Hi I want to convert a 2 digit year date in string format to an Date object. But I an not sure what to use as the format
For 4 digit Dates it works fine, i.e
using Dates
t1 = "01/01/2017"
Date(t1, "dd/mm/yyyy")
# Out > 2017-01-01
But for 2 digit year
t2 = "27/01/17
Date(t2, "dd/mm/yy")
# Out > 0017-01-27
Any idea what to use as the formatting?
This seems not be implemented (yet). See the discussion here or the (open) pull request to implement it here.
It is a debated topic, as the default in other languages is to assign years >68 to the twenty century and those <=68 to the twenty-first century, that is a bit subjective, so the Julia developers preferred to go for the explicit way that the missing digits must be explicitly added.
So for now just add 2000 years:
t2 = "27/01/17"
Date(t2, "dd/mm/yy") + Dates.Year(2000)
Related
I am working on the transformation of week based dates to month based dates.
When checking my work, I found the following problem in my data which is the result of a simple call to as.Date()
as.Date("2016-50-4", format = "%Y-%U-%u")
as.Date("2016-50-5", format = "%Y-%U-%u")
as.Date("2016-50-6", format = "%Y-%U-%u")
as.Date("2016-50-7", format = "%Y-%U-%u") # this is the problem
The previous code yields correct date for the first 3 lines:
"2016-12-15"
"2016-12-16"
"2016-12-17"
The last line of code however, goes back 1 week:
"2016-12-11"
Can anybody explain what is happening here?
Working with week of the year can become very tricky. You may try to convert the dates using the ISOweek package:
# create date strings in the format given by the OP
wd <- c("2016-50-4","2016-50-5","2016-50-6","2016-50-7", "2016-51-1", "2016-52-7")
# convert to "normal" dates
ISOweek::ISOweek2date(stringr::str_replace(wd, "-", "-W"))
The result
#[1] "2016-12-15" "2016-12-16" "2016-12-17" "2016-12-18" "2016-12-19" "2017-01-01"
is of class Date.
Note that the ISO week-based date format is yyyy-Www-d with a capital W preceeding the week number. This is required to distinguish it from the standard month-based date format yyyy-mm-dd.
So, in order to convert the date strings provided by the OP using ISOweek2date() it is necessary to insert a W after the first hyphen which is accomplished by replacing the first - by -W in each string.
Also note that ISO weeks start on Monday and the days of the week are numbered 1 to 7. The year which belongs to an ISO week may differ from the calendar year. This can be seen from the sample dates above where the week-based date 2016-W52-7 is converted to 2017-01-01.
About the ISOweek package
Back in 2011, the %G, %g, %u, and %V format specifications weren't available to strptime() in the Windows version of R. This was annoying as I had to prepare weekly reports including week-on-week comparisons. I spent hours to find a solution for dealing with ISO weeks, ISO weekdays, and ISO years. Finally, I ended up creating the ISOweek package and publishing it on CRAN. Today, the package still has its merits as the aforementioned formats are ignored on input (see ?strptime for details).
As #lmo said in the comments, %u stands for the weekdays as a decimal number (1–7, with Monday as 1) and %U stands for the week of the year as decimal number (00–53) using Sunday as the first day. Thus, as.Date("2016-50-7", format = "%Y-%U-%u") will result in "2016-12-11".
However, if that should give "2016-12-18", then you should use a week format that has also Monday as starting day. According to the documentation of ?strptime you would expect that the format "%Y-%V-%u" thus gives the correct output, where %V stands for the week of the year as decimal number (01–53) with monday as the first day.
Unfortunately, it doesn't:
> as.Date("2016-50-7", format = "%Y-%V-%u")
[1] "2016-01-18"
However, at the end of the explanation of %V it sais "Accepted but ignored on input" meaning that it won't work.
