I need to parse dates and have a cases like "31/02/2018":
library(lubridate)
> dmy("31/02/2018", quiet = T)
[1] NA
This makes sense as the 31st of Feb does not exist. Is there a way to parse the string "31/02/2018" to e.g. 2018-02-28 ? So not to get an NA, but an actual date?
Thanks.
We can write a function assuming you would only have dates which could be higher than the actual date and would have the same format always.
library(lubridate)
get_correct_date <- function(example_date) {
#Split vector on "/" and get 3 components (date, month, year)
vecs <- as.numeric(strsplit(example_date, "\\/")[[1]])
#Check number of days in that month
last_day_of_month <- days_in_month(vecs[2])
#If the input date is higher than actual number of days in that month
#replace it with last day of that month
if (vecs[1] > last_day_of_month)
vecs[1] <- last_day_of_month
#Paste the date components together to get new modified date
dmy(paste0(vecs, collapse = "/"))
}
get_correct_date("31/02/2018")
#[1] "2018-02-28"
get_correct_date("31/04/2018")
#[1] "2018-04-30"
get_correct_date("31/05/2018")
#[1] "2018-05-31"
With small modification you can adjust the dates if they have different format or even if some dates are smaller than the first date.
Related
I am trying to convert numeric values into times and dates. I am working with a data set so it would be appreciated if you should show an example using a dataset.
Here are some examples, converting 93537 into 09:35:57 (HH:MM:SS). Additionally, I need to convert 220703 into 22-07-03 (YY:MM:DD).
I will add an example of my code below:
CPLF_data$HMS <- substr(as.POSIXct(sprintf("%04.0f", CPLF_data$StartTime), format='%H%M%S'), 12, 16)
CPLF_data$YMD <- as.POSIXct(CPLF_data$Date, tz="UTC", origin ="1970-01-01", format ="%Y-%M-%D")
The first line is correct however, it does not show seconds.
The second line is incorrect.
Thank you.
I want my final product to be a new column with the times and dates in the correct format with their own columns.
Use chron times class to get the times or if a character string is wanted use as.character on that. Use as.Date to get a Date class object. The sub puts colons between the parts of the time after which we can convert it to times class. The sprintf pads the date with 0 on the left if it is only 5 characters and otherwise leaves it as 6 characters and then we convert that to Date class.
library(chron)
time <- 93537
date <- 220703
tt <- times(sub("(..)(..)$", ":\\1:\\2", time))
tt
## [1] "09:35:37"
as.character(tt)
## [1] "09:35:37"
dd <- as.Date(sprintf("%06d", date), "%y%m%d")
dd
## [1] "2022-07-03"
as.character(dd)
## [1] "2022-07-03"
Try the ymd_hms function in the lubridate package.
output$datetime <- ymd_hms(paste(input$year, input$month, input$day,
input$HH, input$MM, input$SS, sep="-"))
You can enter 00 if you don't have seconds, for example ....
Base R does not have a class for just "time" (of day), as.POSIXct doesn't deal with "times", it deals with "date-times". The lubridate:: package does give number-like HMS values, which may be relevant, but since each row has both date and time, it seems relevant to combine them instead of putting them into separate columns.
CPLF_data |>
transform(
StartTime = as.numeric(StartTime),
Date = as.numeric(Date)
) |>
transform(
DateTime = ISOdate(
2000 + Date %/% 10000, (Date %% 10000) %/% 100, Date %% 100,
StartTime %/% 10000, (StartTime %% 10000) %/% 100, StartTime %% 100)
)
# StartTime Date DateTime
# 1 93537 220703 2022-07-03 09:35:37
Note: I'm assuming that all years are 2-digits and at/after 2000. If this is not true, it's not difficult to work around it with some custom code. Also, over to you if you want to set the timezone of this timestamp by adding tz="US/Mountain" or whichever is more appropriate for the data.
