我有一个名为df的dataFrame,如:
temp1 temp2
4 3
3 6
4 5
如何找到整个R dataFrame的均值和标准差?我试过这个
mean(as.matrix(df))
但是我收到了这个错误:
In mean.default(data) : argument is not numeric or logical: returning NA
答案 0 :(得分:1)
在评论的两条建议中,as.matrix
方法看起来很优秀:
microbenchmark(unlist = mean(unlist(iris[,1:4])),
as.matrix = mean(as.matrix(iris[,1:4])))
## Unit: microseconds
## expr min lq mean median uq max neval cld
## unlist 236.82 237.961 245.47638 239.102 240.812 374.807 100 b
## as.matrix 58.54 60.440 67.38185 61.961 68.613 273.694 100 a