Error using Bar, X must be same Length as Y.
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I am trying to plot on a bar chart through app designer and i'm getting the following error: Error using bar (line 172)
X must be same length as Y.
My code is as below:
bar(app.UIAxesPositionBar, T_BarChart.UndlyDate, T_BarChart{:,{'TotalPosition'}}, 0.8,'FaceAlpha',0.5);
hold(app.UIAxesPositionBar,'on');
bar(app.UIAxesPositionBar, T_BarChart.UndlyDate, T_BarChart{:,{'FuturesPosition','DeltaEquivPosition'}}, 0.8,'FaceAlpha', 0.5);
The second call to bar throws the error, when i inspect the table i'm using you can see its 1x4, the first call to bar works just fine.
You can see the table contents/variables below:
K>> bar(app.UIAxesPositionBar, T_BarChart.UndlyDate, T_BarChart{:,{'FuturesPosition','DeltaEquivPosition'}}, 0.8,'FaceAlpha', 0.5);
Error using bar (line 172)
X must be same length as Y.
K>> T_BarChart.UndlyDate
ans =
datetime
01-Dec-2019
K>> T_BarChart{:,{'FuturesPosition','DeltaEquivPosition'}}
ans =
1.0e+02 *
0 -1.887165000000000
K>> T_BarChart
T_BarChart =
1×4 table
UndlyDate FuturesPosition DeltaEquivPosition TotalPosition
___________ _______________ __________________ _____________
01-Dec-2019 0 -188.7165 -188.7165
Any ideas on why i am getting the "must be the same length" error?
4 Kommentare
Nyaz Jamel
am 25 Mai 2020
1500
Error using stem (line 43)
X must be same length as Y.
Error in untitled4 (line 6) stem(n,x) Please help My common:
n=input('input length of the sine seq'); L= input('up sampling factor'); fi= input('input signal freq'); t=0:n-1; x=sin(2*pi*fi*t); y=zeros(1,L*length(x)); y=x([1:L:length(x)]); subplot(2,1,1); stem(n,x) title('input sequence'); xlabel('time'); ylabel('amplitude'); subplot(2,1,2); stem(n,y(1:length(x))) title(['output sequence',num2str(L)]); xlabel('time'); ylabel('amplitude');
Nyaz Jamel
am 25 Mai 2020
Help me bro
Nyaz Jamel
am 25 Mai 2020
Note : Sample value=50, upsampling=2 input signal freq=1500
Adam Danz
am 25 Mai 2020
With
stem(app.UIAxes, X, Y)
length(X) must equal length(Y). According to your error message, this is not the case for your values. It makes sense, right? X and Y are pairs of values so each X needs to have a Y value and vise versa.
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Weitere Antworten (2)
Matt Glowacki
am 24 Jun. 2019
1 Kommentar
Adam Danz
am 25 Jun. 2019
Please reply in the comment section under my answer.
Matt Glowacki
am 25 Jun. 2019
1 Kommentar
Adam Danz
am 25 Jun. 2019
Please reply in the comment section under my answer.
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