The date is approaching (in fact, 2022-01-08), when Unix time turns 19000-days old. Currently, there is a way (namely, %s) to display Unix second in dates, but there seem to be no way to display Unix day in date formatting. How does one display Unix day in dates? If not possible, then how to introduce new date formatting symbol (%?) symbol for it?
Why?
Epoch date is easy to compute just dividing Unix seconds / 86400, and works well as natural decimal time, where:
1 dYear = 1000 days, 1 dMonth = 100 days, 1 dWeek = 10 days,
1 dHour = 0.1 days, 1 dMinute = 0.001 days, 1 dSecond = 0.00001 days.
Unix day digits, unlike Unix second's digits, approximate lengths of ordinary concepts of time with less than 1 order of magnitude, thus work well as decimal time and date, where:
For example, now is Unix day = 18993.94830052069 means:
18993.94830052069
\ \\\ \ \ 30.052069 th decimal second
\ \\\ \ 48 th decimal minute
\ \\\ 9 th decimal hour
\ \\ 3 rd day of week
\ \ 9 th decimal week
\ 9 th decimal month
18 th decimal year
Unix day, interpreted this way, has the benefit of convenience of use as natural support for decimal time, so some people use it and need it.
Ultimately, I'd like to know a path for incorporating this display of time as one of the supported ways of displaying date and time in *nix systems.
date +%s displays the Unix second, is there a way to display Unix day with it?
No, it is not possible.
how to introduce new date formatting symbol (%?) symbol for it?
Download the sources of the date utility implementation that you are using and patch it with a custom modifier. GNU date (if you are using GNU date...) is part of GNU coretuils, see https://www.gnu.org/software/coreutils/ . Date utility follows POSIX specification.
I do not have ever encountered any need for a "number of days since epoch with fractional part" format. I do not believe it has any practical usage. I do not believe such a formatting specifier should be added to date. I see no value in it over "number of seconds since epoch with fractional part" (date +%s.%N), it's just a change of unit. I wouldn't want those hard-working GNU developers that are giving their work to the world for free, for them to spend time on something that has no practical application. I would like them to they spent their time with their families or on other more important things. As such, I encourage you to not take developers time, unless you can strongly prove that the community requires such a formatting specifier for practical reasons.
Related
Instant.FromUnixTimeSeconds(-100100000000).ToDateTimeUtc1
Once the date gets too ancient this doesn't work anymore, for example, BC dates.
Is there any easy way to convert NodaTime instant values to years months days, that works for the entire range of supported Instant values (aka 27000 BCE to 31000 CE) ?
I don't mind what data type, I am just looking to easily extract the regular time periods from the Instant values.
It's been a year or more since I've used Nodatime, but [this page in the user guide][1] says
Additionally, all calendars are restricted to four digit formats, even
in year-of-era representations, which avoids ever having to parse
5-digit years. This leads to a Gregorian calendar from 9999 BCE to
9999 CE inclusive, or -9998 to 9999 in "absolute" years.
You're question could be read to mean you didn't think BC dates worked at all. When you get more than a few thousand years from the present, strange things start to happen, such as the changing rotation rate of the Earth means there are different kinds of days; those that would be counted from sunrises vs. the kind of time used in radioactive decay, or calculation of planetary positions. It might be helpful if you mentioned your application.
[1]: https://nodatime.org/3.1.x/userguide/range
Trying to convert 1504865618099.00 Unix time into a readable date time.
I tried this:
=(UNIX + ("1/1/1970"-"1/1/1900"+1)*86400) / 86400
But it's not working.
To convert the epoch(Unix-Time) to regular time like for the below timestamp
Ex: 1517577336206
First convert the value with the following function like below
=LEFT(A1,10) & "." & RIGHT(A1,3)
The output will be like below
Ex: 1517577336.206
Now Add the formula like below
=(((B1/60)/60)/24)+DATE(1970,1,1)
Now format the cell like below or required format(Custom format)
m/d/yyyy h:mm:ss.000
Now example time comes like
2/2/2018 13:15:36.206
The three zeros are for milliseconds
=A1/(24*60*60) + DATE(1970;1;1) should work with seconds.
=(A1/86400/1000)+25569 if your time is in milliseconds, so dividing by 1000 gives use the correct date
Don't forget to set the type to Date on your output cell. I tried it with this date: 1504865618099 which is equal to 8-09-17 10:13.
TLDR
=(A1/86400)+25569
...and the format of the cell should be date.
If it doesn't work for you
If you get a number you forgot to format the output cell as a date.
If you get ##### you probably don't have a real Unix time. Check your
timestamps in https://www.epochconverter.com/. Try to divide your input by 10, 100, 1000 or 10000**
You work with timestamps outside Excel's (very extended) limits.
