Question

我有这段代码来删除字符串中的前导零，我想知道是否可以提高其速度

rmleadingzeros <- function(x){
x <- as.numeric(sapply(x,gsub,pattern="00",replacement="0"))
x <- as.numeric(sapply(x,gsub,pattern="01",replacement="1"))
x <- as.numeric(sapply(x,gsub,pattern="02",replacement="2"))
x <- as.numeric(sapply(x,gsub,pattern="03",replacement="3"))
x <- as.numeric(sapply(x,gsub,pattern="04",replacement="4"))
x <- as.numeric(sapply(x,gsub,pattern="05",replacement="5"))
x <- as.numeric(sapply(x,gsub,pattern="06",replacement="6"))
x <- as.numeric(sapply(x,gsub,pattern="07",replacement="7"))
x <- as.numeric(sapply(x,gsub,pattern="08",replacement="8"))
x <- as.numeric(sapply(x,gsub,pattern="09",replacement="9"))
return(x)
}

datainOKdate$mo <- rmleadingzeros(datainOKdate$mo)

非常感谢

Answer 1

x <- paste0(0, 1:9)
x
# [1] "01" "02" "03" "04" "05" "06" "07" "08" "09"

as.numeric(x)
# [1] 1 2 3 4 5 6 7 8 9

更新以下评论
如果x是一个因素，我们必须更加小心。请参阅?factor

中的“警告”部分

x <- as.factor(paste0(0, c(1, 11, 22, 3)))
x 
# [1] 01  011 022 03 
# Levels: 01 011 022 03

as.numeric(x)
# [1] 1 2 3 4
# not what we want

# This is the way according to ?factor
as.numeric(as.character(x))
# [1]  1 11 22  3

as.numeric(levels(x))[x]
# [1]  1 11 22  3

如何提高速度去除前导零

1 个答案: