Let's say I have some date variable which I convert into a string in given format:
mydate <- as.Date("2021-01-01")
myformat <- "%b-%Y"
formatted_date <- format(mydate, myformat)
formatted_date
[1] "Jan-2021"
Now that I've converted it to a string, I then try to read it back to date with the same format, but it does not succeed:
strptime(formatted_date, myformat)
[1] NA
As per the R documentation:
For strptime the input string need not specify the date completely: it is assumed that unspecified seconds, minutes or hours are zero, and an unspecified year, month or day is the current one.
Adding the day at the beginning with %d or similar will make it work, but the docs say it shouldn't be necessary.
Am I missing something?
The complete doc reads :
For strptime the input string need not specify the date completely: it is assumed that unspecified seconds, minutes or hours are zero, and an unspecified year, month or day is the current one. (However, if a month is specified, the day of that month has to be specified by %d or %e since the current day of the month need not be valid for the specified month.) Some components may be returned as NA (but an unknown tzone component is represented by an empty string).
Specifically, the important part here is (However, if a month is specified, the day of that month has to be specified by %d or %e since the current day of the month need not be valid for the specified month.)
Related
Getting NA trying to parse this date:
fasttime::fastPOSIXct("1967-01-01", required.components = 3)
How come?
It is mentioned in the ?fastPOSIXct description
fastPOSIXct converts timestamps in textual (string) form into POSIXct objects. It interprets sequences of digits separated by non-digits as a timestamp in GMT. The order of interpretation is fixed: year, month, day, hour, minute, second. Note that only true (positive) POSIX dates (since 1970-01-01 00:00:00) are supported and fastPOSIXct accepts dates up to year 2199.
I am trying to use a dataset that uses Day of Year (DOY) and want to calculate dates:
To convert DOY format to a date I use:
as.Date(DOY, origin = "%Y-01-01")
But I seem to be getting a problem. For example: DOY = 121 for 2003
as.Date(121, origin = "2003-01-01")
[1] "2003-05-02"
This should be: "2003-05-01" (see here: https://asd.gsfc.nasa.gov/Craig.Markwardt/doy2003.html)
The code works for a leap year:
as.Date(121, origin = "2004-01-01")
[1] "2004-05-01"
(and shows up correctly here: https://asd.gsfc.nasa.gov/Craig.Markwardt/doy2004.html)
Problem persists for 2005.
Am I missing something obvious here or is there a problem in the code missing leap years?
as.Date will accept numeric data (the number of days since an epoch), but only if origin is supplied.
https://www.rdocumentation.org/packages/base/versions/3.6.1/topics/as.Date
So if you convert an integer to a date, the integer will count the number of days that have passed since the origin, which means the date will be the day after the last day has passed
An alternative would be to use the Julian date
as.Date('2005-121', '%Y-%j')
would yield
[1] "2005-05-01"
I have this UTC date in a Google spreadsheet: 2018-10-18T08:55:13Z and would like to convert it to Unix timestamp (1539852913). I tried this formula, but it's unable to recognize the timevalue:
=DATEVALUE(MID(A1;1;10)) + TIMEVALUE(MID(A1;12;8))
If I can get a valid date and time, I can use this formula to convert to Unix timestamp:
=(A1-$C$1)*86400
Does anyone have a solution for this?
Simpler:
=86400*(left(substitute(A1,"T"," "),19))-2209161600
Replaces T with space and cuts off Z, leaving what's left recognisable as date and time in arithmetical calculations. Convert day and time index into seconds and adjust for the offset.
Assuming your date has proceeding zeros for single digit days and month, pull each date string part and drop it into the DATE formula as follows:
Year
=LEFT(A1,4)
Month
=MID(A1,6,2)
Day
=MID(A1,9,2)
Use the date formula
=DATE(year,month,day)
=DATE(LEFT(A1,4),MID(A1,6,2),MID(A1,9,2))
A similar process can be used for TIME
Hour
=MID(A1,12,2)
Minutes
=MID(A1,15,2)
Seconds
=MID(A1,18,2)
Time
=TIME(Hour,Minutes,Seconds)
=TIME(MID(A1,12,2),MID(A1,15,2),MID(A1,18,2))
1) There are other methods
2) The formulas will need to be adapted if you do not have leading 0 for each unit. In that case you would need to use FIND to identify the position of key characters and measure the distance between them to determine if there was a single digit unit or double digit unit.
