04-04-2011 01:57 PM
Hi All,
I am fairly new to LabView, and I have read several threads that are similar to my problem, but I do not know how to implement it yet. I basically have a large array that stores raw data read from several text files. This large array's size would vary based on the number of text files generated by a data acquisition system. I would like to plot each row of the array on a new plot. Instead of wiring it manually, is there an easier way around?
Thank you,
Lynniz
Solved! Go to Solution.
04-04-2011 03:07 PM
Lynniz,
When you say plot are you talking about a graph? As you can see, it affects how you assemble your data....
Eric
04-05-2011 05:25 AM
Yes. I would like to have each row of the arry on a separate graph. The number of graphs will change based on the size of the array. I am thinking about maybe a stacked chart or something.
04-05-2011 06:39 AM
Hi Lynniz,
If you connect 2D array to a Waveform chart then each row will be treated as individual plot.
So what you have to do is, combine data from text files, build 2 D array and use Waveform graph.
P.S. I assume element size in each row will be same, otherwise extra elemnet will be considered as 0.
04-05-2011 08:12 AM
Thanks for the response Gak. I forgot to mention that I also have different time segments to graph with each row of the array. I am currently bundling the time and each row of the array manually to a XY graph. And to make things more complicated, element size in each row is different. I just found this out, so I need to find a way around that too.
04-05-2011 10:05 AM
Lynniz,
If you need a separate graph for each plot then can you at least determine beforehand what the maximum number of plots could be and show/hide your graphs programmatically based on the data in. All the graphs could be in a subVI that appears separate from your main application window or you could use a subpanel.
I'm just brainstorming here. There might be a more elegant way to do this, but showing/hiding graphs that already exist is the only thing I can think of right now.
Eric
04-08-2011 08:18 AM
Thanks for the reply. I am going to try your approach.