Index exceeds the number of array elements (0).

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Chameleon 2020-11-23

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https://ww2.mathworks.cn/matlabcentral/answers/659308-index-exceeds-the-number-of-array-elements-0

评论： Chameleon 2020-12-3

The code makes a new vector of values in the second column depending on the value 1-4 in the first column, for each iteration. outt is just for testing, in reality outt would contain different values in column 2 each time.

I would like it to work for all these scenarios:

outt = [1,100;2,200;3,300;4,400];
outt = [1,100;2,200;3,300];
outt = [1,100;2,200;3,300];
outt = [1,100];

but it only works for this scenario currently:

outt = [1,100;2,200;3,300;4,400];

here is the code:

%dessa ska vara utanför för att inte skriva över
    statV_F = []
    statV_A = []
    statV_B = []
    statV_D = []
outt = [1,100;2,200;3,300;4,400]
kolumn1 = outt(:,1)
%hitta om det finns 1-4:
    no1 = find(1 == kolumn1);
    no2 = find(2 == kolumn1);
    no3 = find(3 == kolumn1);
    no4 = find(4 == kolumn1);
%plocka ut raderna:
    NO1 = outt(no1,:);
    NO2 = outt(no2,:);
    NO3 = outt(no3,:);
    NO4 = outt(no4,:);
%göra vektorer som uppdaterar varje itertion:
    statV_F = [statV_F NO3(2)]
    statV_A = [statV_A NO3(2)]
    statV_B = [statV_B NO3(2)]
    statV_D = [statV_D NO4(2)]
   

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Chameleon 2020-11-23

the outt is an example of a typical input but the second column varies each time. I want to create 4 vectors that collects the data in the second column, the numbers in column one represents a category of data (1-4). Think of column 1 as one data type and column 2 as a value for that data type. The output for each datatype 1-4 should then be [iteration1_datacategory1 iteration2_datacategory1 iteration3_datacategory1] etc

Chameleon 2020-11-23

Sometimes there is only one datatype sometimes there are two...three..or four. But I check with find since datatype 3 might be missing etc

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Answer 1

Shubham Khatri 2020-12-3

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https://ww2.mathworks.cn/matlabcentral/answers/659308-index-exceeds-the-number-of-array-elements-0#answer_564513

在 MATLAB Online 中打开

Hello,

I have rewritten the code in the form of an if else statement.

clc
clear
%defining variables
    statV_F = []
    statV_A = []
    statV_B = []
    statV_D = []
outt = [1,100;2,200;3,300;4,400]
kolumn1 = outt(:,1)
k=length(outt)
    if k==4
        %finding value
    no1 = find(1 == kolumn1);
    no2 = find(2 == kolumn1);
    no3 = find(3 == kolumn1);
    no4 = find(4 == kolumn1);
    
    %storing value
    NO1 = outt(no1,:);
    NO2 = outt(no2,:);
    NO3 = outt(no3,:);
    NO4 = outt(no4,:);
    
    %Returning results
    statV_F = [statV_F NO1(2)]
    statV_A = [statV_A NO2(2)]
    statV_B = [statV_B NO3(2)]
    statV_D = [statV_D NO4(2)]
    
    else if k==3
    no1 = find(1 == kolumn1);
    no2 = find(2 == kolumn1);
    no3 = find(3 == kolumn1);
    NO1 = outt(no1,:);
    NO2 = outt(no2,:);
    NO3 = outt(no3,:);
    statV_F = [statV_F NO1(2)]
    statV_A = [statV_A NO2(2)]
    statV_B = [statV_B NO3(2)]
   
        else if k==2
    no1 = find(1 == kolumn1);
    no2 = find(2 == kolumn1);
    NO1 = outt(no1,:);
    NO2 = outt(no2,:);
    statV_F = [statV_F NO1(2)]
    statV_A = [statV_A NO2(2)]
    
            else k==1
    no1 = find(1 == kolumn1);
    NO1 = outt(no1,:);
    statV_F = [statV_F NO1(2)]
                end
        end
    end