIN OF SIGNAL FUNCTIONS>];

[output subscribeNext:^(id x) { NSLog(@"(%@)",x); }];

[input sendNext: @"A,2"]; // Expect no output
[input sendNext: @"B,4"]; // Expect no output
[input sendNext: @"B,2"]; // Expect no output
[input sendNext: @"B,1"]; // Expect no output
[input sendNext: @"A,1"]; // Expect output: (A,1) (A,2) (B,1) (B,2) 
// Note: (A,1) (B,1) (B,2) (A,2) would *not* be right because A,2 appeared on the input before B,1
[input sendNext: @"C,1"]; // Expect output: (C,1)
[input sendNext: @"B,3"]; // Expect output: (B,3) (B,4)
[input sendNext: @"C,3"]; // Expect no output
[input sendNext: @"C,2"]; // Expect output: (C,2) (C,3)

还应该“急切地”产生输出.如果我必须等到输入信号完成才能看到输出(当然,除非排序规则规定是这种情况,即如果A,1出现在最后),那么它是没有用的

有任何想法吗？

以命令式方式编写它,您可能会使用一些累加器变量,然后循环输入值并根据需要操作累加器.

函数式编程中最接近的并行是扫描(在ReactiveCocoa中表示为-scanWithStart：reduce :).扫描允许您通过流“线程化”状态,并在新输入值到达时使用它.

结果看起来非常类似于命令式累积,除了任何突变都不会逃离扫描块：

RACSignal *output = [[[[input
    map:^(NSString *combo) {
        NSArray *components = [combo componentsSeparatedByString:@","];
        NSInteger number = [components[1] integerValue];

        return RACTuplePack(components[0], @(number));
    }]
    // We need four state parameters:
    // 1. The letter we're waiting for.
    // 2. The number we're waiting for.
    // 3. Values received that cannot be forwarded until a certain
    //    letter/number.
    // 4. The values to forward at each step.
    scanWithStart:RACTuplePack(@"A", @1, @[], @[]) reduce:^(RACTuple *state, RACTuple *letterAndNumber) {
        NSString *waitingLetter = state[0];
        NSNumber *waitingNumber = state[1];
        NSArray *queuedValues = state[2];

        // Enqueue this value until we're ready to send it (which may or may not
        // occur on this step of the scan).
        queuedValues = [queuedValues arrayByAddingObject:letterAndNumber];

        if ([letterAndNumber.first isEqual:waitingLetter] && [letterAndNumber.second isEqual:waitingNumber]) {
            // Determine the next letter and number.
            waitingLetter = …;
            waitingNumber = @(waitingNumber.integerValue + 1);

            // Sort queuedValues lexically and numerically.
            NSArray *forwardValues = …;

            // We should no longer have any values queued, since we want to
            // forward them all.
            return RACTuplePack(waitingLetter, waitingNumber, @[], forwardValues);
        } else {
            // No values should escape the scan yet. Just pass on our queued
            // values.
            return RACTuplePack(waitingLetter, waitingNumber, queuedValues, @[]);
        }
    }]
    map:^(RACTuple *state) {
        // Convert the array of values into a signal.
        NSArray *forwardValues = state.last;
        return forwardValues.rac_sequence.signal;
    }]
    // Forward values from each inner signal in the correct, sorted order.
    concat];

为简洁起见,我省略了一些排序逻辑,但是很容易填写算法的确切细节.