Data
CPLF_data <- data.frame(StartTime = "93537", Date = "220703")
I imported Excel data into R and I have a problem to convert dates.
In R, my data are character and look like :
date<-c('1971-02-00 00:00:00', '1979-06-00 00:00:00')
I would like to convert character into date (MM/YYYY) but the '00' value used for days poses a problem and 'NA' are returned systematically.
It works when I manually replace '00' with '01' and then use as.yearmon, ymd and format. But I have lots of dates to change and I don't know how to change all my '00' into '01' in R.
# data exemple
date1<-c('1971-02-00 00:00:00', '1979-06-00 00:00:00')
# removing time -> doesn't work because of the '00' day
date1c<-format(strptime(date1, format = "%Y-%m-%d"), "%Y/%m/%d")
date1c<-format(strptime(date1, format = '%Y-%m'), '%Y/%m')
# trying to convert character into date -> doesn't work either
date1c<-ymd(date1)
date1c<-strptime(date1, format = "%Y-%m-%d %H:%M:%S")
date1c<-as.Date(date1, format="%Y-%m-%d %H:%M:%S")
date1c<as.yearmon(date1, format='%Y%m')
# everything works if days are '01'
date2<-c('1971-02-01 00:00:00', '1979-06-01 00:00:00')
date2c<-as.yearmon(ymd(format(strptime(date2, format = "%Y-%m-%d"), "%Y/%m/%d")))
date2c
If you have an idea to do it or an another idea to solve my problem, I would be thankful!
Use gsub to replace -00 with -01.
date1<-c('1971-02-01 00:00:00', '1979-06-01 00:00:00')
date1 <- gsub("-00", "-01", date1)
date1c <-format(strptime(date1, format = "%Y-%m-%d"), "%Y/%m/%d")
> date1c
[1] "1971/02/01" "1979/06/01"
Another possibility could be:
as.Date(paste0(substr(date1, 1, 9), "1"), format = "%Y-%m-%d")
[1] "1971-02-01" "1979-06-01"
Here it extracts the first nine characters, pastes it together with 1 and then converts it into a date object.
These alternatives each accept a vector input and produce a vector as output.
Date output
These all will accept a vector as input and produce a Date vector as the output.
# 1. replace first occurrence of '00 ' with '01 ' and then convert to Date
as.Date(sub("00 ", "01 ", date1))
## [1] "1971-02-01" "1979-06-01"
# 2. convert to yearmon class and then to Date
library(zoo)
as.Date(as.yearmon(date1, "%Y-%m"))
## [1] "1971-02-01" "1979-06-01"
# 3. insert a 1 and then convert to Date
as.Date(paste(1, date1), "%d %Y-%m")
## [1] "1971-02-01" "1979-06-01"
yearmon output
Note that if you really are trying to represent just months and years then yearmon class directly represents such objects without the kludge of using an unused day of the month. Such objects are internally represented as a year plus a fraction of a year, i.e. year + 0 for January, year + 1/12 for February, etc. They display in a meaningful way, they sort in the expected manner and can be manipulated, e.g. take the difference between two such objects or add 1/12 to get the next month, etc. As with the others it takes a vector in and produces a vector out.
library(zoo)
as.yearmon(date1, "%Y-%m")
## [1] "Feb 1971" "Jun 1979"
character output
If you want character output rather than Date or yearmon output then these variations work and again accept a vector as input and produce a vector as output:
# 1. replace -00 and everything after that with a string having 0 characters
sub("-00.*", "", date1)
## [1] "1971-02" "1979-06"
# 2. convert to yearmon and then format that
library(zoo)
format(as.yearmon(date1, "%Y-%m"), "%Y-%m")
## [1] "1971-02" "1979-06"
# 3. convert to Date class and then format that
format(as.Date(paste(1, date1), "%d %Y-%m"), "%Y-%m")
## [1] "1971-02" "1979-06"
# 4. pick off the first 7 characters
substring(date1, 1, 7)
## [1] "1971-02" "1979-06"
I would like to create a vector of dates between two specified moments in time with step 1 month, as described in this thread (Create a Vector of All Days Between Two Dates), to be then converted into factors for data visualization.
