I import the csv file:
{casesRaw<-read.csv('CasesRaw.csv')
tail(casesRaw$Date)
my result:
"12/28/2020" "12/29/2020" "12/30/2020" "12/31/2020" "1/1/2021" "1/2/2021"
after conversion:
casesRaw$Date<-as.Date(casesRaw$Date,"%m/%d/%y")
tail(casesRaw$Date)
my result is:
[1] "2020-12-28" "2020-12-29" "2020-12-30" "2020-12-31" "2020-01-01" "2020-01-02"
as you can see still I have 2020-01-01 , ....
Any Idea?
We need %Y for 4-digit year instead of %y which is for 2-digit year
casesRaw$Date <- as.Date(casesRaw$Date, "%m/%d/%Y")
Here is another base R option using gsub
casesRaw$Date <- as.Date(gsub("(.*)/(.*)", "\\2/\\1", casesRaw$Date))
Related
So I have this long dataset, where in one column I have a date specified as character in format YYYMMM, but month abbreviated. So for example 1995MAR, 1995APR and so on. How can I transform that to date format?
I tried as.Date but it obviously hasn't worked, and with lubridate::ymd which hasn't worked as well.
Using parse_date_time from lubridate
date <- "1995MAR"
library(lubridate)
parse_date_time(date, order = "Yb")
Output:
[1] "1995-03-01 UTC"
Alternatively using zoo
library(zoo)
as.Date(as.yearmon(date, '%Y%b'))
Output:
"1995-03-01"
str(as.Date(as.yearmon(date, '%Y%b')))
Date[1:1], format: "1995-03-01"
In Base R, add a day number to parse:
date <- "1995MAR"
as.Date(paste(date, "01"), format = "%Y%b %d")
#[1] "1995-03-01"
I am trying to the date format 2019-07-04 14:01 +0000 to mm/dd/yyyy format.
I am using this:
as.Date(strptime(d <- Twitter$time, "%b %d %Y %H:%M %p"))
I've also tried:
ymd_hms(Twitter$time)
However it returns NA values. Is there any way to convert this format to MM/dd/yyyy in R?
As we are not interested in the time component convert the column to Date class with as.Date (here the format is not required as the input is in the default format mode) and use format to change the format
format(as.Date(str1), "%m/%d/%Y")
#[1] "07/04/2019"
data
str1 <- "2019-07-04 14:01 +0000"
There are always two steps: parse, and format.
You can use as.Date() as shown or anydate() from the anytime package (which will also work for different input formats as shown here):
R> inp <- anytime::anydate(c("2019-07-04 14:01 +0000", "04-Jul-2019 14:02"))
R> inp
[1] "2019-07-04" "2019-07-04"
R> format(inp, "%m/%d/%Y")
[1] "07/04/2019" "07/04/2019"
R>
I have the following dataset with dates (YYYY-MM-DD):
> dates
[1] "20180412" "20180424" "20180506" "20180518" "20180530" "20180611" "20180623" "20180705" "20180717" "20180729"
I want to convert them in:
DD-MMM-YYYY but with the month being text. For example 20180412 should become 12Apr2018
Any suggestion on how to proceed?
M
You can try something like this :
# print today's date
today <- Sys.Date()
format(today, format="%B %d %Y") "June 20 2007"
where The following symbols can be used with the format( ) function to print dates 1
You need to first parse the text strings as Date objects, and then format these Date objects to your liking to have the different text output:
R> library(anytime) ## one easy way to parse dates and times
R> dates <- anydate(c("20180412", "20180424", "20180506", "20180518", "20180530",
+ "20180611", "20180623", "20180705", "20180717", "20180729"))
R> dates
[1] "2018-04-12" "2018-04-24" "2018-05-06" "2018-05-18" "2018-05-30"
[6] "2018-06-11" "2018-06-23" "2018-07-05" "2018-07-17" "2018-07-29"
R>
R> txtdates <- format(dates, "%d%b%Y")
R> txtdates
[1] "12Apr2018" "24Apr2018" "06May2018" "18May2018" "30May2018"
[6] "11Jun2018" "23Jun2018" "05Jul2018" "17Jul2018" "29Jul2018"
R>
You could use the as.Date() and format() functions:
dts <- c("20180412", "20180424", "20180506", "20180518", "20180530",
"20180611", "20180623")
format(as.Date(dts, format = "%Y%m%d"), "%d%b%Y")
More information here
Simply use as.POSIXct and as.format:
dates <- c("20180412", "20180424", "20180506")
format(as.POSIXct(dates, format="%Y%m%d"),format="%d%b%y")
Output:
