我正在Sub Dosomething()
...
For Each xSh In Worksheets
DeleteColumns ws:=xSh
Next
End Sub
Sub DeleteColumns(ByVal ws as Worksheet)
...
For Each V In Split(ColumnsToDelete, "|")
Dim rng as Range
Set rng = ws.Rows(1).Find(What:=Trim(V), LookAt:=xlWhole)
If Not rng Is Nothing Then
rng.EntireColumn.Delete
Else
Debug.Print V & " was not found on " & ws.Name
End If
Next
End Sub
中拟合模型。当我在函数调用中拟合模型时，它不起作用吗？
我愿意：
rstanarm其中mod <- stan_glm(form, data, weights=data[['weight']])
是data，而data.frame是公式。
和模型适合。
我愿意：
form我明白了
f <- function(data){
return(fit.model(form, data, weights=data$weight))
}
编辑：
我无法创建可复制的示例，但是通过在函数中调用Error in eval(extras, data, env) : object 'data' not found找到了解决方法。
assign