How to plot 17 subplots?
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I am trying to plot a large number of subplots, but want to make sure formartting is such that it takes the most compact form (instead of having 1 row with 17 plots, 2 rows, with 13 and 14 plots, etc). So in this case, I would like 5 rows of plots with 4 plots in each row and 1 in the last row. If 16 plots are selected, then 4 rows with 4 plots, etc.
2 件のコメント
Adam
2017 年 6 月 30 日
Depending on what you are plotting you are probably better of creating your own axes and positioning them in the desired pattern than using subplot. If you don't want x and y tick labels for every plot subplot creates a lot of wasted space around the axes.
採用された回答
John BG
2017 年 6 月 29 日
Hi Ibro
let be N the amount of images, then the most compact, square-like lay-out is achieved with
L=ceil(N^.5)
for k=1:1:N
subplot(L,L,k)
plot(..) or stem(..) or any other plot type
end
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thanks in advance
John BG
1 件のコメント
John BG
2017 年 6 月 30 日
Ibro
the thing with subplot and a large amount if images is that the area for each image is reduced, therefore losing visual resolution.
One way around would be to plot 17 separate figures
for k=1:1:17
hf=figure(k)
plot(..) or stem(..) or any other plot type
hf.Position=[left down width height]
end
Usually top left corner of the screen is [0 0].
capturing the figure handle you can save further space, away from things not needed, to give more space for the images
hf=figure
hf.ToolBar='none'
Another way would be to group the figures in groups of 4, reducing the amount of needed figures to
N=17
for k=1:1:floor(N/4)
figure(k)
for s=1:1:4
subplot(2,2,s)
plot(..) or stem(..) or any other plot type
end
end
for k=(N-floor(N/4)):1:N
figure(k)
for s=1:1:(N-floor(N/4))
subplot(2,2,s)
plot(..) or stem(..) or any other plot type
end
end
regards
John BG
その他の回答 (1 件)
Jan
2017 年 6 月 30 日
編集済み: Jan
2017 年 6 月 30 日
This adds a new row only if required:
N = 17;
figure;
nS = sqrt(N);
nCol = ceil(nS);
nRow = nCol - (nCol * nCol - N > nCol - 1);
for k = 1:N
subplot(nRow, nCol, k);
plot(1:10, rand(1, 10));
end
Comparison to the pure square solution: Saving an empty row helps to get the "most compact form", which is not a 5x5 arrangement:
See also:
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