Rxjava - 用批处理迭代列表

Question

我有一些字符串的列表，我需要在其中批量迭代列表。
示例：

val list = ["a","b","c","d","e","f"]
Observable.fromIteratable(list)
          .map{
          //here i need to get 4 items at a time
          //like in first iteration i need to get ["a","b","c","d"]
          //in second iteration i need to get ["e","f"]
        }

是否有任何选项可以执行此操作？

Answer 1

用户buffer

periodically gather items emitted by an Observable into bundles and emit these bundles rather than emitting the items one at a time

 val list = arrayOf("1", "2", "3", "4", "5")
 Observable.fromIterable(list.asIterable())
     .buffer(4)
     .map { stringList: MutableList<String> ->
       println("insideMap -> $stringList")
       return@map "wow $stringList"
     }
     .subscribe { value: String -> println("OnResult $value")}

//Output
insideMap -> [1, 2, 3, 4]
OnResult wow [1, 2, 3, 4]
insideMap -> [5]
OnResult wow [5]

The Buffer operator transforms an Observable that emits items into an Observable that emits buffered collections of those items. There are a number of variants in the various language-specific implementations of Buffer that differ in how they choose which items go in which buffers.

注意，如果源 Observable 发出 onError 通知，Buffer 将立即传递此通知，而不会首先发出它正在组装的缓冲区，即使该缓冲区包含源 Observable 在发出错误通知之前发出的项目。

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