将整数转换为"20160119"到"day""year""month"的不同列

Question

如何将整数列转换为日期:

       DATE PRCP
1: 19490101   25
2: 19490102    5
3: 19490118   18
4: 19490119  386
5: 19490202   38

到这样的表:

days   month   years   PRCP

Answer 1

我们可以用 extract

library(tidyr)
extract(df, DATE, into=c('YEAR', 'MONTH', 'DAY'), 
         '(.{4})(.{2})(.{2})', remove=FALSE)
#       DATE YEAR MONTH DAY PRCP
#1 19490101 1949    01  01   25
#2 19490102 1949    01  02    5
#3 19490118 1949    01  18   18
#4 19490119 1949    01  19  386
#5 19490202 1949    02  02   38

Answer 2

这是使用正则表达式的另一种方式:

df <- read.table(header=T, stringsAsFactors=F, text="
DATE PRCP
19490101   25
19490102    5
19490118   18
19490119  386
19490202   38")
dates <- as.character(df$DATE)
res <- t(sapply(regmatches(dates, regexec("(\\d{4})(\\d{2})(\\d{2})", dates)), "[", -1))
res <- structure(as.integer(res), .Dim=dim(res)) # make them integer values
cbind(df, setNames(as.data.frame(res), c("Y", "M", "D"))) # combine with original data frame
#       DATE PRCP    Y  M  D
# 1 19490101   25 1949 01 01
# 2 19490102    5 1949 01 02
# 3 19490118   18 1949 01 18
# 4 19490119  386 1949 01 19
# 5 19490202   38 1949 02 02

Answer 3

我建议您使用 lubridate 包：

require(lubridate)
df[, DATE := ymd(DATE)]
df[, c("Day", "Month", "Year") := list(day(DATE), month(DATE), year(DATE))]
df[, DATE := NULL]

Answer 4

另一种选择是separate从tidyr包中使用：

library(tidyr)
separate(df, DATE, c('year','month','day'), sep = c(4,6), remove = FALSE)

这导致：

       DATE year month day PRCP
1: 19490101 1949    01  01   25
2: 19490102 1949    01  02    5
3: 19490118 1949    01  18   18
4: 19490119 1949    01  19  386
5: 19490202 1949    02  02   38

基础 R 中的两个选项：

1）与substr作为所述通过在注释中@coffeinjunky：

df$year <- substr(df$DATE,1,4)
df$month <- substr(df$DATE,5,6)
df$day <- substr(df$DATE,7,8)

2）用as.Date和format：

df$DATE <- as.Date(as.character(df$DATE),'%Y%m%d')
df$year <- format(df$DATE, '%Y')
df$month <- format(df$DATE, '%m')
df$day <- format(df$DATE, '%d')