Let's say there is a list like this:
date
17-Dec-19
7/26/2018
02/01/2019
02-Mar-18
I suppose I could do several ifelse statements but is there a way I could just get them all formatted at once to look like:
date
2020-12-19
2018-07-26
2019-02-01
2018-03-02
It may be easier to convert to date with anydate from anytime
library(anytime)
df1$date <- anydate(df1$date)
df1$date
#[1] "2019-12-17" "2018-07-26" "2019-02-01" "2018-03-02"
Or with parse_date_time
library(lubridate)
as_date(parse_date_time(df1$date, c("dmy", "mdy")))
Related
date
05-06-2016
05-07-2016
4/13/2016
4/14/2016
I want to format the column to date format using below code
td3 <- read.csv("Book2.csv")
td3$date <- as.Date(td3$date, "%m-%d-%y")
when i run the code the last 2 rows return NA
as.Date.character(gsub("/", "-",td3$date), '%m-%d-%Y')
[1] "2016-05-06" "2016-05-07" "2016-04-13" "2016-04-14"
Here is a solution with parse_date_time from lubridate package:
library(lubridate)
as.Date(parse_date_time(df$date, orders = c('mdy', 'dmy')))
[1] "2016-05-06" "2016-05-07" "2016-04-13" "2016-04-14"
I have a date column in a dataframe. I have read this df into R using openxlsx. The column is 'seen' as a character vector when I use typeof(df$date).
The column contains date information in several formats and I am looking to get this into the one format.
#Example
date <- c("43469.494444444441", "12/31/2019 1:41 PM", "12/01/2019 16:00:00")
#What I want -updated
fixed <- c("2019-04-01", "2019-12-31", "2019-12-01")
I have tried many work arounds including openxlsx::ConvertToDate, lubridate::parse_date_time, lubridate::date_decimal
openxlsx::ConvertToDateso far works best but it will only take 1 format and coerce NAs for the others
update
I realized I actually had one of the above output dates wrong.
Value 43469.494444444441 should convert to 2019-04-01.
Here is one way to do this in two-step. Change excel dates separately and all other dates differently. If you have some more formats of dates that can be added in parse_date_time.
temp <- lubridate::parse_date_time(date, c('mdY IMp', 'mdY HMS'))
temp[is.na(temp)] <- as.Date(as.numeric(date[is.na(temp)]), origin = "1899-12-30")
temp
#[1] "2019-01-04 11:51:59 UTC" "2019-12-31 13:41:00 UTC" "2019-12-01 16:00:00 UTC"
as.Date(temp)
#[1] "2019-01-04" "2019-12-31" "2019-12-01"
You could use a helper function to normalize the dates which might be slightly faster than lubridate.
There are weird origins in MS Excel that depend on platform. So if the data are imported from different platforms, you may want to work woth dummy variables.
normDate <- Vectorize(function(x) {
if (!is.na(suppressWarnings(as.numeric(x)))) # Win excel
as.Date(as.numeric(x), origin="1899-12-30")
else if (grepl("A|P", x))
as.Date(x, format="%m/%d/%Y %I:%M %p")
else
as.Date(x, format="%m/%d/%Y %R")
})
For additional date formats just add another else if. Format specifications can be found with ?strptime.
Then just use as.Date() with usual origin.
res <- as.Date(normDate(date), origin="1970-01-01")
# 43469.494444444441 12/31/2019 1:41 PM 12/01/2019 16:00:00
# "2019-01-04" "2019-12-31" "2019-12-01"
class(res)
# [1] "Date"
Edit: To achieve a specific output format, use format, e.g.
format(res, "%Y-%d-%m")
# 43469.494444444441 12/31/2019 1:41 PM 12/01/2019 16:00:00
# "2019-04-01" "2019-31-12" "2019-01-12"
format(res, "%Y/%d/%m")
# 43469.494444444441 12/31/2019 1:41 PM 12/01/2019 16:00:00
# "2019/04/01" "2019/31/12" "2019/01/12"
To lookup the codes type ?strptime.
I have searched but I could not find out how to convert a date from a character string formatted as follows:
date <- "07-21-2015-09:30AM"
I wanted to use as.Date, but I have not manage to. All I get is the following:
as.Date(date, format="%m-%d-%y-%hAM")
NA
as.Date(dates, format="%m-%d-%y-%h")
NA
If we need the 'date' and 'time', one option is as.POSIXct
as.POSIXct(date, format='%m-%d-%Y-%I:%M%p')
#[1] "2015-07-21 09:30:00 EDT"
You can also use the lubridate package like this:
library('lubridate')
date <- "07-21-2015-09:30AM"
mdy_hm(date)
# "2015-07-21 09:30:00 UTC"
I like strptime for this:
strptime(date, format="%m-%d-%Y-%R%p")
#[1] "2015-07-21 09:30:00 EDT"
And in the case that you needed to see the date in the same format as entered, you can call the related strftime. It doesn't change the internal storage of the variable, rather it changes the format only.
strftime(xx, format="%m-%d-%Y-%R%p")
#[1] "07-21-2015-09:30AM"
I want convert numeric values to time without the date for the data like 1215,1423,1544,1100,0645,1324 in R.
These data has to read like 12:15,14:23,15:44.
I was trying as.POSIXct.
We can use strptime with format
format(strptime(sprintf("%04d", v1), "%H%M"), "%H:%M")
The above output is character class, but if we needed a times class, then we can use times from chron on a "HH:MM:SS" format created with sub or from the above code
library(chron)
times(sub("(.{2})(.{2})","\\1:\\2:", sprintf("%04d00", v1)))
#[1] 12:15:00 14:23:00 15:44:00 11:00:00 06:45:00 13:24:00
Or
times(format(strptime(sprintf("%04d", v1), "%H%M"), "%H:%M:%S"))
data
v1 <- c( 1215,1423,1544,1100,0645,1324)
I have searched but I could not find out how to convert a date from a character string formatted as follows:
date <- "07-21-2015-09:30AM"
I wanted to use as.Date, but I have not manage to. All I get is the following:
as.Date(date, format="%m-%d-%y-%hAM")
NA
as.Date(dates, format="%m-%d-%y-%h")
NA
If we need the 'date' and 'time', one option is as.POSIXct
as.POSIXct(date, format='%m-%d-%Y-%I:%M%p')
#[1] "2015-07-21 09:30:00 EDT"
You can also use the lubridate package like this:
library('lubridate')
date <- "07-21-2015-09:30AM"
mdy_hm(date)
# "2015-07-21 09:30:00 UTC"
I like strptime for this:
strptime(date, format="%m-%d-%Y-%R%p")
#[1] "2015-07-21 09:30:00 EDT"
And in the case that you needed to see the date in the same format as entered, you can call the related strftime. It doesn't change the internal storage of the variable, rather it changes the format only.
strftime(xx, format="%m-%d-%Y-%R%p")
#[1] "07-21-2015-09:30AM"