将列表转换为字符向量后，make.names返回意外结果

Question

我正在尝试从字符串列表中创建一个字符向量。我的最终目标是将其传递给make.names()以创建数据框的列名。

temp<-data.frame(col1=c("use this row as name"),col2=c("use this row as name 2"))
make.names(temp[1,]) #fails
make.names(c("hi","xxx!@#$")) #works as expected
typeof(temp[1,])
typeof(c("hi","xxx!@#$"))

typeof(as.character(temp[1,]))
make.names(as.character(temp[1,]))

输出：

> temp<-data.frame(col1=c("use this row as name"),col2=c("use this row as name 2"))
> make.names(temp[1,])
[1] "X1" "X1"
> make.names(c("hi","xxx!@#$"))
[1] "hi"      "xxx...."
> typeof(temp[1,])
[1] "list"
> typeof(c("hi","xxx!@#$"))
[1] "character"
> typeof(as.character(temp[1,]))
[1] "character"
> make.names(as.character(temp[1,]))
[1] "X1" "X1"
> temp[1,]
                  col1                   col2
1 use this row as name use this row as name 2

目标：

>colnames(temp) 
"use this row as name" "use this row as name 2" 

以某种形式。只要列名称是半可理解的，间距和其他格式就不重要了。

Answer 1

您可以unlist数据框：

make.names(unlist(temp))
[1] "use.this.row.as.name"   "use.this.row.as.name.2"

请注意，您的某些尝试失败了，因为temp实际上是两列因素，当您在列表上调用as.character时，会出现一些稍微不直观的行为。 （见?as.character）。