I have a Table with 2 different set of experiments. each experiments have 2 subexperiments. each subexperiment have 5 sets of data. How to plot a graph of each subexperiment results? I need a plot of all subexperiments.

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Peter Perkins
Peter Perkins am 18 Mai 2017

2 Stimmen

You don't need table2array.
Your example data don't seem to match either description, so I'm just guess what they look like. Also you don't say what kind of plot.
>> type = [1;1;1;1;1;2;2;2;2;2;3;3;3;3;3];
>> subexperiment = [1;2;3;4;5;1;2;3;4;5;1;2;3;4;5];
>> result = randn(15,1);
>> t = table(type,subexperiment,result)
t =
15×3 table
type subexperiment result
____ _____________ __________________
1 1 -0.204966058299775
1 2 -0.124144348216312
1 3 1.48969760778546
1 4 1.40903448980048
1 5 1.41719241342961
2 1 0.67149713360808
2 2 -1.20748692268504
2 3 0.717238651328838
2 4 1.63023528916473
2 5 0.488893770311789
3 1 1.03469300991786
3 2 0.726885133383238
3 3 -0.303440924786016
3 4 0.293871467096658
3 5 -0.787282803758638
>> i1 = (t.type == 1);
>> plot(t.subexperiment(i1),t.result(i1))
That plots all the type 1 results. Put that in a loop, perhaps using subplot. It's also pretty simple to do something like
rowfun(@plot,t,'GroupingVariable','type','NumOutputs',0)
but actually you'd need to write a two-line function that made a new figure window or paused or something, otherwise you'll only see the last plot.

4 Kommentare

Pratheek Manangi
Pratheek Manangi am 18 Mai 2017
Bearbeitet: Pratheek Manangi am 19 Mai 2017
Thanks for ur reply. But what if the elements are cell type?
Undefined operator '==' for input arguments of type 'cell'.
Pratheek Manangi
Pratheek Manangi am 22 Mai 2017
Bearbeitet: Pratheek Manangi am 22 Mai 2017
@Guillaume or @Peter perkins could u please reply
Peter Perkins
Peter Perkins am 23 Mai 2017
Create a short example of actual code that you are using, using a small sample of exactly the kind of data you have, and show exactly what you are trying to do, and where you are running into problems.
Pratheek Manangi
Pratheek Manangi am 24 Mai 2017
I figured it out. Thanks

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