I have a numeric vector as follows
aa <- c(1022011, 2022011, 13022011, 23022011) (this vector is just a sample, it is very long)
Values are written in such a way that first value is day then month and then year.
What I am doing right now is
as.Date(as.character(aa), %d%m%Y")
but,
it is causing problems (returning NA) in case of single digits day numbers. (i.e. 1022011, 2022011).
so basically
as.Date("1022011", "%d%m%Y") does not work
but
as.Date("01022011", "%d%m%Y") (pasting '0' ahead of the number) works.
I want to avoid pasting '0' in such cases. Is there any other (direct) alternative to convert numeric values to dates at once?
It could be rearranged using sub in which case a plain as.Date with no format works:
x <- c(1022011, 11022011) # test data
pat <- "^(..?)(..)(....)$"
as.Date(sub(pat, "\\3-\\2-\\1", x))
giving:
[1] "2011-02-01" "2011-02-11"
Depending on your platform, you could use sprintf in order to add a zero at the beginning. It seems that Mac is OK with this, but not windows 7 given the discussion with the OP.
aa <- c(1022011, 2022011, 13022011, 23022011)
as.Date(sprintf("%08s", aa), format = "%d%m%Y")
[1] "2011-02-01" "2011-02-02" "2011-02-13" "2011-02-23"
UPDATE
#CathyG kindly mentioned that sprintf("%08i",aa) works on Windows 7.
You can use dmy in lubridate:
library(lubridate)
aa <- c(1022011, 2022011, 13022011, 23022011)
> dmy(aa)
[1] "2011-02-01 UTC" "2011-02-02 UTC" "2011-02-13 UTC" "2011-02-23 UTC"
and if you don't want the timezone just wrap it in as.Date:
> as.Date(dmy(aa))
[1] "2011-02-01" "2011-02-02" "2011-02-13" "2011-02-23"
Thank you #Ben Bolker,
> as.Date(mdy(aa))
[1] "2011-01-02" "2011-02-02" "2012-01-02" "2011-01-02"
I know you don't want to add a "0" but still, in base R, this works :
as.Date(sapply(aa,function(x){ifelse(nchar(x)==8,x,paste("0",x,sep=""))}),format = "%d%m%Y")
Related
I am working with an external package that's converting columns of a dataframe with the lubridate date type Date into numeric type. (Confirmed by running as.numeric() on the columns).
I'm wondering if there's a way to convert it back?
For example, if I have the date "O1-01-2021" then running as.numeric on it returns -719143. How can I turn that back into "O1-01-2021" ?
Note that Date class is part of base R, not lubridate.
You probably assumed that the data was year/month/day by mistake. Using base R to eliminate lubridate as a problem we can replicate the question's result like this:
as.numeric(as.Date("01-01-2021", "%Y-%m-%d"))
## [1] -719143
Had we used day/month/year we would have gotten:
as.numeric(as.Date("01-01-2021", "%d-%m-%Y"))
## [1] 18628
or using lubridate
library(lubridate)
as.numeric(dmy("01-01-2021"))
## [1] 18628
It would be best if you fix the mistake that resulted in -719143 but if you don't control that and are faced with an input of
-719143 and want to get as.Date("2021-01-01") as the output then:
# input x is numeric; result is Date class
fixup <- function(x) as.Date(format(.Date(x), "%y-%m-%d"), "%d-%m-%y")
fixup(-719143)
## [1] "2020-01-01"
Note that we can't tell from the question whether 01-01-2020 is supposed to represent day-month-year or month-day-year so we assumed the first but if it is to represent the second then it should be obvious at this point how to proceed.
EDIT #2: It looks like the original data is being parsed as Jan 20, year 1, which might happen if the year-month-day columns were jumbled while being parsed:
as.numeric(as.Date("01-01-2021", format = "%Y-%m-%d", origin = "1970-01-01"))
[1] -719143
as.numeric(as.Date("0001-01-20", origin = "1970-01-01"))
[1] -719143
Is there a way to share an example of the raw data as you have it? e.g. dput(MY_DATA[1:10, DATE_COL])
EDIT: -719143 is about 1970 years of days, which can't be a coincidence, given that many date/time formats use 1970 as a baseline. I wonder if 01-01-2021 is being interpreted as the numeric formula equal to -2021 and so we're looking at perhaps -2021 seconds/days/[?] before year zero, which would be about -1970 years before the epoch...
-719143/(365)
[1] -1970.255
For instance, we can get something close with:
as.numeric(as.Date("0000-01-01", origin = "1970-01-01"))
[1] -719528
Original answer:
R treats a string describing a date as text:
x <- "01-01-2021"
class(x)
[1] "character"
We can convert it to a Date data type using these two equivalent commands:
base_dt <- as.Date(x, "%m-%d-%Y") # base R version
lubridt <- lubridate::mdy(x) # convenience lubridate function
identical(base_dt, lubridt)
[1] TRUE
Under the hood, a Date object in R is a numeric value with a flag telling R it's a date:
> typeof(lubridt) # What general type of data is it?
[1] "double" # --> numeric, stored as a double
> as.numeric(lubridt)
[1] 18628
> class(lubridt) # Does it have any special class attributes?
