我有一个发布者发出字符串，而许多订阅者可能使用相同的Map函数来创建具有不同过滤器的模型。
出版商：

val publisher: Flux<String> = ...

订户#1

val sub1 = publisher.map{veryExpensiveConverter.convert(it)}
                    .filter(it.metric<10)

订户#2

val sub2 = publisher.map{veryExpensiveConverter.convert(it)}
                    .filter(it.metric>5)

订户#3

val sub3 = sub2.map{cheapConverter.convert(it)}
                    .filter(it.metric>8)

订户#4

val sub4 = sub3.map{yetAnotherConverter.convert(it)}
                    .filter(it.metric>80)

最后我订阅了所有的流量

Flux.merge(sub1, sub2, sub3, ..., subn)
     .map{//some logic for following data of subscribers}
     .subscribe()

问题是：veryexpensiveconverter对于每个订阅服务器的同一个已发布记录执行多次。执行流看起来

Input1 -> veryExpensiveConverter -> filter1 -> output1
       -> veryExpensiveConverter -> filter2 -> output2
       -> veryExpensiveConverter -> cheapConverter -> filter3 -> output3

我也想要

Input1 -> veryExpensiveConverter -> filter1 -> output1  
                                 -> filter2 -> output2
                                 -> cheapConverter -> filter3 -> output3

什么样的模式最适合避免对每个订户执行相同的Map？