Vectors Must be the Same Length

Question

Rafael 2017-12-10

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https://ww2.mathworks.cn/matlabcentral/answers/372119-vectors-must-be-the-same-length

评论： Rafael 2017-12-12

I want to plot this but the problem is that I need to somehow have the vectors be the same length.

Year=input('Enter the year ');
Hour=input('Enter the hour ');
Index=find(time(:,1)==Year & time(:,16)==Hour);
Max=max(time(Index,2:13))
Mean=mean(mean(time(Index,2:13)));
X=(1:365)';
Y=Max';
y=Mean*ones(1,365);
cla
hold on
plot(X,y,'b') %This plots the mean value with x.
plot(X,Y)

The objective is to plot max and mean when x goes from 1 to 365. Mean has to be a single line because it's a single value. Max has to be 12x365 and it should return only one value for each column but the problem is that when I run that code it gives me the error.

    Day=input('Enter the year ');
    Year=input('Enter the hour ');
    Index=find(time(:,1)==Year & time(:,16)==Hour);
    Max=max(time(Index,2:13),1)
    Mean=mean(mean(time(Index,2:13)));
    X=(1:365)';
    Y=Max';
    y=Mean*ones(1,365);
    cla
    hold on
    plot(X,y,'b') %This plots the mean value with x.
    plot(X,Y)

If I run the above code it works fine. It doesn't return any error. Now the problem with the second one is that it returns more than one value for each column and the plot shows different colors in the graph. I want it to show only one color and hopefully run the code which returns one value for each column for max.

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Walter Roberson 2017-12-11

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When you use

Max=max(time(Index,2:13),[],1)

you will get one result for each entry in Index. Then

    X=(1:365)';
    Y=Max';
    plot(X, Y)

assumes that Y is 365 entries, which assumes that your Index happened to match exactly 365 locations every time. That seems unlikely to be the case as your Index is found by selecting on Day and Hour, so it seems unlikely that you would just happen to have 365 years to match the Day and Hour to.

If you have one year of data taken once per hour, it would make more sense for Index to be based upon Hour alone, giving you one result per day.

Question: what are you doing about leap year?

Question: Are the times local timezone or UTC? Because if they are local then you have to worry about any daylight savings time in effect.

Rafael 2017-12-11

Sorry, Days is supposed to be Years. I fixed it.

I have to plot X that is 1:365 (which is days) vs Y has to be the Max from maximum value of each day for the inputted year. That is supposed to give me something like this,

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Answer 1

Michal Dobai 2017-12-12

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https://ww2.mathworks.cn/matlabcentral/answers/372119-vectors-must-be-the-same-length#answer_295831

编辑：Michal Dobai 2017-12-12

在 MATLAB Online 中打开

It's a lot more clear to me now. Or at least I hope so :) (reference to your comment )

Let's say that you have these input data:

%Year  < < <               this is your data            > > >  hour (day)
%-----------------------------------------------------------------------
time = ...
[2016  76  78  87  33  33  81  58  49   0  11  81  54   5  82   6;  % 1
54  98  26   8  94  42  96  72  37  91  70  57  42  37   7;  % 1
59  22  10  98  95  49  92  39  64   1  54   5  25  74   6;  % 2
95  20  79  12  33  67  21  14  23  58  71  61  19  16   7;  % 2
84  19  57  45  89  25   6  59   1  81  30  66  60  74   6;  % 3
49   4  50  36  22  30  91  46  69   2  78  26  26  83   7;  % 3
36   8  60  62  91  38  30  54  56  56  99  87  80   2   6;  % 4
78  75  51  34  37  50  68  16  99  95  87  99  64  43   7;  % 4
68  56  46  53  80  10   2  93  61  81  79  41  40  25   6;  % 1
70  21  34  22  17   8  62  64   0  56  85  16  25  84   7;  % 1
12  86  63  95  11   9   8  53  99  15  51  84  71  90   6;  % 2
28  18  36  14  76  86  54  44  40   7  90  79  26  41   7;  % 2
97  12  63  57  89  37  28  14  89  21  45  69  34  20   6;  % 3
86  82   8  57  91  28  64  18   6  33  61  58  47  45   7;  % 3
79  11  99  30  56  96   8  37  14  28  25  74   7  34   6;  % 4
19  97  70  93  64  29  85  34  78  97  16  92  72  87   7]; % 4

In this example I will assume that year has only four days and that you have somehow collected data for hours 6 and 7 for each day.

Now, when you input 2017 as year and 6 as hour, this

Index=find(time(:,1)==Year & time(:,16)==Hour);

will select exactly four rows from time matrix. (well, at least in my case, because I 'collected' values for every single day of the year 2017 at 6'o clock).

So now we can create selection of size 4x12 (in your case 365x12) like you already did: time(Index,2:13) Maximum of each day (each row of this selection) can be calculated like this:

Max = max(time(Index,2:13),[],2);

Why 2? From Documentation for function max:

M = max(A,[],dim) returns the largest elements along dimension dim. For example, if A is a matrix, then max(A,[],2) is a column vector containing the maximum value of each row.

