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I have on matrix such as A=[ 2.8866 2.5708 2.9381 2.6234 2.7873 2.8535 3.5818 4.1817 3.6535 4.6312 4.2028 4.4109 3.6008 4.4279 3.8003 3.7995 4.3140 4.4061 5.2382 4.5073 4.2002 3.9283 3.1146 3.0891]. A is one matrix 24*1.each element of matrix is value of losses for each hour a day.for example for hour between 0 ,1 losses is 2.8866 for hour between 1 ,2 the losses is 2.5708. how can i plot losses for 24 hours?
thanks

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Star Strider
Star Strider 2019-8-19

0 个投票

Try this:
A=[2.8866 2.5708 2.9381 2.6234 2.7873 2.8535 3.5818 4.1817 3.6535 4.6312 4.2028 4.4109 3.6008 4.4279 3.8003 3.7995 4.3140 4.4061 5.2382 4.5073 4.2002 3.9283 3.1146 3.0891];
Hours = 0.5:23.5;
figure
stairs(Hours, A)
axis([0.5 23.5 0 6])
set(gca, 'XTick',Hours, 'XTickLabel',(0:23))
grid
xlabel('Hours')
ylabel('Loss')

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abtin irani
abtin irani 2019-8-19
thanks. it is perfect but has one problem. it doesn't cover for hour 24 . it shows until 23.can you help me?
Star Strider
Star Strider 2019-8-19
My pleasure.
Try this:
A=[ 2.8866 2.5708 2.9381 2.6234 2.7873 2.8535 3.5818 4.1817 3.6535 4.6312 4.2028 4.4109 3.6008 4.4279 3.8003 3.7995 4.3140 4.4061 5.2382 4.5073 4.2002 3.9283 3.1146 3.0891];
Hours = 0.5:23.5;
figure
stairs(Hours, A)
axis([0.5 23.5 0 6])
xtl = sprintfc('%.0f', linspace(0, 24, 24));
set(gca, 'XTick',Hours, 'XTickLabel',xtl)
grid
xlabel('Hours')
ylabel('Loss')
That should do what you want.
abtin irani
abtin irani 2019-8-19
thanks.but last value that it shows is element of 23 matrix.in reality for hour=24 the losses shoud be the last element of A.
THIS IS my figure.
please help me.
Star Strider
Star Strider 2019-8-19
I have no idea what you want.
Try this:
figure
stairs(Hours, A)
axis([0.5 23.5 0 6])
xtl = sprintfc('%.0f', linspace(1, 24, 24));
set(gca, 'XTick',Hours, 'XTickLabel',xtl)
grid
xlabel('Hours')
ylabel('Loss')
abtin irani
abtin irani 2019-8-19
no it doesn't work. is there any other solution?
Star Strider
Star Strider 2019-8-19
I have no idea what ‘doesn’t work’ means.
Try this:
figure
stairs([0:24], [A A(end)])
axis([0 24 0 6])
xtl = sprintfc('%.0f', linspace(0, 24, 25));
set(gca, 'XTick',(0:24), 'XTickLabel',xtl)
grid
xlabel('Hours')
ylabel('Loss')
abtin irani
abtin irani 2019-8-19
first and second answer were better . they just didn't have the value of last element of A.now each 1.5 hour changes losses.in reality each hour(1 hour) should change losses.because each element of matrix A is for one hour.the first answer just didn't support for hour 24 .the last element of A wasn't in figure.thanks
abtin irani
abtin irani 2019-8-19
the last answer worked.thanks million
Star Strider
Star Strider 2019-8-19
My pleasure.
Slight variation:
A=[ 2.8866 2.5708 2.9381 2.6234 2.7873 2.8535 3.5818 4.1817 3.6535 4.6312 4.2028 4.4109 3.6008 4.4279 3.8003 3.7995 4.3140 4.4061 5.2382 4.5073 4.2002 3.9283 3.1146 3.0891];
Hours = 1:24;
figure
stairs([0:24], [A A(end)])
axis([0 24 0 6])
xtl = sprintfc('%.0f', linspace(0, 24, 25));
set(gca, 'XTick',(0:24), 'XTickLabel',xtl)
grid
xlabel('Hours')
ylabel('Loss')
I am out of ideas.
abtin irani
abtin irani 2019-8-19
excuse me . thanks for your answerS . i have five matrix .how can i have five matrix in one figure. each matrix should be in specific color.for example first element of each matrix is for first hour, second element of each matrix is for second hour, .....