You can circumvent this behavior as follows to get the correct dates:
# create a vector of dates
d <- c("2016-50-4","2016-50-5","2016-50-6","2016-50-7", "2016-51-1")
# convert to the correct dates
as.Date(paste0(substr(d,1,8), as.integer(substring(d,9))-1), "%Y-%U-%w") + 1
which gives:
[1] "2016-12-15" "2016-12-16" "2016-12-17" "2016-12-18" "2016-12-19"
The issue is because for %u, 1 is Monday and 7 is Sunday of the week. The problem is further complicated by the fact that %U assumes week begins on Sunday.
For the given input and expected behavior of format = "%Y-%U-%u", the output of line 4 is consistent with the output of previous 3 lines.
That is, if you want to use format = "%Y-%U-%u", you should pre-process your input. In this case, the fourth line would have to be as.Date("2016-51-7", format = "%Y-%U-%u") as revealed by
format(as.Date("2016-12-18"), "%Y-%U-%u")
# "2016-51-7"
Instead, you are currently passing "2016-50-7".
Better way of doing it might be to use the approach suggested in Uwe Block's answer. Since you are happy with "2016-50-4" being transformed to "2016-12-15", I suspect in your raw data, Monday is counted as 1 too. You could also create a custom function that changes the value of %U to count the week number as if week begins on Monday so that the output is as you expected.
#Function to change value of %U so that the week begins on Monday
pre_process = function(x, delim = "-"){
y = unlist(strsplit(x,delim))
# If the last day of the year is 7 (Sunday for %u),
# add 1 to the week to make it the week 00 of the next year
# I think there might be a better solution for this
if (y[2] == "53" & y[3] == "7"){
x = paste(as.integer(y[1])+1,"00",y[3],sep = delim)
} else if (y[3] == "7"){
# If the day is 7 (Sunday for %u), add 1 to the week
x = paste(y[1],as.integer(y[2])+1,y[3],sep = delim)
}
return(x)
}
And usage would be
as.Date(pre_process("2016-50-7"), format = "%Y-%U-%u")
# [1] "2016-12-18"
I'm not quite sure how to handle when the year ends on a Sunday.
I imported date variables as strings from SQL (date1) into Stata and then created a new date variable (date2) like this:
gen double date2 = clock(date1, "YMDhms")
format date2 %tc
However, now I want to calculate the number of days between two dates (date3-date2), formatted as above, but I can't seem to do it.
I don't care about the hms, so perhaps I should strip that out first? And then deconstruct the date into YYYY MM DD as separate variables? Nothing I seems to do is working right now.
It sounds like by dates you actually mean timestamp (aka datetime) variables. In my experience, there's usually no need to cast dates/timestamps as strings since ODBC and Stata will handle the conversion to SIF td/tc formats nicely.
But perhaps you exported to a text file and then read in the data instead. Here are a couple solutions.
tc timestamps are in milliseconds since 01jan1960 00:00:00.000, assuming 1000*60*60*24=86,400 seconds/day (that is, ignoring leap seconds). This means that you need to divide your difference by that number to get elapsed days.
For example, 2016 was a leap year:
. display (tc(01jan2017 00:00:00) - tc(01jan2016 00:00:00))/(1000*60*60*24)
366
You can also use the dofc() function to make dates out of timestamps and omit the division:
. display (dofc(tc(01jan2018 00:00:00)) - dofc(tc(01jan2016 00:00:00)))
731
2017 is not a leap year, so 366 + 365 = 731 days.
You can use generate with all these functions, though display is often easier for debugging initial attempts.
I wish to convert following value which is a chararray in pig
2016-05-11 23:59:57.628197
to
2016-05-11T23:59:57.628-05:00
How can I do it ?
Following is what I tried considering alias 'a2' contains list of datetime values in chararray in the column named 'input_date_value'
FOREACH a2 GENERATE input_date_value AS input_date:chararray,
ToDate(input_date_value,'YYYY-MM-DD HH:mm:ss.SSSSSS') AS modification_datetime:datetime;
For input -
2002-07-11 16:58:40.249764
Output is -
2002-01-11T16:58:40.249-05:00
The month values like '07' are not getting picked up,
The created timestamp has month set to 01' i.e. January everytime for all dates.
