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RVhydrA
Level III

Add values into X number of lag rows

I have a continuous dataset with measurements every 5 minutes over many years. Certain time periods have been designated and assigned and ID# (e.g. 1, 2, 3,... 500). For each of these periods of time that received an ID I want to extend the column ID backwards in time for 4 hours (ie 48 rows back). So for example If the period categorized as ID# 200 starts on 1/1/2020 at 12:00PM, then I currently have empty cells in the previous 48 rows leading up to that timestamp. I want to fill those rows (1/1/2020 8:00AM - 1/1/2020 11:55AM) with the ID value of 200. I figured there is some sort of indexing (row i - lag row i-48) way to do this but not sure how to execute that. There is a very very clunky way to execute this using a ton of if,then statements but i'm trying to avoid that because i'd need 48 of those to make this work and the formula would be ridiculous.

1 ACCEPTED SOLUTION

Accepted Solutions
txnelson
Super User

Re: Add values into X number of lag rows

Here is a simple script that when it finds an ID#, it populates the previous 48 rows with the current value of ID#

Names Default To Here( 1 );

// Create a table with ID# every 49 rows
dt = New Table( "Example",
	add rows( 490 ),
	New Column( "ID#",
		character,
		formula(
			x = "";
			If( Mod( Row(), 49 ) == 0,
				x = Char( Row() / 49 * 100 )
			);
			x;
		)
	)
);
dt << rerun formulas;
dt:ID# << delete formula;

// Observe the data table for 5 seconds
Wait( 5 );

// Pass across the data table, and when an ID# is found,
// Populate the previous 48 columns
For( i = 49, i <= N Rows( dt ), i++,
	If( :ID#[i] != "",
		:ID#[(i - 48) :: (i - 1) ] = :ID#[i]
	)
);
Jim

View solution in original post

2 REPLIES 2
jthi
Super User

Re: Add values into X number of lag rows

I think it possible to do it with Lag() but I went this time without by using rounding to closest 48. Maybe something like this would work:

dt = New Table("Untitled",
	Add Rows(192),
	Compress File When Saved(1),
	New Column("time",
		Numeric,
		"Continuous",
		Format("Best", 12),
		Set Values(
			[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
			21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
			39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56,
			57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74,
			75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92,
			93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108,
			109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122,
			123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136,
			137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150,
			151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164,
			165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178,
			179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192]
		)
	),
	New Column("id",
		Numeric,
		"Continuous",
		Format("Best", 12),
		Set Values(
			[., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., .,
			., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., 1,
			., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., .,
			., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., 2,
			., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., .,
			., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., 2,
			., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., .,
			., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., ., 3]
		),
		Set Display Width(47)
	)
);

tempcol = dt << New column("formula", 
	Formula(
		If(IsMissing(:id),
			:id[Ceiling(Row()/48)*48], //Ceiling rounding to closest 48 and get value from :id
			:id
		);
	);
);
-Jarmo
txnelson
Super User

Re: Add values into X number of lag rows

Here is a simple script that when it finds an ID#, it populates the previous 48 rows with the current value of ID#

Names Default To Here( 1 );

// Create a table with ID# every 49 rows
dt = New Table( "Example",
	add rows( 490 ),
	New Column( "ID#",
		character,
		formula(
			x = "";
			If( Mod( Row(), 49 ) == 0,
				x = Char( Row() / 49 * 100 )
			);
			x;
		)
	)
);
dt << rerun formulas;
dt:ID# << delete formula;

// Observe the data table for 5 seconds
Wait( 5 );

// Pass across the data table, and when an ID# is found,
// Populate the previous 48 columns
For( i = 49, i <= N Rows( dt ), i++,
	If( :ID#[i] != "",
		:ID#[(i - 48) :: (i - 1) ] = :ID#[i]
	)
);
Jim