Modifying a structure array
Show older comments
Hi. I am working with a structure array, and I want to modify the structure based on a condition on the field entries.
For example,
Here is the input,
S(1).f1=[1:100] and S(2).f2=[2:101];
S(2).f1=[40:120] and S(2).f2=[60:140]
S(3).f1=...... and S(3).f2=.....
S(4).f1=.... and S(4).f2=.....
.
.
S(i).f1=.... and S(i).f2=....
Can anyone suggest me some smart way to do this.
Conditions on input Output
S(1).f1>=0 and S(1).f1<=50 then ------> S(1).f1=[1:50] and S(1).f2=[2:51]
S(1).f1>50 and S(1).f1<=100 then ------>S(2).f1=[51:100] and S(2).f2=[52:101]
.... so on
Now, the same process for original/input S(2).f1, S(3).f1 .... S(i).f1 to re-form the structure.
After, modification the final step is to delete S(i) if the size of S(i).f1 is < 5.
Accepted Answer
More Answers (1)
Since you don't provide any indication of how you want this to work for struct array input, you'll have to modify this code yourself.
S=struct;
S(1).f1=[2,3,4,5,6,7,8,10,14,16,18,20];
S(1).f2=[3,6,9,12,15,18,21,24,27,30,33,36];
L=S(1).f1<=10;
if sum(L)>=5
S(2).f1=S(1).f1(L);
S(2).f2=S(1).f2(L);
end
S(1).f1(L)=[];S(1).f2(L)=[];
Edit:
%create input struct
S=struct;
S(1).f1=5*[2,3,4,5,6,7,8,10,14,16,18,20];
S(1).f2=5*[3,6,9,12,15,18,21,24,27,30,33,36];
S(2).f1=40:120;S(2).f2=60:140;
S(3).f1=20:100;S(3).f2=40:120;
%create the cell array that will hold the struct elements
c=cell(2,numel(S));
for n=1:numel(S)
c{1,n}=S(n);
L=S(n).f1<=50;
if sum(L)>=5
c{2,n}.f1=S(n).f1(L);
c{2,n}.f2=S(n).f2(L);
end
c{1,n}.f1(L)=[];c{1,n}.f2(L)=[];
end
%reshape back to a struct
S=[c{:}];
4 Comments
Rik
on 12 Aug 2019
That is not working because you use && instead of &. The single operator should be used for arrays, the double operator for scalars.
For numbered fields you could use a loop like this:
for n=1:5
S.(sprintf('f%d',n)) = rand;
end
Rik
on 12 Aug 2019
The .() syntax works for char array inputs (and scalar strings).
Categories
Find more on Structures in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!