How can I count the number of data in this particular figure?

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smo
smo 2016 年 4 月 14 日
コメント済み: Star Strider 2016 年 4 月 15 日
Hi, Can someone please help me with this: I want to count the number of data in each of the 5'groups' as shown in the figure. Hoping to get a result as a vector with 1row 5 column and each number in the column shows the number of counts.
What's the best way of doing it?
---------------
Edit info: I have a set of signal data, which gives me a plot with 5dips. I am trying to set a threshold (say, -6E-3), and count the number of data at each of these dips.
I now get the index of these data and they looks something like this:
[620, 621,...,670,671,2323, 2324,...,2374,4023,....,4068, (etc)], and I was trying to plot them to show in here, that the data are distributed in 5 'area', and I am hoping to count the number of data in each of these area. -------------------------------------
Thank you.

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Star Strider
Star Strider 2016 年 4 月 14 日
編集済み: Star Strider 2016 年 4 月 14 日
You obviously have the original data or you could not have plotted that figure. I would use the histcounts2 function:
y = randi([0 9], 5, 30);
x = repmat([1:5]', 1, size(y,2));
figure(1)
plot(x, y, 'x')
axis([0 6 ylim])
[N, xe, ye] = histcounts2(x,y);
  2 件のコメント
smo
smo 2016 年 4 月 15 日
編集済み: smo 2016 年 4 月 15 日
sorry I didnt explain myself more clear.
I have a set of signal data, which gives me a plot with 5dips. I am trying to set a threshold (say, -6E-3), and count the number of data at each of these dips.
I now get the index of these data and they looks something like this: [620, 621,...,670,671,2323, 2324,...,2374,4023,....,4068, (etc)], and I was trying to plot them to show in here, that the data are distributed in 5 'area', and I am hoping to count the number of data in each of these area. ===========================
Thank you for the hint!! I used 'histcounts' and it works!
Star Strider
Star Strider 2016 年 4 月 15 日
My pleasure!
I had to simulate your data, so it was something of a guess on my part as to what the best approach would be.

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