Separate a wave into two new waves

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Tue, 02/10/2015 - 09:59 am

I would like to separate the black wave on the attached graph into two new waves to display them using different colors and Y-offsets. How can I do it?

Thanks in advance
I assume you refer to the black wave at the bottom in your image (and not the one with the red data points).
What would be the criterion to split the waves?

Maybe this helps you already:
displayhelptopic "Subrange Display Syntax"


ModifyGraph offset ={x,y} is used for offsets.
HJ

February 10, 2015 at 10:14 am - Permalink

Yes, it's the black one on the bottom.
As it can be seen on it, it has only two values (let's say 2 or 1). So I want to separate them in two new waves, one with the "2" values and the other one with the "1" values in order to plot as I explained in my first message. But they need to keep their relative positions with respect to the "2 theta" wave, this is important.

February 10, 2015 at 10:27 am - Permalink

In this case you have to duplicate the wave (I call it "Int") and assign only the relevant values for "High"(2) and "Low"(1):
duplicate Int, High,Low
High= (Int==2) ? 2 : NaN
Low = (Int==1) ? 1 : NaN


NaN might be replaced with zero.
And, of course, these two waves have to be appended to the graph...
HJ

February 10, 2015 at 10:34 am - Permalink

One way to colorize on a point-by-point basis would be to use Color as f(z). Since it appears the wave is binary, simply use the Y wave as the color wave. Pick a color table that has the colors you want, and adjust the parameters of the colorizing to get the particular colors you want out of the color table.

John Weeks
WaveMetrics, Inc.
support@wavemetrics.com

February 10, 2015 at 12:26 pm - Permalink

HJDrescher wrote:
In this case you have to duplicate the wave (I call it "Int") and assign only the relevant values for "High"(2) and "Low"(1):
duplicate Int, High,Low
High= (Int==2) ? 2 : NaN
Low = (Int==1) ? 1 : NaN


NaN might be replaced with zero.
And, of course, these two waves have to be appended to the graph...
HJ



Thank you very much. It was what I needed :)

February 11, 2015 at 07:00 am - Permalink