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Sysy13
Level II

Use for loop variable in a formula

Trying with this script but variable "l" is not working, only variable "k" is working. Anyway to solve this?

For( k = 40, k <= 54, k += 2,
	x1 = 1;
	x2 = 8;
	y1 = 40;
	y2 = 54;
	m = (y2 - y1) / (x2 - x1);
	c = y2 - (m * x1);
	l = (k - c) / m;
	dt << New Column( "Test" || Char( k ),
		Numeric,
		"Continuous",
		Format( "Best", 12 ),
		Formula(
			If( :Strobe == 1,
				If(
					:Tested Variable == 0 | :Tested Plane == 1,
						If( Left( Right( :"x"n, Column[l] ) ), 1 ) == "1",
					1, Left( Right( :"x"n, (l) ), 1 ) == "0",
					0,
				), 

			)
		)
	);
)
1 ACCEPTED SOLUTION

Accepted Solutions
txnelson
Super User

Re: Use for loop variable in a formula

The below script appears to work.  Your supplied script used \\ as comment initiators, where // is what the correct syntax is.

The real working formula really boiled down to using the Substr() function.

The addition of the Eval(Substitute(Expr())) section just changed the value of your l variable into the actual value of l, rather than having the value referenced through the variable.

Names Default To Here( 1 );
dt = Current Data Table();
//k start 40, For each loop will add 2
For( k = 40, k <= 54, k += 2,
	x1 = 1;
	x2 = 8;
	y1 = 40;
	y2 = 54;
	m = (y2 - y1) / (x2 - x1);
	c = y2 - (m * x2);
	l = (k - c) / m;
	
	//l is calculated based on k\
	//Tested variable & Column x & Strobe are the column in my data set
	//If k = 40;, then the readout will be 1st bit from the right of column x;
	//If k = 40;,then the readout will be 8th bit from the right of column x;
	
	// The variable l needs to be fully parsed in the formula, so that if the 
	// value of l changes, it does not change previous formulas
	Eval(
		Substitute(
				Expr(
					dt << New Column( "Test" || Char( k ),
						Numeric,
						"Continuous",
						Format( "Best", 12 ),
						Formula(
							theVal = .;
							If( :Strobe == 1,
								If( :Tested Variable == 0 | :Tested Variable == 1,
									theVal = Num( Substr( :x, _l_ , 1 ) )
								)
							);
							theVal;
						)
					)
				),
			Expr( _l_ ), l
		)
	);
);
Jim

View solution in original post

7 REPLIES 7
txnelson
Super User

Re: Use for loop variable in a formula

A couple of issues

The 

Column[ l ]

you specified is incorrect syntax.  Column is a function (), not a matrix [] therefore it needs to be specified as

Column( l )

Next, the formula will have an issue with the value you are assigning.  Given your formula, there are cases where no value will be assigned.  Since a JMP formula returns the last value assigned, you need to explicitly assign a value to the formula for every case.  In my version of your formula, I am using a variable  X to assign your logic to.  I initialize the value to a missing value, and then X is assigned by your If() logic.  I then assure that the calculated value of X is the last value processed, by simply adding 

x;

as the last statement in the formula.

For( k = 40, k <= 54, k += 2,
	x1 = 1;
	x2 = 8;
	y1 = 40;
	y2 = 54;
	m = (y2 - y1) / (x2 - x1);
	c = y2 - (m * x1);
	l = (k - c) / m;
	dt << New Column( "Test" || Char( k ),
		Numeric,
		"Continuous",
		Format( "Best", 12 ),
		Formula(
			x = .;
			If( :Strobe == 1,
				If(
					:Tested Variable == 0 | :Tested Plane == 1,
						If( Left( Right( :"x"n, Column(l) ) ),
							1
						) == "1",
					x = 1, 
					Left( Right( :"x"n, (l) ), 1 ) == "0",
					x = 0,
				), 

			);
			x;
		)
	);
)

Finally, when entering JSL into a Discussion Question, please use the txnelson_0-1702341399738.png icon so the JSL is easier to read by the Community Members

 

Jim
Sysy13
Level II

Re: Use for loop variable in a formula

Ok sure sure, will use the icon for next time. Btw the script is not working. Maybe I should further elaborate what are the problems I faced

I have a column with 8bit binary number, for example x="00101101"

 

Then when my variable k is different, the bit read out from column x should be different

 

 

