reactive-programming - RX 中没有主题的反馈循环

Question

我有以下运动方程

move = target_position - position
position = position + move

其中 target_position 是一个流，位置初始化为零。我想有一个职位流。我尝试过这样的事情(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)

Answer 1

您可以使用 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);