This question already has answers here:
Converting year and month ("yyyy-mm" format) to a date?
(9 answers)
Closed 1 year ago.
I have a date that is in the this format:
chr [1:56] "Sep-2016" "Oct-2016" "Nov-2016" "Dec-2016" "Jan-2017" "Feb-2017" "Mar-2017" "Apr-2017"
I tried as.Date(Dates, "%b-%Y") and got NA for all the values. For some reason I have tried multiple ways and using different format but it is still not working. I am looking to get it into either 09-2016, 10-2016 or just simply turn it into a date format.
Any help is much appreciated!
Date class needs a day as well. Easiest is to convert to yearmon class from zoo and then coerce it to Date, which adds a dummy day
library(zoo)
as.Date(as.yearmon(Dates, '%b-%Y'))
or in base R, paste a day and convert
as.Date(paste0(Dates, '-01'), '%b-%Y-%d')
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:
date functions in R return wrong year
(2 answers)
Closed 3 years ago.
Within R, I'm trying to convert a text string into a Date variable type using lubridate's as.Date function.
I have a vector of values such as:
Dates
11/28/2019
11/29/2019
I am attempting to convert these to standard date variables using this as.Date function:
as.Date(Dates, "%m/%d/%y")
I do not receive an error message, and it correctly interprets the month and date, but for some reason it's outputting the wrong year - one year ahead:
"2020-11-28"
"2020-11-29"
I have no earthly idea why it is incorrectly interpreting the year in this way. Any help is appreciated!
We need to use %Y for 4 digit year as %y refers to only 2 digit
as.Date(Dates, "%m/%d/%Y")
Or using lubridate, this would be resolved
library(lubridate)
mdy(Dates)
Or with anydate from anytime
library(anytime)
anydate(Dates)
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$date, format='%m/%Y') #df is the data frame
The expected results should have the observed dates and the missing dates replaced with NA. I greatly appreciate your help.
You need to add a date component to make it a complete date. Once you do that it is easy to convert it into an actual date object
as.Date(paste0("1/", "8/2019"), "%d/%m/%Y")
#[1] "2019-08-01"
Or using dmy from lubridate
lubridate::dmy(paste0("1/", "8/2019"))
This question already has answers here:
Converting year and month ("yyyy-mm" format) to a date?
(9 answers)
Get the difference between dates in terms of weeks, months, quarters, and years
(9 answers)
Closed 4 years ago.
I transformed a date in the yyyy-mm-dd format to the yyyy-mm format using the following command:
format(as.Date(DATE, "%Y-%m-%d"), "%Y-%m")
This works but it returns a character. I want to further use this as a date, so I want to transform this character back to a date class. Using the as.Date() function gives me the error:
Error in charToDate(x) : character string is not in a standard unambiguous format
Does anyone know how to solve this problem?
Update: In the end I want to determine the number of months between two dates in the format yyyy-mm. Does anyone knows how to do this without transforming the characters back to date class?
I had once the same problem and unfortunately the only solution I found was to keep format YYYY-mm-dd with dd==01...
Here is part of the code if you want:
DATE <- str_c(DATE,"-01") df$date <- as.Date(DATE,format="%Y-%m-%d")