Now max will be the column vector of size 4 (one value for each day of the year). X is also column vector of size 4, so we can plot it without any problem. (and without error Vectors Must be the Same Length)

Here is the whole code:

      Year=input('Enter the year ');
      Hour=input('Enter the hour ');
      Index=find(time(:,1)==Year & time(:,16)==Hour);
      Max = max(time(Index,2:13),[],2);
      Mean = mean(mean(time(Index,2:13)));
      X=(1:4)';          % In this example - year has only 4 days
      Y=Max';
      y=Mean*ones(1,4);  % In this example - year has only 4 days
      cla
      hold on
      plot(X,y,'b') %This plots the mean value with x.
      plot(X,Y)
      ylim([0,100]);

And here is the plot this code produce for input values Year=2017 and Hour=6:

Now, as Walter Roberson already mentioned in his comment , there can be cases when your code doesn't select exactly 365 rows. You should consider these cases and handle them properly.

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Michal Dobai 2017-12-12

...Plot the hourly maximum and hourly mean time (minutes)...

Hourly means hourly. So you should plot mean time the same way as maximum value.

Now, couple more things:

First of all, I'm glad that we managed to solve your problem, BUT. It took me a really long time to fully understand what you are trying to achieve. I understand, that this is some kind of homework or maybe your school project, and that you was maybe trying to avoid post here anything else other than your code, but next time try to give an example of input and output that you are hoping for. It doesn't have to be exact same data from your assignment, it can be and should be something simple that sufficiently demonstrate your intentions. If your program can't plot anything (because of error), you can always open paint and draw how should your plot look ;)
Try to fully understand basic concept of MATLAB Matrices and Arrays. Use documentation. MATLAB has awesome documentation with lots of examples you can just copy-paste and try. It can save you - and everyone who is starting to learn MATLAB - a lot of time.
And last thing. I personally think Cody can help beginners a lot. "Cody™ is a MATLAB Central game that challenges and expands your knowledge of MATLAB and Simulink®." You can try it. There are some really fun and interesting problems from complete basics to advanced.

That's all. If something is still not clear to you, you can ask here, I'll try to explain it to you in further detail. If you tried this solution, and it works as expected, you could consider to accept this answer :)

Michal Dobai 2017-12-12

在 MATLAB Online 中打开

I already written this comment, so I'll just post it, maybe it will help the others...

And as for the ' line 117'. It says: Error in mean (line 117)

mean is MATLAB function. It has it's own code stored in it's own file. In fact, when you click on text (line 117) in your command window, MATLAB opens it for you.

Whole error info looks like this:

Error using sum
Invalid option. Option must be 'double', 'native', 
'default', 'omitnan' or 'includenan'.
Error in mean (line 117)
        y = sum(x, dim, flag)/size(x,dim);
Error in something (line 41)    % 'something' is name of my file
      Mean = mean(time(Index,2:13),[],2);

MATLAB is trying o tell you:

Error using sum - I was trying to sum elements of this matrix for you, but I couldn't do it, because: Invalid option. Option must be 'double', 'native', 'default', 'omitnan' or 'includenan'. (you don't have to fully understand what happened here, this error happened INSIDE mean code)
Error in mean (line 117) - MATLAB is saying: 'It happened here!, Right on this line (117 :) )', when I was trying to do this: y = sum(x, dim, flag)/size(x,dim);. This is NOT your code. It's MATLAB's. ...and there is nothing wrong with it. So that means error has to be caused by something else.
Error in something (line 41) - THIS is your code. (or mine, to be accurate). It means that in your code, you tried to do this: Mean = mean(time(Index,2:13),[],2);, MATLAB tried to execute this command, called function mean(), and got an error on line 117 (in that function), where it was trying to execute sum.

Michal Dobai 2017-12-12

I'm glad you managed to solve your problem :)

Rafael 2017-12-12

Hey, thank you! You are a master!

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Answer 2

Jos (10584) 2017-12-11

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在 MATLAB Online 中打开

Is this what you want?

X = 1:50 ;
Y = rand(size(X)) ;
Ymean = mean(Y(:)) ; % get single value; prefered over mean(mean(..))
plot(X, Y, 'ro') ; 
hold on ;
plot(X([1 end]), Ymean([1 1]), 'b-') ;

There are also utilities on the File Exchange to plot horizontal and vertical lines at a specific value, like my own GRIDXY function, available here: https://uk.mathworks.com/matlabcentral/fileexchange/9973-gridxy--v2-2-feb-2008-

plot(X, Y, 'ro')
gridxy([], mean(Y(:)))

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Rafael 2017-12-11

There has to be two plot.

X has to go from 1:365
Y is the Max
The other "y" is Mean and that plot is working. The one not working is with the Max.

Rafael 2017-12-12

在 MATLAB Online 中打开

Yup, I remember now. The mean has to look like Max. I can't be a straight line. I have to set Mean like this,

Mean=mean(data_2(Index,2:13))
y=Mean.*ones(1,365);

The problem is that it returns Matrix dimensions must agree.

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Vectors Must be the Same Length

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