Graph Builder: Fit of line -disable y offset?

[#1]

In GraphBuilder, for fit of line, is it possible to set the y-offset fixed to 0?

[#2]

Hi @hogi ,

No, I don't think that it is. I think that you are asking to constrain the intercept to zero. For more complex modelling options like this, you need to use Fit Y by X from the Analyze menu. From the red triangle menu for Bivariate Fit you can select Fit Special... and then you can select Constrain Intercept to 0:

Phil_Kay_0-1680079981890.png

I hope this helps,

Phil

[#3]

You can fit a model as @Phil_Kay describes, save the prediction formula, and add that column to the Y axis as a function graphics element.

[#4]

Hm, I was searching for a one-click solution to get a "1:1" correlation line *) into the Graph Builder Plot, one which goes through  0/0 and runs close to the data points **)  - I guess I will write a script which just calculates the center of gravity and draws the line.

*)
edit:
along the idea:
- 0 % -> 0%
- 100% -> 100%
For all the cases where it's 100% clear that X and Y correlate without offset,

like a conversion of scales: any jitter around the line can just be a measurement artifact (either on the X or Y axis - or on both).

 **) By the way - actually I thought that a fit with constrained Y intercept goes somewhere close to the center of gravity:
no freedom for the Intercept, 1 degree of freedom to get close to the data points.

hogi_0-1680105442245.png

I was a bit surprised to see (red line: y not constrained, green line y=0)

hogi_3-1680105610009.png

hogi_4-1680105676128.png

hogi_5-1680105793566.png

So, calculating the center of gravity and drawing the line seems to be the better approach for what I want to do ...

[#5]

When you say ""1:1" correlation line", do you want a 45-degree line i.e. y=x : if so, you can do that in Graph Builder 

right-click> customise> add a script using the Y Function template and set the function of x as x:

Y Function(x, x);

(ok, not a "one-click" solution, but very effective if you want a reference line for a calibration curve, or a reference line for actual versus predicted)

[#6]

Thanks I will use the Y function to add the line.

But it will not be y = ONE times x.

Let's say X is measured in cm and Y is measured in inch.
Same holds for measurements of the same property at different tools - or at different times during production process.
(if I know that 0 is measured as 0)

[#7]

You can create a new variable inside the platform with a formula equal to the x variable (right click on x variable, formula,, and okay), then add that as a y variable and draw a function element on the chart.  Remove that from the other graph elements and only select the transformed variable in the function. It does take a few clicks and necessitates custom y axis and chart labels:

ih_1-1680112439902.png

ih_0-1680112280954.png

Script to reproduce charts:

View more...

Names Default To Here( 1 );

dt = Open( "$Sample_data/iris.jmp" );

dt << Graph Builder(
	Transform Column( "Transform[Sepal length]", Formula( :Sepal length ) ),
	Size( 491, 506 ),
	Variables(
		X( :Sepal length ),
		Y( :Petal length ),
		Y( :"Transform[Sepal length]"n, Position( 1 ) )
	),
	Elements( Points( X, Y( 1 ), Legend( 6 ) ), Formula( X, Y( 2 ), Legend( 8 ) ) ),
	SendToReport(
		Dispatch( {"Points"}, "", OutlineBox, {Close( 0 )} ),
		Dispatch( {"Formula"}, "", OutlineBox, {Close( 0 )} )
	)
);

dt << Graph Builder(
	Transform Column( "Transform[Sepal length]", Formula( :Sepal length ) ),
	Size( 443, 462 ),
	Show Control Panel( 0 ),
	Variables(
		X( :Sepal length ),
		Y( :Petal length ),
		Y( :"Transform[Sepal length]"n, Position( 1 ) )
	),
	Elements( Points( X, Y( 1 ), Legend( 6 ) ), Formula( X, Y( 2 ), Legend( 8 ) ) ),
	SendToReport(
		Dispatch(
			{},
			"400",
			ScaleBox,
			{Legend Model(
				8,
				Level Name( 0, "1:1 Line", Item ID( "Transform[Sepal length]", 1 ) )
			)}
		),
		Dispatch(
			{},
			"graph title",
			TextEditBox,
			{Set Text( "Petal length is usually smaller than Sepal length" )}
		),
		Dispatch( {}, "Y title", TextEditBox, {Set Text( "Petal length" )} )
	)
);

[#8]

Argh, I just thought again - and remembered a wonderful micro-"lecture" that @jules gave in the DataViz breakout room at Discovery Summit Europe 2023 - how easy you can solve puzzles with regression - by showing the squared residuals in the plot via 

Open( "$SAMPLE_DATA/../Scripts/InteractiveDemos/DemonstrateRegression.jsl" );

After dragging the points and lines one can sees that my puzzles with the constrained fit were actually the ordinary thing with fit outliers - and the squared weighting of the residuals - and a fit line which definitely doesn't hit the center of gravity:

hogi_0-1680168055422.png
hogi_1-1680168078398.png