Convert integer as "20160119" to different columns of "day" "year" "month" - r

Question

How can I convert a column of integers as dates:
DATE PRCP
1: 19490101 25
2: 19490102 5
3: 19490118 18
4: 19490119 386
5: 19490202 38
to a table like this:
days month years PRCP

Answer 1

We can use 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

Here's another way using regular expressions:
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

I would advise you to use the lubridate package:
require(lubridate)
df[, DATE := ymd(DATE)]
df[, c("Day", "Month", "Year") := list(day(DATE), month(DATE), year(DATE))]
df[, DATE := NULL]

Answer 4

Another option would be to use separate from the tidyr package:
library(tidyr)
separate(df, DATE, c('year','month','day'), sep = c(4,6), remove = FALSE)
which results in:
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
Two options in base R:
1) with substr as said by #coffeinjunky in the comments:
df$year <- substr(df$DATE,1,4)
df$month <- substr(df$DATE,5,6)
df$day <- substr(df$DATE,7,8)
2) with as.Date and 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')

Answer 5

First I would convert the DATE column to Date type using as.Date(), then build the new data.frame using calls to format():
df <- data.frame(DATE=c(19490101,19490102,19490118,19490119,19490202),PRCP=c(25,5,18,386,38),stringsAsFactors=F);
df$DATE <- as.Date(as.character(df$DATE),'%Y%m%d');
data.frame(day=as.integer(format(df$DATE,'%d')),month=as.integer(format(df$DATE,'%m')),year=as.integer(format(df$DATE,'%Y')),PRCP=df$PRCP);
## day month year PRCP
## 1 1 1 1949 25
## 2 2 1 1949 5
## 3 18 1 1949 18
## 4 19 1 1949 386
## 5 2 2 1949 38