假设我有一个Observable<Player>，我想将它映射到另一个Observable<Integer>，其中Integer等于player.height，但有一个条件：我想映射所有玩家，但第一个和最后一个（我们应该再检查一个他们）。所以在迭代编程中它会像这样：
heights = []
num_of_players = len(players)
for idx in len(num_of_players):
if (idx == 0 or idx == num_of_players - 1):
if (isGoodEnough(players[idx]):
heights.append(player.height)
else:
heights.append(player.height)
return height
我如何以 Rx 的方式重写它（你应该假设我得到的是 Observable 而不是 List）？
鉴于：
Observable<Player> players;
Single<Integer> playerHeight(int playerId);
您必须使用publish(Function)s将序列拆分为 first、middle 和 last，然后将它们组合在一起：
players
.publish(sharedPlayers -> {
return Observable.merge(
// work only on the very first player
sharedPlayers.take(1)
.filter(firstPlayer -> isGoodEnough(firstPlayer))
.flatMapSingle(firstPlayer -> playerHeight(firstPlayer.playerId)),
// work with not the first and not the last
sharedPlayers.skip(1).skipLast(1)
.flatMapSingle(midPlayers -> playerHeight(midPlayers .playerId)),
// work with the last which shouldn't be the first again
sharedPlayers.skip(1).takeLast(1)
.filter(lastPlayer -> isGoodEnough(lastPlayer))
.flatMapSingle(lastPlayer-> playerHeight(lastPlayer.playerId))
);
})
.subscribe(/* ... */);
请根据需要调整此解决方案。
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