Can someone help. What am I doing wrong ?
https://pig.apache.org/docs/r0.11.1/func.html#to-date ToDate takes SimpleDateFormat only supports milliseconds http://docs.oracle.com/javase/6/docs/api/java/text/SimpleDateFormat.html
The -05:00 you see is the time zone ToDate is actually truncating to 3 digits as it supports only milliseconds
Use lowercase character d instead of uppercase D for parsing date values.
Now, I have managed to fix it myself on (In Pig 0.11)
Apparently Pig 0.11 does not support the date format components I used earlier for parsing the month and date.
I found below inference which hints on the incompatibility as mentioned https://www.w3.org/TR/NOTE-datetime
Use:
'YYYY-MM-dd HH:mm:ss.SSSSSS'
instead of 'YYYY-MM-DD HH:mm:ss.SSSSSS'
It now gives correct output.
Input:
2001-11-28 16:04:49.22388
Output:
2001-11-28T16:04:49.223-05:00
The %tw format in Stata has the form: 1960w1 which has no equivalent in R.
Therefore %tw dates must be post-processed.
Importing a .dta file into R, the date is an integer like 1304 (instead of 1985w5) or 1426 (instead of 1987w23). If it was a simple time series you could set a starting date as follows:
ts(df, start= c(1985,5), frequency=52)
Another possibility would be:
as.Date(Camp$date, format= "%Yw%W" , origin = "1985w5")
But if each row is not a single date, then you must convert it.
The package ISOweek is based on ISO-8601 with the form "1985-W05" and does not process the Stata %tw.
The Lubridate package does not work with this format. The week() returns the number of complete seven day periods that have occurred between the date and January 1st, plus one. week function
In Stata week 1 of any year starts on 1 January, whatever day of the week that is. Stata Documentation on Dates
In the format %W of Date in R the week starts as Monday as first day of the week.
From strptime %V is
the Week of the year as decimal number (00--53) as defined in ISO
8601. If the week (starting on Monday) containing 1 January has four or more days in the new year, then it is considered week 1. Otherwise,
it is the last week of the previous year, and the next week is week 1.
(Accepted but ignored on input.) Strptime
Larmarange noted on Github that Haven doesn't interpret dates properly:
months, week, quarter and halfyear are specific format from Stata,
respectively %tm, %tw, %tq and %th. I'm not sure that there are
corresponding formats available in R. So far they are imported as
integers.
Is there a way to convert Stata %tw to a date format R understands?
Here is an Stata file with dates
This won't be an answer in terms of R code, but it is commentary on Stata weeks that can't be fitted into a comment.
Strictly, dates in Stata are not defined by the display formats that make them intelligible to people. A date in Stata is always a numeric variable or scalar or macro defined with origin the first instance in 1960. Thus it is at best a shorthand to talk about %tw dates, etc. We can use display to see the effects of different date display formats:
. di %td 0
01jan1960
. di %tw 0
1960w1
. di %tq 0
1960q1
. di %td 42
12feb1960
. di %tw 42
1960w43
. di %tq 42
1970q3
A subtle point made explicit above is that changing the display format will not change what is stored, i.e. the numeric value.
Otherwise put, dates in Stata are not distinct data types; they are just integers made intelligible as dates by a pertinent display format.
The question presupposes that it was correct to describe some weekly dates in terms of Stata weeks. This seems unlikely, as I know no instance in which a body outside StataCorp uses the week rules of Stata, not only that week 1 always starts on 1 January, but also that week 52 always includes either 8 or 9 days and hence that there is never a week 53 in a calendar year.
So, you need to go upstream and find out what the data should have been. Failing some explanation, my best advice is to map the 52 weeks of each year to the days that start them, namely days 1(7)358 of each calendar year.
Stata weeks won't map one-to-one to any other scheme for defining weeks.
More in this article on Stata weeks
It's not completely clear what the question is but the year and week corresponding to 1304 are:
wk <- 1304
1960 + wk %/% 52
## [1] 1985
wk %% 52 + 1
## [1] 5
so assuming that the first week of the year is week 1 and starts on Jan 1st, the beginning of the above week is this date:
as.Date(paste(1960 + wk %/% 52, 1, 1, sep = "-")) + 7 * (wk %% 52)
## [1] "1985-01-29"
How to convert between year,month,day and dates in R?