[1] "Date" # --> yes, it's a Date
> dput(lubridt) # How would we construct it from scratch?
structure(18628, class = "Date") # --> by giving 18628 a Date attribute
In R, a Date is encoded as the number of days since 1970 began:
> as.Date("1970-01-1") + as.numeric(lubridt)
[1] "2021-01-01"
We could convert it back to the original text using:
format(base_dt, "%m-%d-%Y")
[1] "01-01-2021"
identical(x, format(base_dt, "%m-%d-%Y"))
[1] TRUE
i'm fairly new to R
i have these values:
> results$batterydate[file_i]
[1] "2014-05-20 EDT"
> L[[13]]
[1] 1400558400
I'm trying to do:
results$batterydate[file_i] <- L[[13]]
But end up with this error:
Error in as.POSIXct.numeric(value) : 'origin' must be supplied
I'm just trying to replace the value in results$batterydate[file_i] with L[[13]], what's going on?
try,
results$batterydate[file_i] <- as.POSIXct(L[[13]],origin="1970-01-01")
1400558400 is 2014-05-20 06:00:00 CEST in POSIXct, i.e.,
as.POSIXct(1400558400,origin="1970-01-01")
you can use unclass() to extract the raw value, if that's what you are looking to do,
(z <- Sys.time())
#> [1] "2018-02-02 21:45:47 CET"
unclass(z)
#> [1] 1517604347
if you can provide a complete minimal reproducible example to go along with your question we can work from it would be possible to show specifically how it might be possible to solve your issue.
I have three data tables in R. Each one has a date column. The tables are vix_data,gold_ohlc_data,btc_ohlc_data. They are formatted as follows:
head(vix_data$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"
I am trying to convert the string "2013-JAN-14" into a Date as follow :
sdate1 <- "2013-JAN-14"
ddate1 <- as.Date(sdate1,format="%Y-%b-%d")
ddate1
but I get :
[1] NA
What am I doing wrong ? should I install a package for this purpose (I tried installing chron) .
Works for me. The reasons it doesn't for you probably has to do with your system locale.
?as.Date has the following to say:
## This will give NA(s) in some locales; setting the C locale
## as in the commented lines will overcome this on most systems.
## lct <- Sys.getlocale("LC_TIME"); Sys.setlocale("LC_TIME", "C")
x <- c("1jan1960", "2jan1960", "31mar1960", "30jul1960")
z <- as.Date(x, "%d%b%Y")
## Sys.setlocale("LC_TIME", lct)
Worth a try.
This can also happen if you try to convert your date of class factor into a date of class Date. You need to first convert into POSIXt otherwise as.Date doesn't know what part of your string corresponds to what.
Wrong way: direct conversion from factor to date:
a<-as.factor("24/06/2018")
b<-as.Date(a,format="%Y-%m-%d")
You will get as an output:
a
[1] 24/06/2018
Levels: 24/06/2018
class(a)
[1] "factor"
b
[1] NA
Right way, converting factor into POSIXt and then into date
a<-as.factor("24/06/2018")
abis<-strptime(a,format="%d/%m/%Y") #defining what is the original format of your date
b<-as.Date(abis,format="%Y-%m-%d") #defining what is the desired format of your date
You will get as an output:
abis
[1] "2018-06-24 AEST"
class(abis)
[1] "POSIXlt" "POSIXt"
b
[1] "2018-06-24"
class(b)
[1] "Date"
My solution below might not work for every problem that results in as.Date() returning NA's, but it does work for some, namely, when the Date variable is read in in factor format.
Simply read in the .csv with stringsAsFactors=FALSE
data <- read.csv("data.csv", stringsAsFactors = FALSE)
data$date <- as.Date(data$date)
After trying (and failing) to solve the NA problem with my system locale, this solution worked for me.
Is there an easy way in R for me to itemize all valid days that occurred between two specified dates? For instance, I'd like the following inputs:
itemizeDates(startDate="12-30-11", endDate="1-4-12")
To produce the following dates:
"12-30-11" "12-31-11", "1-1-12", "1-2-12", "1-3-12", "1-4-12"
I'm flexible on classes and formatting of the dates, I just need an implementation of the concept.
You're looking for seq
> seq(as.Date("2011-12-30"), as.Date("2012-01-04"), by="days")
[1] "2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03"
[6] "2012-01-04"
Or, you can use :
> as.Date(as.Date("2011-12-30"):as.Date("2012-01-04"), origin="1970-01-01")
[1] "2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03"
[6] "2012-01-04"
Note that with : "Non-numeric arguments are coerced internally". Thus, we convert back to class Date, using as.Date method for class 'numeric' and provide origin.
Here's a function to meet your specific request
itemizeDates <- function(startDate="12-30-11", endDate="1-4-12",
format="%m-%d-%y") {
out <- seq(as.Date(startDate, format=format),
as.Date(endDate, format=format), by="days")
format(out, format)
}
> itemizeDates(startDate="12-30-11", endDate="1-4-12")
[1] "12-30-11" "12-31-11" "01-01-12" "01-02-12" "01-03-12" "01-04-12"
I prefer using the lubridate package to solve datetime problems. It is more intuitive and easier to understand and use once you know it.
library(lubridate)
#mdy() in lubridate package means "month-day-year", which is used to convert
#the string to date object
>start_date <- mdy("12-30-11")
>end_date <- mdy("1-4-12")
#calculate how many days in this time interval
>n_days <- interval(start_date,end_date)/days(1)
>start_date + days(0:n_days)
[1]"2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03" "2012-01-04"
#convert to original format
format(start_date + days(0:n_days), format="%m-%d-%y")
[1] "12-30-11" "12-31-11" "01-01-12" "01-02-12" "01-03-12" "01-04-12"
Reference:
Dates and Times Made Easy with lubridate
2 similar implementations in lubridate:
library(lubridate)
as_date(mdy("12-30-11"):mdy("1-4-12"))
# OR
seq(mdy("12-30-11"), mdy("1-4-12"), by = "days")
These don't format your dates in month-day-year but you can fix the formatting if you want. But year-month-day is a bit easy to work with when analyzing.