You didn't replace A1 with the cell containing the timestamp ;-p
Explanation
Unix system represent a point in time as a number. Specifically the number of seconds* since a zero-time called the Unix epoch which is 1/1/1970 00:00 UTC/GMT. This number of seconds is called "Unix timestamp" or "Unix time" or "POSIX time" or just "timestamp" and sometimes (confusingly) "Unix epoch".
In the case of Excel they chose a different zero-time and step (because who wouldn't like variety in technical details?). So Excel counts days since 24 hours before 1/1/1900 UTC/GMT. So 25569 corresponds to 1/1/1970 00:00 UTC/GMT and 25570 to 2/1/1970 00:00.
Now if you also note that we have 86400 seconds per day (24 hours x60 minutes x60 seconds) and you will understand what this formula does: A1/86400 converts seconds to days and +25569 adjusts for the offset between what is zero-time for Unix and what is zero-time for Excel.
By the way DATE(1970,1,1) will helpfully return 25569 for you in case you forget all this so a more "self-documenting" way to write our formula is:
=A1/(24*60*60) + DATE(1970,1,1)
P.S.: All these were already present in other answers and comments just not laid out as I like them and I don't feel it's OK to edit the hell out of another answer.
*: that's almost correct because you should not count leap seconds
**: E.g. in the case of this question the number was milliseconds since the the Unix epoch.
If you have ########, it can help you:
=((A1/1000+1*3600)/86400+25569)
+1*3600 is GTM+1
in case the above does not work for you. for me this did not for some reasons;
the UNIX numbers i am working on are from the Mozilla place.sqlite dates.
to make it work : i splitted the UNIX cells into two cells : one of the first 10 numbers (the date) and the other 4 numbers left (the seconds i believe)
Then i used this formula, =(A1/86400)+25569 where A1 contains the cell with the first 10 number; and it worked
You are seeing the date as ######## most likely because by definition the EPOCH times is in seconds - https://en.wikipedia.org/wiki/Unix_time. This means the number should be 10 characters long. Your number has 13 characters (see 1504865618099) and it is most likely in milliseconds (MS). In order to fix the formula just divide the number by 1000. Just keep in mind this way you'll loose the MS precision, but in most cases this is OK. So the final formula should be:
=A1/(86400 * 1000) + DATE(1970,1,1)
Just point and shoot.
Replace the C2 with your cell no. No need to format your Excel cell.
Also, you can use this unixtimestamp website to verify your data.
International format (ISO 8601):
=TEXT(C2/(1000*60*60*24)+25569,"YYYY-MM-DD HH:MM:SS")
2022-10-20 00:04:22
2022-10-20 00:05:20
2022-10-20 00:14:58
US format:
=TEXT(C2/(1000*60*60*24)+25569,"MM/DD/YYYY HH:MM:SS")
10/20/2022 00:04:22
10/20/2022 00:05:20
10/20/2022 00:14:58
Europe format:
=TEXT(C2/(1000*60*60*24)+25569,"DD.MM.YYYY HH:MM:SS")
20.10.2022 00:04:22
20.10.2022 00:05:20
20.10.2022 00:14:58
If you only need the date, remove the 'HH:MM:SS'.
=TEXT(C2/(1000*60*60*24)+25569,"YYYY-MM-DD")
I am doing some work in Excel and am running into a bit of a problem. The instruments I am working with save the date and the time of the measurements and I can read this data into Excel with the following format:
A B
1 Date: Time:
2 12/11/12 2:36:25
3 12/12/12 1:46:14
What I am looking to do is find the difference in the two date/time stamps in mins so that I can create a decay curve from the data. So In Excel, I am looking to Make this (if the number of mins in this example is wrong I just calculated it by hand quickly):
A B C
1 Date: Time: Time Elapsed (Minutes)
2 12/11/12 2:36:25 -
3 12/12/12 1:46:14 1436.82
I Have looked around for a bit and found several methods for the difference in time but they always assume that the dates are the same. I exaggerated the time between my measurements some but that roll over of days is what is causing me grief. Any suggestions or hints as to how to go about this would be great. Even If I could find the difference between the date and times in hrs or days in a decimal format, I could just multiple by a constant to get my answer. Please note, I do have experience with programming and Excel but please explain in details. I sometimes get lost in steps.
time and date are both stored as numerical, decimal values (floating point actually). Dates are the whole numbers and time is the decimal part (1/24 = 1 hour, 1/24*1/60 is one minute etc...)
Date-time difference is calculated as:
date2-date1
time2-time1
which will give you the answer in days, now multiply by 24 (hours in day) and then by 60 (minutes in hour) and you are there:
time elapsed = ((date2-date1) + (time2-time1)) * 24 * 60
or
C3 = ((A3-A2)+(B3-B2))*24*60
To add a bit more perspective, Excel stores date and times as serial numbers.
