This question already has answers here:
Converting to date in a character column that contains two date formats
(4 answers)
Closed 5 years ago.
I have various data sources and each data source has a different convention on naming dates.
Example:
Some are:
May 16,2017 as Character type
While others are:
16/05/2017 as Date type
Is it possible to write a code that would coerce everything including those in character to dd/mm/yyyy date format?
We could use the parse_date_time from lubridate which takes multiple formats
library(lubridate)
parse_date_time(str1, c("mdy", "dmy"))
#[1] "2017-05-16 UTC" "2017-05-16 UTC"
data
str1 <- c("May 16,2017", "16/05/2017")
Related
This question already has answers here:
Convert string to date, format: "dd.mm.yyyy"
(4 answers)
What are the "standard unambiguous date" formats for string-to-date conversion in R?
(8 answers)
Closed 3 months ago.
For my dataset, the Date variable has dates in the format of this example: 19-Feb-03
I want to change the above character format dates in the column to a Date format. (As I have to do time series analysis later on)
I tried using the as.Date() method but it didn't work.
x <- '19-Feb-03'
lubridate::ymd(x)
"2019-02-03"
Not sure whether 19 is year or day. You can try lubridate package
x<-"19-Feb-03"
library(lubridate)
ymd(x)
dmy(x)
This question already has answers here:
How to convert date to format "yyyy-mm-dd" in R when input value can be of different formats
(3 answers)
Change Date print format from yyyy-mm-dd to dd-mm-yyyy
(2 answers)
Closed 2 years ago.
I need to find the difference in days between 2 date columns, but one of them is in the format of "6/16/2019" and the other is in the format of "2019-02-25". Not sure which one would be easier to convert to which, but would like to get end result in days. Which I know how to do. I would appreciate help converting the second yyyy-mm-dd to mm-dd-yyyy.
We can use functions from the lubridate package to convert the different formats to dates, and then subtract.
rawData <- "date1,date2
2002-05-15,6/16/2019
2019-12-31,4/15/2020"
data <- read.csv(text = rawData,stringsAsFactors = FALSE)
library(lubridate)
mdy(data$date2) - ymd(data$date1)
...and the output:
> mdy(data$date2) - ymd(data$date1)
Time differences in days
[1] 6241 106
>
This question already has answers here:
Convert date-time string to class Date
(4 answers)
Convert string to date, format: "dd.mm.yyyy"
(4 answers)
Closed 2 years ago.
still fairly new to r. I have searched through the forum on different solutions that don't work on the data set that i have and what output i'm trying to achieve.
I have a column containing dates in the format dd/mm/yyyy (WorkoutDay) and would like to format it to dd-mm-yyyy.
running as.date function through the column gives me this output:
## example date in on of the columns "20/03/2020"
rd$WokoutDay <- as.Date(rd$WorkoutDay)
##output 0031-03-20
I tried to run a format function through but I get this instead which is also not what I'm after.
rd$WorkoutDay <- as.Date(rd$WorkoutDay) %>%
format("%d-%m-%y")
## output 20-03-31
I've read that maybe strptime or as.POSIXct might help but not sure how to use them
thanks in advance :)
This question already has answers here:
Converting year and month ("yyyy-mm" format) to a date?
(9 answers)
Closed 3 years ago.
I have encountered a difficulty, trying to use the as.Date function (in R) on a data frame to preserve date format. The date column consists of blank cells (i.e. missing dates) and observed dates in the format month/year (e.g. 8/2019).
As mentioned earlier, I have tried using the as.Date function but the column for the dates turns blank completely (i.e. no dates are reported). Below is the code I am using:
df$date <- df$date<- as.Date(df$date, format='%m/%Y') #df is the data frame
The expected results should have the observed dates and the missing dates replaced with NA. I greatly appreciate your help.
You need to add a date component to make it a complete date. Once you do that it is easy to convert it into an actual date object
as.Date(paste0("1/", "8/2019"), "%d/%m/%Y")
#[1] "2019-08-01"
Or using dmy from lubridate
lubridate::dmy(paste0("1/", "8/2019"))
This question already has an answer here:
Generating a date from a string with a 'Month-Year' format
(1 answer)
Closed 4 years ago.
I have a list of program dates as character strings in the following format
program.date.have <-c('Sep-14','Aug-14','Sep-16')
I am assuming that all these programs started on the first day of each month, and I want the program.date to end up like
program.date.want<-c('2014-09-01', '2014-08-01, '2016-09-01') or in YYYY-MM-DD format.
To start somewhere I have decided to covert the character strings into the date format in the following way
program.date.have<-c('Sep-14','Aug-14','Sep-16')
betterDates <- as.Date(program.date,
format = "%m-%y")
But even that does not seem to work. how do I use values in program.date variable to be converted into format I want in program.date.want
We can use as.yearmon from zoo, specify the format, and wrap with as.Date which automatically generates the 'day' as the first of the month.
library(zoo)
as.Date(as.yearmon(program.date.have, "%b-%y"))
#[1] "2014-09-01" "2014-08-01" "2016-09-01"
Or a base R option is to paste the '01' at the start or end and then specify the appropriate format in as.Date
as.Date(paste0(program.date.have, "-01"), "%b-%y-%d")
#[1] "2014-09-01" "2014-08-01" "2016-09-01"