Please consider the following code:
请考虑以下代码:
test <- function(x,n){
selection<-names(x)[n]
graph <- ggplot(x, aes(factor(selection)))
graph + geom_bar()
}
test(mtcars,1)
It throws an error cause R can't find selection. I also played around with substitute, eval and get without success. I found this similar question and thought I understood Joris' answer but can't use the same trick for arguments of ggplot as well.
它抛出一个错误，因为R找不到选择。我也用替代品，eval和get而不是成功。我发现了这个类似的问题，我想我理解了Joris的答案，但也不能用同样的技巧来解释ggplot的论点。
1 个解决方案
#1
9
you can use aes_string for this purpose. So test should be like this:
您可以为此目的使用aes_string。所以测试应该是这样的:
test <- function(x,n){
graph <- ggplot(x, aes_string(x = names(x)[n]))
graph + geom_bar()
}
#1
9
you can use aes_string for this purpose. So test should be like this:
您可以为此目的使用aes_string。所以测试应该是这样的:
test <- function(x,n){
graph <- ggplot(x, aes_string(x = names(x)[n]))
graph + geom_bar()
}