使用 Rx 运行直方图流 [英] Running histogram stream with Rx

问题描述

我有以下单字母流

A 
B 
C 
A
D 
B 
A 
C 
D

从这个流中，我想要一个每个字母的运行计数流

And from this stream, I would like a stream of running count per letter

(A,1)
(A,1), (B,1)
(A,1), (B,1), (C,1)
(A,2), (B,1), (C,1)
(A,2), (B,1), (C,1), (D,1)
(A,2), (B,2), (C,1), (D,1)
(A,3), (B,2), (C,1), (D,1)     
(A,3), (B,2), (C,2), (D,1) 
(A,3), (B,2), (C,2), (D,2)     

，即在每个新字母处，更新和发出总数.

, i.e at each new letter, the totals are updated and emitted.

我猜这个问题与语言无关，所以请毫不犹豫地用您选择的语言提出解决方案.

I guess this problem is fairly language agnostic, so don't hesitate to propose a solution in your language of choice.

推荐答案

这是如何使用 RxJava 完成的:

This is how it can be done using RxJava:

final Observable<String> observable = Observable.just("A", "B", "C", "A", "D", "B", "A", "C", "D");
final Observable<LinkedHashMap<String, Integer>> histogram = observable.scan(new LinkedHashMap<>(), (state, value) -> {
  if (state.containsKey(value)) {
    state.put(value, state.get(value) + 1);
  } else {
    state.put(value, 1);
  }

  return state;
});

histogram.subscribe(state -> {
  System.out.println(state);
});

输出:

{}
{A=1}
{A=1, B=1}
{A=1, B=1, C=1}
{A=2, B=1, C=1}
{A=2, B=1, C=1, D=1}
{A=2, B=2, C=1, D=1}
{A=3, B=2, C=1, D=1}
{A=3, B=2, C=2, D=1}
{A=3, B=2, C=2, D=2}

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