However, I'd like to have the dates in the YYYY-Mon, ie. 2010-Feb, format. But so far I managed only to have the dates in the standard format 2010-02-01, using a code like this:
require(lubridate)
first <- ymd_hms("2010-02-07 15:00:00 UTC")
start <- ymd(floor_date(first, unit="month"))
last <- ymd_hms("2017-10-29 20:00:00 UTC")
end <- ymd(ceiling_date(last, unit="month"))
> start
[1] "2010-02-01"
> end
[1] "2017-11-01"
How can I change the format to YYYY-Mon?
You can use format():
start %>% format('%Y-%b')
To create the vector, use seq():
seq(start, end, by = 'month') %>% format('%Y-%b')
Obs: Use capital 'B' for full month name: '%Y-%B'.
Hey I have some data aggregated at quarter level and there is a column contains data like this:
> unique(data$fiscalyearquarter)
[1] "2012Q3" "2010Q3" "2012Q1" "2011Q4" "2012Q4" "2008Q1" "2008Q2" "2010Q4" "2010Q1"
[10] "2009Q2" "2012Q2" "2011Q3" "2013Q2" "2013Q1" "2011Q2" "2013Q4" "2009Q4" "2009Q3"
[19] "2011Q1" "2010Q2" "2013Q3" "2008Q4" "2009Q1" "2014Q1" "2008Q3" "2014Q2"
I am thinking about writing a function that turn a string into a timestamp.
Something like this, split the the string to be year and quarter and then force the quarter to be converted to be month(the middle of the quarter).
convert <- function(myinput = "2008Q2"){
year <- substr(myinput, 1, 4)
quarter <- substr(myinput, 6, 6)
month <- 3 * as.numeric(quarter) - 1
date <- as.Date(paste0(year, sprintf("%02d", month), '01'), '%Y%m%d')
return(date)
}
I have to convert those strings to date format and then analyze it from there.
> convert("2010Q3")
[1] "2010-08-01"
Is there any way beyond my hard coding solution to analyze time series problem at quarterly level?
If x is your vector and you're okay having the date be the first day of the quarter:
library(zoo)
as.Date(as.yearqtr(x))
If you want the date to be the first day of the second month of the quarter like your example, you could hack together something like this:
as.Date(format(as.Date(as.yearqtr(x))+40, "%Y-%m-01"))
Currently my dataframe has dates displayed in the 'Date' column as 01/01/2007 etc I would like to convert these into a week/year value i.e. 01/2007. Any ideas?
I have been trying things like this and getting no where...
enviro$Week <- strptime(enviro$Date, format= "%W/%Y")
You have to first convert to date, then you can convert back to the week of the year using format, for example:
### Converts character to date
test.date <- as.Date("10/10/2014", format="%m/%d/%Y")
### Extracts only Week of the year and year
format(test.date, format="Week number %W of %Y")
[1] "Week number 40 of 2014"
### Or if you prefer
format(date, format="%W/%Y")
[1] "40/2014"
So, in your case, you would do something like this:
enviro$Week <- format(as.Date(enviro$Date, format="%m/%d/%Y"), format= "%W/%Y")
But remember that the part as.Date(enviro$Date, format="%m/%d/%Y") is only necessary if your data is not in Date format, and you also should put the right format parameter to convert your character to Date, if that is the case.
What is the class of enviro$Date? If it is of class Date there is probably a better way of doing this, otherwise you can try
v <- strsplit(as.character(enviro$Date), split = "/")
weeks <- sapply(v, "[", 2)
years <- sapply(v, "[", 3)
enviro$Week <- paste(weeks, years, sep = "/")