\\k start 40, For each loop will add 2
For( k = 40, k <= 54, k += 2,
	x1 = 1;
	x2 = 8;
	y1 = 40;
	y2 = 54;
	m = (y2 - y1) / (x2 - x1);
	c = y2 - (m * x2);
	l = (k - c) / m;
\\l is calculated based on k\
\\Tested variable & Column x & Strobe are the column in my data set
\\If k=40, then the readout will be 1st bit from the right of column x;
\\If k=40, then the readout will be 8th bit from the right of column x;
	dt << New Column( "Test" || Char( k ),
		Numeric,
		"Continuous",
		Format( "Best", 12 ),
		Formula(
			If( :Strobe == 1,
				If(
					:Tested Variable == 0 | :Tested Variable == 1,
						If(  Right( :"x"n, Column(l) ) ) == "1",
					1, Right( :"x"n, (l) )== "0",
					0,
				), 

			)
		)
	);
)

The issue I faced is jmp cannot recognize what is the bit that I want to read out no matter I put Column(l), l  in the for loop

txnelson
Super User

Re: Use for loop variable in a formula

It would be helpful to if you could attach a sample data table.

With your further explanation, try changing 

If(  Right( :"x"n, Column(l) ) ) == "1"

to

If(  Right( :"x"n, as Column( "Column" || char(l) )== "1"
Jim
Sysy13
Level II

Re: Use for loop variable in a formula

Hi Jim, here's data table, I put what I expect to see in data table. Column data describe what is the "l" value and which value of "x" it is coming from

txnelson
Super User

Re: Use for loop variable in a formula

The below script appears to work.  Your supplied script used \\ as comment initiators, where // is what the correct syntax is.

The real working formula really boiled down to using the Substr() function.

The addition of the Eval(Substitute(Expr())) section just changed the value of your l variable into the actual value of l, rather than having the value referenced through the variable.

Names Default To Here( 1 );
dt = Current Data Table();
//k start 40, For each loop will add 2
For( k = 40, k <= 54, k += 2,
	x1 = 1;
	x2 = 8;
	y1 = 40;
	y2 = 54;
	m = (y2 - y1) / (x2 - x1);
	c = y2 - (m * x2);
	l = (k - c) / m;
	
	//l is calculated based on k\
	//Tested variable & Column x & Strobe are the column in my data set
	//If k = 40;, then the readout will be 1st bit from the right of column x;
	//If k = 40;,then the readout will be 8th bit from the right of column x;
	
	// The variable l needs to be fully parsed in the formula, so that if the 
	// value of l changes, it does not change previous formulas
	Eval(
		Substitute(
				Expr(
					dt << New Column( "Test" || Char( k ),
						Numeric,
						"Continuous",
						Format( "Best", 12 ),
						Formula(
							theVal = .;
							If( :Strobe == 1,
								If( :Tested Variable == 0 | :Tested Variable == 1,
									theVal = Num( Substr( :x, _l_ , 1 ) )
								)
							);
							theVal;
						)
					)
				),
			Expr( _l_ ), l
		)
	);
);
Jim
Sysy13
Level II

Re: Use for loop variable in a formula

Thx Jim, I think it is working but the print out will be slightly mismatch. It works well when adding ",1" behind  "_l_" in the formula. I attached my full jsl script here

Names Default To Here( 1 );
dt = Current Data Table();
//k start 40, For each loop will add 2
For( k = 40, k <= 54, k += 2,
	x1 = 1;
	x2 = 8;
	y1 = 40;
	y2 = 54;
	m = (y2 - y1) / (x2 - x1);
	c = y2 - (m * x2);
	l = (k - c) / m;
	
	//l is calculated based on k\
	//Tested variable & Column x & Strobe are the column in my data set
	//If k = 40;, then the readout will be 1st bit from the right of column x;
	//If k = 40;,then the readout will be 8th bit from the right of column x;
	
	// The variable l needs to be fully parsed in the formula, so that if the 
	// value of l changes, it does not change previous formulas
	Eval(
		Substitute(
				Expr(
					dt << New Column( "Test" || Char( k ),
						Numeric,
						"Continuous",
						Format( "Best", 12 ),
						Formula(
							theVal = .;
							If( :Strobe == 1,
								If( :Tested Variable == 0 | :Tested Variable == 1,
									theVal = Num( Substr( :x, _l_ ,1) )
								)
							);
							theVal;
						)
					)
				),
			Expr( _l_ ), l
		)
	);
);
txnelson
Super User

Re: Use for loop variable in a formula

You are correct, that is my error.  I will correct my entry.

Jim