Since the date is the integer part (left of the decimal) represents the number of days since 1900/01/01 (with that date being 1) and decimal portion represents time in terms of fraction of a day, to get a full date and time, you would add the date formula to the time formula as follows:
=DATE(LEFT(A1,4),MID(A1,6,2),MID(A1,9,2))+TIME(MID(A1,12,2),MID(A1,15,2),MID(A1,18,2))
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.
This question already has answers here:
Read csv with dates and numbers
(3 answers)
Closed 9 years ago.
I am working on "Localization Data for Person Activity Data Set" dataset from UCI and in this data set there is a column of date and time(both in one column) with following format:
27.05.2009 14:03:25:777
27.05.2009 14:03:25:183
27.05.2009 14:03:25:210
27.05.2009 14:03:25:237
...
I am wondering if there is anyway to convert this column to timestamp using R.
First of all, we need to substitute the colon separating the milliseconds from the seconds to a dot, otherwise the final step won't work (thanks to Dirk Eddelbuettel for this one). Since in the end R will use the separators it wants, to be quicker, I'll just go ahead and substitute all the colons for dots:
x <- "27.05.2009 14:03:25:777" # this is a simplified version of your data
y <- gsub(":", ".", x) # this is your vector with the aforementioned substitution
By the way, this is how your vector should look after gsub:
> y
[1] "27.05.2009 14.03.25.777"
Now, in order to have it show the milliseconds, you first need to adjust an R option and then use a function called strptime, which will convert your date vector to POSIXlt (an R-friendly) format. Just do the following:
> options(digits.secs = 3) # this tells R you want it to consider 3 digits for seconds.
> strptime(y, "%d.%m.%Y %H:%M:%OS") # this finally formats your vector
[1] "2009-05-27 14:03:25.777"
I've learned this nice trick here. This other answer also says you can skip the options setting and use, for example, strptime(y, "%d.%m.%Y %H:%M:%OS3"), but it doesn't work for me. Henrik noted that the function's help page, ?strptime states that the %OS3 bit is OS-dependent. I'm using an updated Ubuntu 13.04 and using %OS3 yields NA.
When using strptime (or other POSIX-related functions such as as.Date), keep in mind some of the most common conversions used (edited for brevity, as suggested by DWin. Complete list at strptime):
%a Abbreviated weekday name in the current locale.
%A Full weekday name in the current locale.
%b Abbreviated month name in the current locale.
%B Full month name in the current locale.
%d Day of the month as decimal number (01–31).
%H Hours as decimal number (00–23). Times such as 24:00:00 are accepted for input.
%I Hours as decimal number (01–12).
%j Day of year as decimal number (001–366).
%m Month as decimal number (01–12).
%M Minute as decimal number (00–59).
%p AM/PM indicator in the locale. Used in conjunction with %I and not with %H.
`%S Second as decimal number (00–61), allowing for up to two leap-seconds (but POSIX-compliant implementations will ignore leap seconds).
%U Week of the year as decimal number (00–53) using Sunday as the first day 1 of the week (and typically with the first Sunday of the year as day 1 of week 1). The US convention.
%w Weekday as decimal number (0–6, Sunday is 0).
%W Week of the year as decimal number (00–53) using Monday as the first day of week (and typically with the first Monday of the year as day 1 of week 1). The UK convention.
%y Year without century (00–99). On input, values 00 to 68 are prefixed by 20 and 69 to 99 by 19
%Y Year with century. Note that whereas there was no zero in the original Gregorian calendar, ISO 8601:2004 defines it to be valid (interpreted as 1BC)