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got an error message

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bahar kh
bahar kh 2017-8-2
评论： bahar kh 2017-8-4
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?
  2 个评论
Stephen23
Stephen23 2017-8-2
编辑：Stephen23 2017-8-2
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
Ganesh Hegade 2017-8-2
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

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采纳的回答

Sanjay Nair
Sanjay Nair 2017-8-2
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

更多回答（1 个）

Steven Lord
Steven Lord 2017-8-2
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) = []
  1 个评论
bahar kh
bahar kh 2017-8-4
编辑：bahar kh 2017-8-4
@ steven lord . it was very fascinating example . thanks alot

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