0

我想在 Shiny 中有一个动态回归 qqplot，它有一个 x 值（预测变量）的下拉菜单。您可以在下面找到我的代码，其中我定义了一个 LM 函数，当我随后运行模型时，该函数将响应变量形式放在第一位，将预测变量从第二位放在第二位。问题是当我运行模型时我必须准确地编写变量名（见下面的代码），但我想要一个 selectInput 函数，它将变量名传递给模型。链接到我的 ShinyApp 以获得更清晰的图片（最后一个仪表板）：https ://schnappi.shinyapps.io/coronavirus/

代码如下：

回归模型函数：

  ggplot(fit$model, aes_string(
    x = names(fit$model)[2],
    y = names(fit$model)[1],
    label = names(fit$model)[3]
  )) +
    geom_point() + labs(x = "Population density (km2)", y = "Cases per 1M") +
    stat_smooth(method = "lm") +
    labs(title = paste(
      "R2 =",
      signif(summary(fit)$r.squared, 2),
      " Slope =",
      signif(fit$coef[[2]], 2),
      " P =",
      signif(summary(fit)$coef[2, 4], 2)
    ))
}```

UI code:

tabItem(
        tabName = "Dens",
        titlePanel(h1(
          "Population density and Covid-19 cases", align = "center"
        )),
        absolutePanel(
          top = 130,
          left = 250,
          draggable = FALSE,
          numericInput(
            "num",
            "x-axis cut off:",
            max(join$Pop_density),
            min = 1,
            max =  max(join$Pop_density)
          )
        ),
        absolutePanel(
          top = 130,
          left = 550,
          draggable = FALSE,
          numericInput(
            "num2",
            "y-axis cut off:",
            max(join$TotCases_1M),
            min = 1,
            max =  max(join$TotCases_1M)
          )
        ),
        absolutePanel(
          top = 130,
          left = 850,
          draggable = FALSE,
          selectInput("con", "Select continent:", choices =
                        join$Continent)
        ),
        absolutePanel(
          top = 130,
          left = 1050,
          draggable = FALSE,
          selectInput("paraM", "Select parameter:", choices = colnames(join))           ## This is the part of the code that defines the predictor variable
        ),
        absolutePanel(
          top = 190,
          left = 220,
          width = 1700,
          height = 500,
          draggable = FALSE,
          mainPanel(plotlyOutput("plotDens"))
        )
      )
This is part which RUN the regression model:

  output$plotDens <- renderPlotly({
      ggplotRegression(lm(
        TotCases_1M ~ input$paraM,                                        ## Here the LM model should recognize "input$paraM" as the column name but this does not work.
        data = subset(join, Continent == input$con | input$con == "") %>%
          filter(input$paraM > input$num &
                   TotCases_1M < input$num2)
      ))
  })

THANK YOU.
4

1 回答 1

0

您没有提供的完整代码ggplotRegression，但我相信您想要：

  ggplotRegression(lm(
    as.formula(paste0("TotCases_1M ~ ", input$paraM)),
    ......
于 2020-07-23T11:24:25.803 回答