R当一个修改另一个时，如何确定最近修改的反应式表达式？

基本上，我有一个多选择输入的层次结构，称它们为a、b和c，大小（a）>大小（b）>大小（c）。b的每个单位都属于a的一个单位，c的每个单位都属于b的一个单位。当你从a中删除某些东西时，它会从b中删除相应的东西，而b又会从c中删除相应的东西

我需要做的是确定用户上次修改的多选输入。我在这里遵循这个结果，但它不起作用，因为如果我修改a，它会修改b，然后修改c，结果会是程序认为用户修改了c

我的问题是，有没有办法区分用户输入和常规更新调用

第二个问题可能是，有没有办法提前确定被动表达式将解决的顺序？例如，我注意到，如果我从a或b中删除很多值，它将解决一次删除一个值的问题，然后继续从较低的层次结构中删除这些值。我之前预计它会交替出现（从b中删除，从c中删除对应项，从b中删除，从c中删除对应项等等），但事实并非如此

下面是一个例子，您可以运行代码，您将看到当您编辑一个或两个选择时，它将快速闪烁“一”或“二”，但它将始终重置为“三”。例外情况是，当您删除列表中的最后一项时，由于某种原因，该项不会更新其下方的级别

library("shiny")
one = vector(mode="list", length = 4)
names(one) = c("a","b","c","d")
two = vector(mode="list", length = 8)
names(two) = c("e","f","g","h","i","j","k","l")
three = vector(mode="list", length = 16)
three = c("m","n","o","p","q","r","s","t","u","v","w","x","y","z","aa","ab")
one[["a"]] = c("e","f")
one[["b"]] = c("g","h")
one[["c"]] = c("i","j")
one[["d"]] = c("k","l")
two[["e"]] = c("m","n")
two[["f"]] = c("o","p")
two[["g"]] = c("q","r")
two[["h"]] = c("s","t")
two[["i"]] = c("u","v")
two[["j"]] = c("w","x")
two[["k"]] = c("y","z")
two[["l"]] = c("aa", "ab")
ui = fluidPage(
    titlePanel("Example"),
    sidebarLayout(
        sidebarPanel(
            selectInput("selectOne", label="Select One:", choices=names(one), selected = names(one), multiple=TRUE),
            selectInput("selectTwo", label="Select Two:", choices=names(two), selected = names(two), multiple=TRUE),
            selectInput("selectThree", label="Select Three:", choices=three, multiple=TRUE),
            actionButton("reset",label = "Reset")
        ),
        mainPanel(
            textOutput("listOfThree"),
            textOutput("lastUpdated"),
            textOutput("updated")
        )
    )
)

server = shinyServer(function(input,output,session){

    output$listOfThree = renderText({
        paste("list of three:", paste(input$selectThree, collapse = ", "), sep = " ")
    })
    output$lastUpdated = renderText({
        paste("last update:",rv$lastUpdate, sep = " ")
    })
    rv = reactiveValues(lastUpdate = "four")
    observe({
        oneList = input$selectOne
        rv$lastUpdate = "one"
        twoList = c()
        for (thing in oneList){
            twoList = c(twoList, one[[thing]])
        }
        twoList = sort(twoList)
        updateSelectInput(session, "selectTwo", selected = twoList)

    })
    observe({
        twoList = input$selectTwo
        rv$lastUpdate = "two"
        threeList = c()
        for (thing in twoList){
            threeList = c(threeList, two[[thing]])
        }
        threeList = sort(threeList)
        updateSelectInput(session, "selectThree", selected = threeList)

    })
    observe({
        input$selectThree
        rv$lastUpdate = "three"
    })
    observeEvent(
        input$reset,
        {
            updateSelectInput(session, "selectOne", selected = names(one))
            updateSelectInput(session, "selectTwo", selected = names(two))
            updateSelectInput(session, "selectThree", selected = three)
            rv$updateTwo = FALSE
            rv$updateThree = FALSE
        }
    )
})

shinyApp(ui, server)

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样本代码？没有示例代码，几乎不可能理解您的意思。在我看来，您确实在危险地试图以有状态的方式使用一个原则上设计为无状态（由于反应性）的工具。你能更深入地谈谈有状态和无状态以及为什么它是危险的吗？