got an error message

i wrote the following code:

a.name=[];
a.age=[];
a.sex=[];
a.ID=[];
a.GPA=[];
a.name={'bahar';'maryam';'ronia';'kaihan';'vahid';'werya';'hana'};
a.age={2;11;21;18;37;38;25}
a.sex={'f'; 'f'; 'f'; 'f'; 'm'; 'm'; 'm'}
a.ID={1234;11344;11344;98764;34786;12984;45295}
a.GPA={11;12;14;17;19;20;13}
for i=1:7
        if a(i).ID==a(i+1).ID

              a(i+1).ID=[];

          end
      end

but i have got this error message:

Index exceeds matrix dimensions.

Error in student11 (line 12)
        if a(i).ID==a(i+1).ID

I dont know why? would you please help me?

[Stephen23]

a is a scalar structure, so clearly a(2), etc, do not exist.

https://www.mathworks.com/help/matlab/matlab_prog/access-data-in-a-structure-array.html

https://www.mathworks.com/help/matlab/matlab_prog/access-multiple-elements-of-a-nonscalar-struct-array.html

You will need to create the structure correctly for it to be non-scaler: read the struct help.

[Ganesh Hegade]

You can use like this. But if you want to delete the duplicate ID's after each loop then below code will not work.

 for i=1:6
        if a.ID{i} == a.ID{i+1}

              a.ID{i+1}=[];

          end
  end

Answer 1

[Sanjay Nair]

Hello Bahar,

I believe the issue here is that you have declared the variable "a" as a scalar struct that contains multiple cell-array variables. From the way you have written your for loop, it appears that what you really want is to do an array of structs containing multiple scalar variables.

As an example, I think you are trying to set up "a" so that the first element is defined as such:

a(1).name = 'bahar';
a(1).age = 2;
a(1).sex = 'f';
a(1).ID = 1234;
a(1).GPA = 11;

You may also find the following documentation from Mathworks helpful: https://www.mathworks.com/help/matlab/matlab_prog/ways-to-organize-data-in-structure-arrays.html

[bahar kh]

hello Sanjay. yes i think so. many thanks

Answer 2

[Steven Lord]

99 (or let's go with 9 to save some time) bottles of beer on the wall. 9 bottles of beer.

bottlesOfBeerOnTheWall = 1:9

Take one down, pass it around, 8 bottles of beer on the wall.

8 bottles of beer on the wall, 8 bottles of beer. Take one down, pass it around, 7 bottles of beer on the wall.

etc.

for k = 1:9
  bottlesOfBeerOnTheWall(k) = []
end

We receive an error when we try to remove the sixth bottle. Note that we have not removed bottles 1, 2, 3, 4, and 5 at that point but have removed bottles 1, 3 (the new second bottle, once bottle 1 was removed), 5 (the new third bottle after 1 and 3 are gone), 7 (the new fourth bottle), and 9 (the new fifth bottle.)

If you're iterating through an array and deleting elements, there are a couple alternatives you can use to avoid this problem.

• Don't delete as you go. Create a vector (of linear or logical indices) of elements to be deleted then delete them all after you've finished iterating through the whole array.

bottlesOfBeer = [1 2 3 3 4 5 5 6];
shouldBeDeleted = false(size(bottlesOfBeer));
for k = 2:length(bottlesOfBeer)
  if bottlesOfBeer(k-1) == bottlesOfBeer(k)
    shouldBeDeleted(k) = true;
  end
end
bottlesOfBeer(shouldBeDeleted) = []

• Start at the end and work your way forward.

bottlesOfBeerOnTheWall = 1:9
for k = 9:-1:1
  bottlesOfBeerOnTheWall(k) = []
end

• Use an approach that avoids iterating. In this case, since you want to compare the diff-erence between adjacent elements, use diff.

bottlesOfBeer = [1 2 3 3 4 5 5 6];
sameAsPrevious = [false diff(bottlesOfBeer)==0]
bottlesOfBeer(sameAsPrevious) = []

[bahar kh]

@ steven lord . it was very fascinating example . thanks alot