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kotlin - 如何结合两个可流动并仅获得相同的排放值？

问题描述

        val flowable01 = Flowable.fromArray(listOf(
            Value(epoch = 1,string ="apple"),
            Value(epoch = 2,string ="apple"), 
            Value(epoch = 3,string ="apple"),
            Value(epoch = 4,string ="apple"), 
            Value(epoch = 5,string ="apple"),
            Value(epoch = 6,string ="apple"), 
            Value(epoch = 7,string ="apple"),
            Value(epoch = 8,string ="apple"), 
            Value(epoch = 9,string ="apple"),
            Value(epoch = 10,string ="apple"), 
            ))
        val flowable02 = Flowable.fromArray(listOf(
            Value(epoch = 2,string ="orange"), 
            Value(epoch = 3,string ="orange"),
            Value(epoch = 5,string ="orange"),
            Value(epoch = 8,string ="orange"), 
            Value(epoch = 10,string ="orange"), 
            ))
            
        flowable01.filter { it.epoch == flowable02.emission.epoch }
                  .map { Value(it.epoch, "juice") }

如何组合两个可流动的并仅从排放中过滤出相同的值并将其映射到另一个数据类？也许使用 combineLatest/zip

//Result:  
Value(epoch = 2,string ="juice")
Value(epoch = 3,string ="juice")
Value(epoch = 5,string ="juice")
Value(epoch = 8,string ="juice")
Value(epoch = 10,string ="juice")

标签： kotlinrx-javarx-java2

解决方案

您可以使用flatMap：

flowable01.flatMap { first ->
  flowable02.filter { second -> first.epoch == second.epoch }
  .map { second -> Value(second.epoch, "Juice") }
}

如果 flowable02 是一个外部数据源，您可能不想每次都重新查询它，以便将其预先缓存：

flowable02cache = flowable02.cache()

flowable01.flatMap { first ->
  flowable02cache.filter { second -> first.epoch == second.epoch }
  .map { second -> Value(second.epoch, "Juice") }
}

如果您想利用每个源中的排序，则在合并步骤中会更加复杂：

Single.zip(flowable01.toList(), flowable02.toList()) { first, second ->

    val result = ArrayList<Value>()

    var i = 0;
    var j = 0;
    while (i < first.length && j < second.length) {
        if (first[i].epoch == second[j].epoch) {
            result.add(Value(first[i].epoch, "Juice"))
            i++;
            j++;
        }
        if (first[i].epoch < second[j].epoch) {
            i++;
        }
        if (first[i].epoch > second[j].epoch) {
            j++;
        }
    }

    result;
}

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