Question

假设我有一个如下所示的数据框：

df1 <- data.frame("long name 1"=c(rnorm(n = 100, mean = 0, sd = 1),NA),
                  treatment=sample(x = c(0,1), size = 101, replace = T))
names(df1) <- c("long name 1", "treatment")

我想用5个参数创建一个函数。现在我有这个：

test.f <- function(data, vname, tname, tvalue=1, cvalue=0) {
  vname<-as.name(vname)
  tname <- as.name(tname)
  mean.Xt <- mean(data$vname[data$tname==tvalue], na.rm = T)
}

当我运行它时，我收到以下错误：

test.f(data = df1, vname = "long name 1", tname = "treatment")
Warning message:
  In mean.default(data$vname[data$tname == tvalue], na.rm = T) :
  argument is not numeric or logical: returning NA

有没有办法做我想做的事？

Answer 1

最好在函数中使用[[代替$来选择列。

test.f <- function(data, vname, tname, tvalue=1, cvalue=0) {
    mean(data[[vname]][data[[tname]]==tvalue], na.rm = T)
 }

test.f(data = df1, vname = "long name 1", tname = "treatment")
#[1] 0.1397585

与

相同

mean(df1$`long name 1`[df1$treatment==1], na.rm=TRUE)
#[1] 0.1397585

创建一个以名称作为参数的R函数？

1 个答案: