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Bar plot with lower values not fixed

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Todd Welti
Todd Welti am 25 Sep. 2023
Kommentiert: Star Strider am 26 Sep. 2023
I'd like to make a plot for 9 item, each having 2 bars spaced close together, showing two related values. I want to have the bottoms of each bar not fixed at 0 or at some fixed offset. i want to show the range of values for each object in other words, max and min. no outliers or whiskers or percentiles. Kind of like as shown in the picture, but with EACH of the bottoms of the bars specifiable in Y value, not fixed..

Akzeptierte Antwort

Star Strider
Star Strider am 25 Sep. 2023
Use the patch function in a loop —
x = (1 : 9).'; % Assume Column Vectors
xd = repmat([-0.1 0.1], numel(x), 1) + x
xd = 9×2
0.9000 1.1000 1.9000 2.1000 2.9000 3.1000 3.9000 4.1000 4.9000 5.1000 5.9000 6.1000 6.9000 7.1000 7.9000 8.1000 8.9000 9.1000
xos = 0.125; % Spatial Separation ('x')
y = rand(numel(x),2) % Y-Values
y = 9×2
0.4302 0.6695 0.0597 0.3789 0.8846 0.3307 0.5977 0.8536 0.3026 0.2960 0.9167 0.9977 0.7873 0.5228 0.9008 0.2920 0.9802 0.0182
VarNames = ["A","B","C","D","E","F","G","H","I"];
figure
hold on
for k = 1:numel(x)
patch([xd(k,:) flip(xd(k,:))]-xos, [-1 -1 1 1]*y(k,1)/2, 'r')
patch([xd(k,:) flip(xd(k,:))]+xos, [-1 -1 1 1]*y(k,2)/2, 'b')
end
hold off
xticks(x)
xticklabels(VarNames)
This sets the bars to be symmetric about the midpoiint. It should not be difficult to adjust the code to use other options, since that simply requires changing the y-argument to the patch calls.
.
  4 Kommentare
Todd Welti
Todd Welti am 26 Sep. 2023
Yes, i suppose the patch would be more flexible.
Star Strider
Star Strider am 26 Sep. 2023
It has its advantages, certainly.

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Weitere Antworten (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov am 25 Sep. 2023
This exercise can be done using bar(), xticks, xticklabels, xtickangle commands:
G = randi([0, 15], 9,2);
Names = {'A(x)'; 'AB(y)'; 'ABC(y)'; 'ABCD(x1)'; 'ABBA(y1)'; 'BAC(z1)'; 'DABA(x,y)'; 'DABC1(w)'; 'ABCD13(u)'};
bar(G), grid on
xticks(1:9)
xticklabels(Names)
xtickangle(45)
ylim([0, 16])
legend('ver: One', 'ver: Two')
  1 Kommentar
Todd Welti
Todd Welti am 26 Sep. 2023
Yeah, that is what I already had. The idea was to be able to also specify the minimum values, not all at 0.

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