million thanks.
A=[2.8968 2.6018 3.1997 3.0090 2.8787 2.6936 3.0979 3.4445 4.4046 3.7958 4.4402 3.9971 3.6959 3.5970 3.8980 4.3530 5.0837 4.5077 5.1276 4.6911 4.2197 3.2374 3.0090 2.7497];
B=[ 2.8895 2.5986 3.2015 3.0089 2.8786 2.6934 3.0965 3.4430 4.4044 3.7932 4.4384 3.9998 3.6971 3.5978 3.8989 4.3542 5.0832 4.5089 5.1280 4.6913 4.2191 3.2375 3.0102 2.7452];
C=[2.8896 2.6003 3.2002 3.0082 2.8791 2.6932 3.0969 3.4459 4.4058 3.7962 4.4436 4.0002 3.6949 3.5975 3.9018 4.3528 5.0833 4.5078 5.1273 4.6919 4.2195 3.2378 3.0097 2.7485];
D=[ 2.8886 2.5998 3.1993 3.0092 2.8799 2.6936 3.0966 3.4464 4.4055 3.7957 4.4396 3.9976 3.6949 3.5977 3.8992 4.3519 5.0834 4.5085 5.1280 4.6922 4.2198 3.2378 3.0101 2.7482];
E=[ 2.8869 2.5997 3.2001 3.0084 2.8791 2.6932 3.0965 3.4453 4.4059 3.7949 4.4380 4.0000 3.6952 3.5987 3.9014 4.3560 5.0837 4.5086 5.1281 4.6898 4.2194 3.2376 3.0100 2.7479];
Star Strider
Star Strider 2019-8-19
As always, my pleasure.
Assuming that my last plot is what you want, this will plot your new vectors with it:
A=[2.8968 2.6018 3.1997 3.0090 2.8787 2.6936 3.0979 3.4445 4.4046 3.7958 4.4402 3.9971 3.6959 3.5970 3.8980 4.3530 5.0837 4.5077 5.1276 4.6911 4.2197 3.2374 3.0090 2.7497];
B=[ 2.8895 2.5986 3.2015 3.0089 2.8786 2.6934 3.0965 3.4430 4.4044 3.7932 4.4384 3.9998 3.6971 3.5978 3.8989 4.3542 5.0832 4.5089 5.1280 4.6913 4.2191 3.2375 3.0102 2.7452];
C=[2.8896 2.6003 3.2002 3.0082 2.8791 2.6932 3.0969 3.4459 4.4058 3.7962 4.4436 4.0002 3.6949 3.5975 3.9018 4.3528 5.0833 4.5078 5.1273 4.6919 4.2195 3.2378 3.0097 2.7485];
D=[ 2.8886 2.5998 3.1993 3.0092 2.8799 2.6936 3.0966 3.4464 4.4055 3.7957 4.4396 3.9976 3.6949 3.5977 3.8992 4.3519 5.0834 4.5085 5.1280 4.6922 4.2198 3.2378 3.0101 2.7482];
E=[ 2.8869 2.5997 3.2001 3.0084 2.8791 2.6932 3.0965 3.4453 4.4059 3.7949 4.4380 4.0000 3.6952 3.5987 3.9014 4.3560 5.0837 4.5086 5.1281 4.6898 4.2194 3.2376 3.0100 2.7479];
Mtx = [A; B; C; D; E];
Hours = 1:24;
figure
hold all
for k = 1:size(Mtx,1)
stairs((0:24), [Mtx(k,:) Mtx(k,end)])
end
hold off
axis([0 24 0 6])
xtl = sprintfc('%.0f', linspace(0, 24, 25));
set(gca, 'XTick',(0:24), 'XTickLabel',xtl)
grid
xlabel('Hours')
ylabel('Loss')
legend('A','B','C','D','E', 'Location','S')
The stairs function apparenlty only wants vectors, so the loop is necessary.
abtin irani
abtin irani 2019-8-20
it is perfect. but it has a problem.beacause the losses of each matrix are near together. the lines overlap.how can i increase accuracy the loss axis.in order to i identify different lines .
thanks man.
Star Strider
Star Strider 2019-8-20
The values are too close together to distinguish them. A plot3 plot is an option, however doing a stairs-type plot with it would probably not be much better, and would definitely be a challenge to write.
Try a bar3 plot instead, to separate them spatially:
figure
bh = bar3(Mtx);
for k = 1:length(bh)
bh(k).CData = bh(k).ZData;
bh(k).FaceColor = 'interp';
end
xlim([0.5 24.5])
zlim([floor(min(Mtx(:))) ceil(max(Mtx(:)))])
xtl = sprintfc('%.0f', linspace(0, 24, 25));
set(gca, 'XTick',(0:24), 'XTickLabel',xtl, 'YTick',(1:5), 'YTickLabel',{'A','B','C','D','E'})
grid
xlabel('Hours')
zlabel('Loss')
grid on
Choose whatever colormap you want. This uses the default.
abtin irani
abtin irani 2019-8-20
thanks you very very much sir.
Star Strider
Star Strider 2019-8-20
As always, my pleasure.

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