使用R中的DT基于另一列设置一列的样式

下面的代码使用变量

test1\u score

和

test2\u score

创建一个数据帧。如果

test2\u分数大于
test1\u分数
，我需要更改
test2\u分数的背景

下面是创建数据帧的代码：

df <- data.frame(
  id = 1:10,
  name = c("Bob", "Ashley", "James", "David", "Jenny", 
           "Hans", "Leo", "John", "Emily", "Lee"), 
  age = c(28, 27, 30, 28, 29, 29, 27, 27, 31, 30),
  grade = c("C", "A", "A", "C", "B", "B", "B", "A", "C", "C"),
  test1_score = c(8.9, 9.5, 9.6, 8.9, 9.1, 9.3, 9.3, 9.9, 8.5, 8.6),
  test2_score = c(9.1, 9.1, 9.2, 9.1, 8.9, 8.5, 9.2, 9.3, 9.1, 8.8),
  final_score = c(9, 9.3, 9.4, 9, 9, 8.9, 9.25, 9.6, 8.8, 8.7),
  registered = c(TRUE, FALSE, TRUE, FALSE, TRUE, TRUE, TRUE, FALSE, FALSE, FALSE),
  stringsAsFactors = FALSE)

你可以做：

library(DT)
df <- data.frame(
  id = 1:10,
  name = c("Bob", "Ashley", "James", "David", "Jenny", 
           "Hans", "Leo", "John", "Emily", "Lee"), 
  age = c(28, 27, 30, 28, 29, 29, 27, 27, 31, 30),
  grade = c("C", "A", "A", "C", "B", "B", "B", "A", "C", "C"),
  test1_score = c(8.9, 9.5, 9.6, 8.9, 9.1, 9.3, 9.3, 9.9, 8.5, 8.6),
  test2_score = c(9.1, 9.1, 9.2, 9.1, 8.9, 8.5, 9.2, 9.3, 9.1, 8.8),
  final_score = c(9, 9.3, 9.4, 9, 9, 8.9, 9.25, 9.6, 8.8, 8.7),
  registered = c(TRUE, FALSE, TRUE, FALSE, TRUE, TRUE, TRUE, FALSE, FALSE, FALSE),
  stringsAsFactors = FALSE)
colors <- with(df, ifelse(test1_score > test2_score, "pink", "blue"))
rgbcolors <- apply(grDevices::col2rgb(colors), 2, 
                   function(rgb) sprintf("rgb(%s)", paste(rgb, collapse=",")))
columns <- c(5,6) # columns test1_score and test2_score
jscode <- 
  paste("function(row, data, index) {",  
        sprintf("var colors=%s;\n%s", 
                sprintf("[%s]", 
                        paste(sprintf("'%s'", rgbcolors), collapse=", ")), 
                paste(sprintf("$(this.api().cell(index, %s).node()).css('background-color', colors[index]);", 
                              columns), collapse="\n")), 
        "}", sep="\n")

datatable(df, escape=FALSE, 
          options = list(rowCallback=JS(jscode))) 

库（DT）
df这里有一个您可以使用的解决方法

#Define a new column for testing
df$Result <- factor(ifelse(df$test2_score > df$test1_score,'More','Not More'),
                    levels=c('More','Not More'))

datatable(df,options = list(columnDefs = list(list(visible=FALSE, targets=c(9))))) %>% #Hide Result using visible=F
   formatStyle('test2_score', valueColumns = 'Result', target = c('cell'),
                backgroundColor = styleEqual(unique(df$Result),  c('gray', 'yellow')))

#定义用于测试的新列
df$Result df$test1_分数，'More'，'Not More'），
级别=c（'More'，'notmore'））
datatable（df，options=list（columnDefs=list）（list（visible=FALSE，targets=c（9 '））%%>%#使用visible=F隐藏结果
formatStyle（'test2_score'，valueColumns='Result'，target=c（'cell'），
backgroundColor=styleEqual（唯一（df$Result），c（'gray'，'yellow'））
这里有一个比我给出的另一个更干净的解决方案（这是很难理解的）

datatable（df，
选项=列表（initComplete=JS(
“功能（设置）{
var table=settings.oInstance.api（）；
对于（var i=0；i cell2.data（））{
cell1.node（）.style.backgroundColor='pink'；
}否则{
cell1.node（）.style.backgroundColor='blue'；
}
}
}")))

#Define a new column for testing
df$Result <- factor(ifelse(df$test2_score > df$test1_score,'More','Not More'),
                    levels=c('More','Not More'))

datatable(df,options = list(columnDefs = list(list(visible=FALSE, targets=c(9))))) %>% #Hide Result using visible=F
   formatStyle('test2_score', valueColumns = 'Result', target = c('cell'),
                backgroundColor = styleEqual(unique(df$Result),  c('gray', 'yellow')))

datatable(df, 
          options=list(initComplete = JS(
          "function(settings) {
            var table = settings.oInstance.api();
            for(var i=0; i<10; i++){
              var cell1 = table.cell(i,5);
              var cell2 = table.cell(i,6);
              if(cell1.data() > cell2.data()){
                cell1.node().style.backgroundColor='pink';
              }else{
                cell1.node().style.backgroundColor='blue';
              }
            }
          }")))