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Adjusting Y-Values in Histograms

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alberto tonizzo
alberto tonizzo 2024 年 7 月 11 日
コメント済み: Umar 2024 年 7 月 12 日
Hello,
This might sound like a simple question, but how do I change the y-values of the histograms in the figure below (bottom panels)? Both the left and right histograms have bins with only 2 or 3 y-values, meaning they are binned. How can I adjust this?
The code is simply:
histogram(-xuncL(:,2), 'BinWidth', BWidth, 'Normalization', 'Probability', 'Facecolor', 'c')
Thank you!
  3 件のコメント
alberto tonizzo
alberto tonizzo 2024 年 7 月 12 日
Thank you!
Umar
Umar 2024 年 7 月 12 日
No problem Alberto, glad to help out. If you have any further questions, please let us know.

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回答 (2 件)

the cyclist
the cyclist 2024 年 7 月 11 日
Your question is not perfectly clear to me. If you don't want the values binned, I think you more likely want to use the bar function rather than the histogram function to plot them. By construction, histograms are for binning!
Plot bars with x positions where the data are, y values of one.
N = 5;
x = rand(N,1); % <-- Made up data
y = ones(N,1);
bar(x,y)
  1 件のコメント
Umar
Umar 2024 年 7 月 11 日
Hi @cyclist,
I do appreciate your contribution and helping out to resolve this problem.

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DGM
DGM 2024 年 7 月 12 日
編集済み: DGM 2024 年 7 月 12 日
The reason that the bar heights have discrete values is simply that the height represents an underlying integer count, and you have relatively few things to count compared to the number of bins (along x) that you have. Consequently, you have only a few counts in each bin. Based on your images, I'm estimating that there are only about 42 items being counted in total.
A = randi([1 100],50,1);
histogram(A, 'BinWidth', 1, 'Normalization', 'Probability', 'Facecolor', 'c')
At least that's my suspicion.

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