- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Report Inappropriate Content
bin each row based on date time
I often need to bin each row based on the date time. However, this is kind of troublesome for the data time. let say I am binning based before 7/22/2020 12:00:00pm as Good and after as bad. I can use the IF formula logic to compare the date time to 7/22/2020 12:00:00pm. But to do that I need to convert to the numeric value of 3678264000. if there any quicker way to do this? thanks
5 REPLIES 5
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Report Inappropriate Content
Re: bin each row based on date time
what you need to use is the Informat() function
if( :date < informat("01/22/2020","m/d/y" ), "Good", "Bad");
Jim
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Report Inappropriate Content
Re: bin each row based on date time
Hi,
Have you considered the "Informat" function, such as used below:
If( :Column 1 < Informat( "07/21/2020 12:00 PM", "MM/dd/yyyy hr:min" ),
1,
0
)
Best,
TS
Thierry R. Sornasse
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Report Inappropriate Content
Re: bin each row based on date time
many thanks both. The informat works perfectly.
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Report Inappropriate Content
Re: bin each row based on date time
Hi,
If you don't mind, could you select our answers as "Solutions": it will let other users know that this question has been addressed
Best,
TS
If you don't mind, could you select our answers as "Solutions": it will let other users know that this question has been addressed
Best,
TS
Thierry R. Sornasse
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Report Inappropriate Content
Re: bin each row based on date time
You can use a date literal in JSL or the formula editor. You just need to use the ddMONyyyy format.
date=22Jul2020;
tomorrow=22Jul2020+in days(1);
show(date, tomorrow, format(tomorrow, "mm/dd/yyyy"));
in the log you get:
date = 22Jul2020; tomorrow = 3678307200; Format(tomorrow, "mm/dd/yyyy") = "07/23/2020";
Read more: Using dates, times, datetimes and durations in JMP
-Jeff