我有一个数字变量DATE,表示最后两个字符为MONTH且前一个或两个字符为DAY的日期.我想将列拆分为MONTH和DAY的单独列.
我可以使用以下R代码执行此操作.虽然我希望有一个更简单的正则表达式解决方案.
my.data <- read.table(text = '
ID DATE VARX
A111 104 0
A111 204 1
A111 1004 4
A111 2004 4
B111 3004 2
C111 3004 3
C111 105 4
C111 1005 4
',header = TRUE,stringsAsFactors = FALSE)
# remove the last two characters of a string
my.data$DAY <- ifelse(nchar(my.data$DATE) == 3,substr(my.data$DATE,nchar(my.data$DATE) - (nchar(my.data$DATE)-1),nchar(my.data$DATE) - (nchar(my.data$DATE)-1)),nchar(my.data$DATE) - (nchar(my.data$DATE)-2)))
# keep the last two characters of a string
my.data$MONTH <- substr(my.data$DATE,(nchar(my.data$DATE)-1),nchar(my.data$DATE))
ID DATE VARX DAY MONTH
1 A111 104 0 1 04
2 A111 204 1 2 04
3 A111 1004 4 10 04
4 A111 2004 4 20 04
5 B111 3004 2 30 04
6 C111 3004 3 30 04
7 C111 105 4 1 05
8 C111 1005 4 10 05
谢谢你的任何建议.
解决方法
要使用正则表达式,您可以尝试：
dat <- c(104,204,1004,2004,3004,105,1005)
day <- gsub("(.*?)(..)","\\1",dat)
day
[1] "1" "2" "10" "20" "30" "30" "1" "10"
mth <- gsub("(.*?)(..)","\\2",dat)
mth
[1] "04" "04" "04" "04" "04" "04" "05" "05"
您还可以尝试sprintf和substr的组合.
在这里,我将这些包装成一个函数,给出你的答案：
foo <- function(x){
dat <- sprintf("%04d",x)
cbind(day=substr(dat,1,2),month=substr(dat,3,4))
}
foo(dat)
foo(dat)
day month
[1,] "01" "04"
[2,] "02" "04"
[3,] "10" "04"
[4,] "20" "04"
[5,] "30" "04"
[6,] "30" "04"
[7,] "01" "05"
[8,] "10" "05"
