强制列表对象键入double

Question

这个问题已经问了很多遍，但是建议的解决方案在某种程度上对我不起作用。

d <- data.frame(var1 = c("ab","ac", "ad"), var2 = c("1.1","2.6","0"),var3 = c("1","5.4","0"))

ff <- as.numeric(as.character(d[ ,2:3]))
fg <- lapply(d[ ,2:3],as.numeric)
fh <- as.numeric(unlist(d[ ,2:3]))
fi <- as.numeric(d[ ,2:3])

var2和var3被分类为因素。我不想重新排列或合并列或单元格。我只希望它们是数字以便能够计算。

Answer 1

要对多列的子集施加强制，lapply是正确的方法。这将产生一个列表，您想要将其分配回子集列。

as.numeric.factor <- function(f) as.numeric(levels(f))[f]    
# as.numeric.factor <- function(f) as.numeric(as.character(f))  ## alternatively

d[,2:3] <- lapply(d[,2:3], as.numeric.factor)

str(d)
# 'data.frame': 3 obs. of  3 variables:
# $ var1: chr  "ab" "ac" "ad"
# $ var2: num  1.1 2.6 0
# $ var3: num  1 5.4 0

如果需要新对象，可以将lapply与replace一起使用。

fg <- replace(d, 2:3, lapply(d[,2:3], as.numeric.factor))
str(fg)
# 'data.frame': 3 obs. of  3 variables:
# $ var1: chr  "ab" "ac" "ad"
# $ var2: num  1.1 2.6 0
# $ var3: num  1 5.4 0

---

数据

d <- structure(list(var1 = c("ab", "ac", "ad"), var2 = structure(c(2L, 
3L, 1L), .Label = c("0", "1.1", "2.6"), class = "factor"), var3 = structure(c(2L, 
3L, 1L), .Label = c("0", "1", "5.4"), class = "factor")), row.names = c(NA, 
-3L), class = "data.frame")