Question

我正在开发一款Shiny应用程序，用于猜测用户输入的文本片段中的下一个单词。为此，我将加载三个数据帧，其中包含对下一个单词的可能猜测，并使用grepl查找用户输入文本末尾的三元或二元匹配。我在“常规”R中构建了这个没有错误或问题，但是当它转换为Shiny时，我收到了这个错误：

错误：评估嵌套太深：无限递归/选项（表达式=）？

正如其他Stack Overflow帖子所建议的那样，我摆弄了选项（expressions =）值，但仍然遇到错误。没有太多其他信息可以指导我，并注意到它没有问题，任何人都可以看到我出错的地方？提前谢谢！

对于后台，我正在运行：Win 7,64位OS | R v3.1.1 | RStudio v0.98.944

ui.R

library(shiny)
shinyUI(fluidPage(
   titlePanel("Capstone: Word Guesser"),
      fluidRow(
         column(12, 
            mainPanel(
               h4('Sentence Fragment'),
               p('Please enter a snippet of text - with this as a start, the app will provide you 5 guesses in ranked order to complete your phrase'),
               textInput('sentence', "Snippet:", value = "Enter some text and have a great"),
               h4('Guesses'),
               p('Below are our top five guesses for the word that completes this snippet'),
               verbatimTextOutput("final.guesses")
            )
         )
      )
))

server.R

options(expressions = 10000)
quad.grams <- readRDS("quadgrams.rds")
tri.grams <- readRDS("trigrams.rds")
kneser.ney <- readRDS("kneserney.rds")
library(shiny)
library(tm)
library(plyr)
library(stringi)
library(RWeka)
library(stringr)
function(input, output, clientData, session) {
   observe({
      sentence <- reactive({as.character(input$sentence)})
      sentence <- reactive({str_replace_all(sentence(), "[[:punct:]]", "")})
      sentence <- reactive({tolower(sentence())})
      sentence <- reactive({scan_tokenizer(sentence())})
      n <- reactive({length(sentence())})
      tri.frag <- reactive({paste(sentence()[n() - 2], sentence()[n() - 1], sentence()[n()])})
      bi.frag <- reactive({paste(sentence()[n() - 1], sentence()[n()])})
      quad.guesses <- reactive({quad.grams[grepl(tri.frag(), quad.grams$n1.Gram) == TRUE, 2]})
      tri.guesses <- reactive({tri.grams[grepl(bi.frag(), tri.grams$n1.Gram) == TRUE, 2]})
      guesses <- reactive({c(quad.guesses(), tri.guesses(), kneser.ney$nGram)})
      output$final.guesses <- renderPrint({guesses()[1:5]})
   })
}

Answer 1

我认为你的问题就在这里（无法进行测试）：

observe({
  sentence <- reactive({as.character(input$sentence)})
  sentence <- reactive({str_replace_all(sentence(), "[[:punct:]]", "")})
  sentence <- reactive({tolower(sentence())})
  sentence <- reactive({scan_tokenizer(sentence())})

这四行都是sentence被动的，每一行依赖于它自己，input$sentence和其他每一行导致闪亮的东西都被混淆了。更好的解决方案是：

observe({
  sentence1 <- reactive({as.character(input$sentence)})
  sentence2 <- reactive({str_replace_all(sentence1(), "[[:punct:]]", "")})
  sentence3 <- reactive({tolower(sentence2())})
  sentence4 <- reactive({scan_tokenizer(sentence3())})

您可以删除sentence1行 - 我认为input$sentence已经是字符。如果你喜欢，可以将其余的行组合起来：

  sentence <- reactive({scan_tokenizer( tolower(
                str_replace_all(input$sentence, "[[:punct:]]", "") ))
             })

Answer 2

冒着引起另一个无限循环的风险，我很高兴地说我可以发布自己的答案。问题似乎是重复使用“句子”变量。我想当使用被动时，你不能反复重复使用变量，即使你可能只是在改变它们。无论如何，这段代码现在有效：

server.R

options(shiny.maxRequestSize = 50*1024^2)
options(expressions = 10000)
quad.grams <- readRDS("quadgrams.rds")
tri.grams <- readRDS("trigrams.rds")
kneser.ney <- readRDS("kneserney.rds")
library(shiny)
library(tm)
library(plyr)
library(stringi)
library(RWeka)
library(stringr)
function(input, output, clientData, session) {
      snippet <- reactive({as.character(input$sentence)})
      clean.snippet <- reactive({str_replace_all(snippet(), "[[:punct:]]", "")})
      lower.snippet <- reactive({tolower(clean.snippet())})
      tokens <- reactive({scan_tokenizer(lower.snippet())})
      n <- reactive({length(tokens())})
      tri.frag <- reactive({paste(tokens()[n() - 2], tokens()[n() - 1], tokens()[n()])})
      bi.frag <- reactive({paste(tokens()[n() - 1], tokens()[n()])})
      quad.guesses <- reactive({quad.grams[grepl(tri.frag(), quad.grams$n1.Gram) == TRUE, 2]})
      tri.guesses <- reactive({tri.grams[grepl(bi.frag(), tri.grams$n1.Gram) == TRUE, 2]})
      guesses <- reactive({c(quad.guesses(), tri.guesses(), kneser.ney$nGram)})
      output$final.guesses <- renderPrint({guesses()[1:5]})
}

R Shiny：无限递归

2 个答案: