我想知道我们是否可以在R Shiny中做到这一点
example
# using styleColorBar
datatable(df) %>% formatStyle(names(df),
background = styleColorBar(range(df), 'lightblue'),
backgroundSize = '98% 88%',
backgroundRepeat = 'no-repeat',
backgroundPosition = 'center')
但是使用正负两种颜色,如果可以的话,将负值作为条形长度的绝对值
泰语,祝你有美好的一天!
最佳答案
您可以使用rowCallback
手动进行操作
如你看到的
sss=datatable(df) %>% formatStyle(names(df),
background = styleColorBar(range(df), 'lightblue'),
backgroundSize = '98% 88%',
backgroundRepeat = 'no-repeat',
backgroundPosition = 'center')
sss$x$options$rowCallback
为每一列生成:
函数(行,数据){var value = data [1];如果(值!==空)
$(this.api()。cell(row,
1).node())。css({'background':isNaN(parseFloat(value))|| value -1.311? '':'线性渐变(90deg,透明'+(2.199-值)/3.51 * 100 +'%,浅蓝色'+(2.199-值)/3.51 * 100 +
'%)','背景大小':'98%
88%','background-repeat':'no-repeat','background-position':'center'});
因此:
1)您需要计算范围变量:
rr=range(df)[2]-range(df)[1]
r1=range(df)[2]
r0=range(df)[1]
2)比在
rowCallback
中使用它 datatable(
df,
options = list(
rowCallback=JS(paste0("function(row, data) {",
paste(lapply(1:ncol(df),function(i){
paste0("var value=data[",i,"];
if (value!==null){
if(value<0){
$(this.api().cell(row,",i,").node()).css({'background':isNaN(parseFloat(value)) || value <=",r0," ? '' : 'linear-gradient(90deg, transparent ' + (",r1," - value)/",rr," * 100 + '%, red ' + (",r1," - value)/",rr," * 100 + '%)','background-size':'98% 88%','background-repeat':'no-repeat','background-position':'center'});
}else{
$(this.api().cell(row,",i,").node()).css({'background':isNaN(parseFloat(value)) || value <=",r0," ? '' : 'linear-gradient(90deg, transparent ' + (",r1," - value)/",rr," * 100 + '%, lightblue ' + (",r1," - value)/",rr," * 100 + '%)','background-size':'98% 88%','background-repeat':'no-repeat','background-position':'center'});
}
} ")
}),collapse="\n"),
"}"))
)
)
在这里,我对颜色进行了硬编码(红色和浅蓝色)
结果
如果您希望正负的长度不同,则需要不同的范围变量,或者将
abs
\ Math.abs
-用作绝对值(对于JS中的var和范围)聚苯乙烯
JS中列的计数从0开始,因此如果没有行名,则需要
0:(ncol(df)-1)
中的lapply
以获得正确的结果更新资料
范围-1:1和绝对值的色标
df = as.data.frame(cbind(matrix(round(runif(50, -1, 1), 3), 10), sample(0:1, 10, TRUE)))
rr=range(abs(df))[2]-range(abs(df))[1]
r1=range(abs(df))[2]
r0=range(abs(df))[1]
datatable(
df,
options = list(
rowCallback=JS(paste0("function(row, data) {",
paste(lapply(1:ncol(df),function(i){
paste0("var value=Math.abs(data[",i,"]);
var value2=data[",i,"];
if (value!==null){
if(value2<0){
$(this.api().cell(row,",i,").node()).css({'background':isNaN(parseFloat(value)) || value <=",r0," ? '' : 'linear-gradient(90deg, transparent ' + (",r1," - value)/",rr," * 100 + '%, red ' + (",r1," - value)/",rr," * 100 + '%)','background-size':'98% 88%','background-repeat':'no-repeat','background-position':'center'});
}else{
$(this.api().cell(row,",i,").node()).css({'background':isNaN(parseFloat(value)) || value <=",r0," ? '' : 'linear-gradient(90deg, transparent ' + (",r1," - value)/",rr," * 100 + '%, lightblue ' + (",r1," - value)/",rr," * 100 + '%)','background-size':'98% 88%','background-repeat':'no-repeat','background-position':'center'});
}
} ")
}),collapse="\n"),
"}"))
)
)