Plot based on bins with binary data

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MiauMiau
MiauMiau el 24 de Nov. de 2016
Comentada: Alexandra Harkai el 25 de Nov. de 2016
Hello
I have two arrays, say array stimDuration - which contains the duration a subject has seen a stimulation - and array answers, which is binary (containing 1 or 0's) to indicate if subject could correctly identify the stimulus or not. The data could look something like that:
stimDuration = [1, 2, 3, 2, 4, 6, 8, 1, 11, 12, 9] answers = [0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0]
I want now to have n bins for the stimDuration (for instance 2 bins, one going from 1 to 6, the other from 7 to 12) as the x axis. The y axis would be the percentage of the correct answers for each bin. For instance for the bin from 1 to 6, we have a total of 7 answers, 5 of which are correct (hence the y value would be 5/7). Is there a straightforward way of doing that? Thanks
  2 comentarios
Alexandra Harkai
Alexandra Harkai el 24 de Nov. de 2016
If you have n bins, how would they be distributed exactly? It seems you want them to be of equal size ranging from the smallest to the largest stimDuration values, but it may not be what you ultimately want.
MiauMiau
MiauMiau el 24 de Nov. de 2016
yes and equal size would be ok, as I calculate a percentage, and will have enough trials such that it won't matter if one of the bins contains a bit more data samples..

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Alexandra Harkai
Alexandra Harkai el 24 de Nov. de 2016
This would do the trick if you bin them from 0 to max(stimDuration):
n = 2; % number of bins
bar(splitapply(@(x) sum(x)/size(x,2), answers, ceil(stimDuration/(max(stimDuration)/n))));
If you have R2016b you could do:
bar(splitapply(@(x) sum(x)/size(x,2), answers, discretize(stimDuration, n)));
  10 comentarios
MiauMiau
MiauMiau el 25 de Nov. de 2016
That's the same thing as you posted above though?
Alexandra Harkai
Alexandra Harkai el 25 de Nov. de 2016
Yes. It gives 2 bins as far as I can see. If you look into the documentation for discretize and linspace, you can see what they do and how you can modify the command depending on what you need.

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