反应编程 – RX中没有主题的反馈循环

2024年1月11日 235次阅读

我有以下运动方程式

move = target_position - position
position = position + move

其中target_position是一个流,位置初始化为零.我想有一个位置流.我尝试过类似的东西(在rx伪代码中)

moves = Subject()
position = moves.scan(sum,0)
target_position.combine_latest(position,diff).subscribe( moves.on_next)

它有效但我已经读过应该避免使用Subject.是否可以在没有主题的情况下计算位置流？

在python中,完整的实现看起来像这样

from pprint import pprint 
from rx.subjects import Subject

target_position = Subject()

moves = Subject()

position = moves.scan(lambda x,y: x+y,0.0)

target_position\
    .combine_latest(position,compute_next_move)\
    .filter(lambda x: abs(x)>0)\
    .subscribe( moves.on_next)

position.subscribe( lambda x: pprint("position is now %s"%x))

moves.on_next(0.0)
target_position.on_next(2.0)
target_position.on_next(3.0)
target_position.on_next(4.0)

最佳答案 你可以使用
expand operator

targetPosition.combineLatest(position, (target, current) => [target, current])
  .expand(([target, current]) => {
    // if you've reached your target, stop
    if(target === current) {
      return Observable.empty()
    }
    // otherwise, calculate the new position, emit it
    // and pump it back into `expand`
    let newPosition = calcPosition(target, current);
    return Observable.just(newPosition)
  })
  .subscribe(updateThings);
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