I know one can do this via strings, but I would prefer to avoid converting to strings, partly because maybe there is a performance hit?, and partly because I worry about regionalization issues, where some of the world uses "year-month-day" and some uses "year-day-month".
It looks like ISODate provides the direction year,month,day -> DateTime , although it does first converts the number to a string, so if there is a way that doesn't go via a string then I prefer.
I couldn't find anything that goes the other way, from datetimes to numerical values? I would prefer not needing to use strsplit or things like that.
Edit: just to be clear, what I have is, a data frame which looks like:
year month day hour somevalue
2004 1 1 1 1515353
2004 1 1 2 3513535
....
I want to be able to freely convert to this format:
time(hour units) somevalue
1 1515353
2 3513535
....
... and also be able to go back again.
Edit: to clear up some confusion on what 'time' (hour units) means, ultimately what I did was, and using information from How to find the difference between two dates in hours in R?:
forwards direction:
lh$time <- as.numeric( difftime(ISOdate(lh$year,lh$month,lh$day,lh$hour), ISOdate(2004,1,1,0), units="hours"))
lh$year <- NULL; lh$month <- NULL; lh$day <- NULL; lh$hour <- NULL
backwards direction:
... well, I didnt do backwards yet, but I imagine something like:
create difftime object out of lh$time (somehow...)
add ISOdate(2004,1,1,0) to difftime object
use one of the solution below to get the year,month,day, hour back
I suppose in the future, I could ask the exact problem I'm trying to solve, but I was trying to factorize my specific problem into generic reusable questions, but maybe that was a mistake?
Because there are so many ways in which a date can be passed in from files, databases etc and for the reason you mention of just being written in different orders or with different separators, representing the inputted date as a character string is a convenient and useful solution. R doesn't hold the actual dates as strings and you don't need to process them as strings to work with them.
Internally R is using the operating system to do these things in a standard way. You don't need to manipulate strings at all - just perhaps convert some things from character to their numerical equivalent. For example, it is quite easy to wrap up both operations (forwards and backwards) in simple functions you can deploy.
toDate <- function(year, month, day) {
ISOdate(year, month, day)
}
toNumerics <- function(Date) {
stopifnot(inherits(Date, c("Date", "POSIXt")))
day <- as.numeric(strftime(Date, format = "%d"))
month <- as.numeric(strftime(Date, format = "%m"))
year <- as.numeric(strftime(Date, format = "%Y"))
list(year = year, month = month, day = day)
}
I forego the a single call to strptime() and subsequent splitting on a separation character because you don't like that kind of manipulation.
> toDate(2004, 12, 21)
[1] "2004-12-21 12:00:00 GMT"
> toNumerics(toDate(2004, 12, 21))
$year
[1] 2004
$month
[1] 12
$day
[1] 21
Internally R's datetime code works well and is well tested and robust if a bit complex in places because of timezone issues etc. I find the idiom used in toNumerics() more intuitive than having a date time as a list and remembering which elements are 0-based. Building on the functionality provided would seem easier than trying to avoid string conversions etc.
I'm a bit late to the party, but one other way to convert from integers to date is the lubridate::make_date function. See the example below from R for Data Science:
library(lubridate)
library(nycflights13)
library(tidyverse)
a <- flights %>%
mutate(date = make_date(year, month, day))
Found one solution for going from date to year,month,day.
Let's say we have a date object, that we'll create here using ISOdate:
somedate <- ISOdate(2004,12,21)
Then, we can get the numerical components of this as follows:
unclass(as.POSIXlt(somedate))
Gives:
$sec
[1] 0
$min
[1] 0
$hour
[1] 12
$mday
[1] 21
$mon
[1] 11
$year
[1] 104
Then one can get what one wants for example:
unclass(as.POSIXlt(somedate))$mon
Note that $year is [actual year] - 1900, month is 0-based, mday is 1-based (as per the POSIX standard)