[1] "12Apr18" "24Apr18" "06May18"
I have three data tables in R. Each one has a date column. The tables are vix_data,gold_ohlc_data,btc_ohlc_data. They are formatted as follows:
head(vix_data$Date)
[1] 1/2/04 1/5/04 1/6/04 1/7/04 1/8/04 1/9/04
3435 Levels: 1/10/05 1/10/06 1/10/07 1/10/08 1/10/11 ... 9/9/16
head(gold_ohlc_data$date)
[1] 8/23/17 8/22/17 8/21/17 8/18/17 8/17/17 8/16/17
2519 Levels: 1/10/08 1/10/11 1/10/12 1/10/13 1/10/14 ... 9/9/16
head(btc_ohlc_data$Date)
[1] "2017-08-23" "2017-08-22" "2017-08-21" "2017-08-20" "2017-08-19"
[6] "2017-08-18"
How can I change the date column in the vix_data and gold_ohlc_data tables to match the btc_ohlc_data format? I have tried several methods, for example using as.Date to transform each column- but this usually messes up the values and inserts a lot of N/A's
An option is to use functions from the package lubridate. The users need to know which one is day and which one is month to select the right function to use, such as dmy or mdy
# Load package
library(lubridate)
# Create example string
date1 <- c("1/2/04", "1/5/04", "1/6/04", "1/7/04", "1/8/04", "1/9/04")
date2 <- c("8/23/17", "8/22/17", "8/21/17", "8/18/17", "8/17/17", "8/16/17")
# Convert to date class
dmy(date1)
# [1] "2004-02-01" "2004-05-01" "2004-06-01" "2004-07-01" "2004-08-01" "2004-09-01"
mdy(date1)
# [1] "2004-01-02" "2004-01-05" "2004-01-06" "2004-01-07" "2004-01-08" "2004-01-09"
mdy(date2)
# [1] "2017-08-23" "2017-08-22" "2017-08-21" "2017-08-18" "2017-08-17" "2017-08-16"
Look into the package lubridate. lubridate::dmy() and ymd() should handle this just fine.
It looks like your data are read in as factors, so first you'll have to change them to characters. Then after that you can convert it to a date and specify the input format where %m represents the numerical month, %d represents the day, and %y represents the 2-digit year.
x <- c('1/2/04', '1/5/04', '1/6/04', '1/7/04', '1/8/04', '1/9/04')
y <- as.Date(x, format = "%m/%d/%y")
y
[1] "2004-01-02" "2004-01-05" "2004-01-06" "2004-01-07" "2004-01-08"
[6] "2004-01-09"
Are you sure you're specifying as.Date correctly? For example, do you have %y, instead of %Y?
I did the following and it worked:
> vix <- c("1/2/04", "1/5/04", "1/6/04", "1/7/04", "1/8/04", "1/9/04")
> vix<- as.factor(vix)
> vix
[1] 1/2/04 1/5/04 1/6/04 1/7/04 1/8/04 1/9/04
Levels: 1/2/04 1/5/04 1/6/04 1/7/04 1/8/04 1/9/04
> as.Date(vix, "%m/%d/%y")
[1] "2004-01-02" "2004-01-05" "2004-01-06" "2004-01-07" "2004-01-08" "2004-01-09"
Apologies for the simple question, but I can't find help for this type of date.
April 5th, 2012 is saved as numeric as "20120405"
How can I convert a vector of such values into usable dates?
You just need the as.Date function:
R> x = "20120405"
R> as.Date(x, "%Y%m%d")
[1] "2012-04-05"
Look at the help file: ?as.Date, but essentially
%Y means year in the form 2012, use %y for 12.
%m is the month.
%d the day.
If your date had separators, say, 2012-04-05, then use something like: %Y-%m-%d. Alternatively, you can use:
R> strptime(x, "%Y%m%d")
[1] "2012-04-05"
In particular, you can pass vectors of dates to these functions, so:
R> y = c("20120405", "20121212")
R> as.Date(y, "%Y%m%d")
[1] "2012-04-05" "2012-12-12"
like this,
(foo <- as.Date("20120405", "%Y%m%d"))
# "2012-04-05"
and maybe you want to format to get the month printed out
format(foo, "%Y %b %d")
# "2012 Apr 05"
You could take a look at this page
With strptime you can convert it to POSIXlt class and with as.Date you can convert it to a Date class using format "%Y%m%d":
strptime( "20120405",format="%Y%m%d")
[1] "2012-04-05"
as.Date( "20120405",format="%Y%m%d")
[1] "2012-04-05"
Edit:
It is not really clear if you have character "20120405" or numeric 20120405. In the latter case you have to convert to character first with as.character(20120405)
You could also use the lubridate package:
library(lubridate)
ymd("20120405")