Here is a Reference material to read up.
I would suggest you to use the following:
Combine date to it's time and then do the difference. So it will not cause you any issues of next day or anything.
Please refer to the image with calculations. You may leave your total minutes cell as general or number format.
MS EXCEL Article: Calculate the difference between two times
Example as per this article
Neat way to do this is:
=MOD(end-start,1)*24
where start and end are formatted as "09:00" and "17:00"
Midnight shift
If start and end time are on the same day the MOD function does not affect anything. If the end time crosses midnight, and the end is earlier then start (say you start 23PM and finish 1AM, so result is 2 hours), the MOD function flips the sign of the difference.
Note that this formula calculates the difference between two times (actually two dates) as decimal value. If you want to see the result as time, display the result as time (ctrl+shift+2).
https://exceljet.net/formula/time-difference-in-hours-as-decimal-value
get n day between two dates, by using days360 function =days360(dateA,dateB)
find minute with this formula using timeA as reference =(timeB-$timeA+n*"24:00")*1440
voila you get minutes between two time and dates
I think =TEXT(<cellA> - <cellB>; "[h]:mm:ss") is a more concise answer. This way, you can have your column as a datetime.
How can I get the day of the year in shell?
date '+%V' will give me the week of the year, which is 15 for today; but I need to find the day of the year!
From the coreutils date manual:
%j day of year (001..366)
Use the date command and the %j option...
doy=$(date +%j)
POSIX mandates the format string %j for getting the day of the year, so if you want it for today's date, you are done, and have a portable solution.
date +%j
For getting the day number of an arbitrary date, the situation is somewhat more complex.
On Linux, you will usually have GNU date, which lets you query for an arbitrary date with the -d option.
date -d '1970-04-01' +%j
The argument to -d can be a fairly free-form expression, including relative times like "3 weeks ago".
date -d "3 weeks ago" +%j
On BSD-like platforms, including MacOS, the mechanism for specifying a date to format is different. You can ask it to format a date with -j and specify the date as an argument (not an option), and optionally specify how the string argument should be parsed with -f.
date -j 04010000 +%j
displays the day number for April 1st 00:00.
The string argument to specify which date to examine is rather weird, and requires the minutes to be specified, and then optionally allows you to prefix with ((month,) day, and) hour, and optionally allows year as a suffix (sic).
date -j -f "%Y-%m-%d" 1970-04-01 +%j
uses -f format date to pass in a date in a more standard format, and prints the day number of that.
There's also the -v option which allows you to specify relative times.
date -j -v -3w +%j
displays the day number of the date three weeks ago.
If you are looking for a proper POSIX-portable solution for getting the day number of arbitrary dates, the least unattractive solution might be to create your own program. If you can rely on Python or Perl (or GNU Awk) to be installed, those make it relatively easy, though it's still a bit of a chore using only their default libraries.
This question already has answers here:
Convert decimal day to HH:MM
(3 answers)
Closed 3 years ago.
I have a dataset that encodes a date-time into two separate variables. Normally, I'd just paste them together inside of an as.POSIXct and carry on. However, the date is provided as a string, and the time of day as a fraction of 24 hours - e.g., 12pm is 0.5, 9:30am is 0.1458333, etc.
It doesn't seem all that tricky to convert the fractional days into clock hours, but I'd prefer to use a pre-existing function if possible. Does something like that exist in base R? A package?
If it's any use, this is an Excel (xlsx) time field imported into R through RODBC.
EDIT
Oddly enough, upon revisiting this problem, the times are now read in as POSIXct. Not sure what to make of that.
The R News 4/1 Help Desk article has a section on reading Excel dates in R.
POSIXct values are simply the number of seconds since midnight GMT 1970-01-01. (So you need to pay attention to your offset from UTC.) You can use the date part and add the number of days times 24*3600 (as.Date(dtval) to your time value * 24*3600. Gabor pointed to the article in R News (which he wrote, thank you, Gabor.)
You didn't give an example of the string. If you are getting your date as a string, then as.Date(strDate) will convert a variable "strDate" to Date class when it is in either "YYYY-MM-DD" or "YYYY/MM/DD" format. Otherwise the formatting codes are on the ?strptime page.
Once you have a POSIXct-classed variable you can just add the number of seconds. This example add 30 minutes to midnight today Feb 1, 2011 (in my time zone which is UTC-5):
> as.POSIXct(as.Date("2011-02-01")) +30*60
[1] "2011-01-31 19:30:00 EST"
And this is your time value added to midnight my time:
> as.POSIXct(as.Date("2011-02-01 00:00", tzone="UTC"))+3600*5 + 3600*24*timeval
[1] "2011-02-01 03:29:59 EST"