how to add labels to the elements of a vector.

hello, everything okay?

Is it possible to label the elements of a vector? To later know in what position is it at a certain moment?

for exemple:

A=[4, 6 ,8 ,14, 90 ] (A contains the vector information set let's say A1 and A2, when A1=[4, 6 ,90] and A2=[8, 14])

Assign the position indices of each element of the vector. LabelA1=[1, 2,5] LabelA2=[3 ,5] LabelA=[ 1, 2,3,4,5]

if I do determine tasks with the vector and it looks like this A=[0 6 8 0 9 ] in another word A(1)=0 and A(4)=0

so A=[ 6 8 9 ] . LabelA1(1)=0 and LabelA2(1)=0 implies that LabelA1=[2,5] LabelA2=[5]

I need to find from the labels that of A1 A2 the position it occupies after updating vector A.

thank you.

Answer 1

[Sindar]

One option: store things in a table.

This particular setup may not be the most efficient, but hopefully it helps you start:

% build a table with columns for the original data, and whether it is part of A1 and/or A2
A_table = table([4; 6; 8; 14; 90],[true;true;false;false;true],[false;false;true;true;false],'VariableNames',["values","A1","A2"])
A_table = 
    values     A1       A2  
    ______    _____    _____
       4      true     false
       6      true     false
       8      false    true 
      14      false    true 
      90      true     false
% access "A" like this
A=A_table{:,1}
A = 5×1    
     4
     6
     8
    14
    90
% and A1 or A2 like this:
A1=A_table{A_table{:,"A1"},1}
A1 = 3×1    
     4
     6
    90
% label A1 can be calculated
LabelA1 = find(A_table{:,"A1"} & A_table{:,1}~=0)
LabelA1 = 3×1    
     1
     2
     5
% update A (silly example algorithm to replace)
A_table{:,1} = A_table{:,1}.^0.*[0 6 8 0 9 ]'
A_table = 
    values     A1       A2  
    ______    _____    _____
      0       true     false
      6       true     false
      8       false    true 
      0       false    true 
      9       true     false
% re-extract A, etc. as before (you could create functions for this)
A=A_table{:,1}
A = 3×1    
     6
     8
     9
A1=A_table{A_table{:,"A1"},1}
A1 = 2×1    
     6
     9
LabelA1 = find(A_table{:,"A1"} & A_table{:,1}~=0)
LabelA1 = 2×1    
     2
     5

An idea that might be useful, but I haven't coded/checked: add columns for LabelA1,LabelA2. These would contain either the original index or NaN. This would allow you to actually discard rows from the table

    values     A1       A2      LabelA1
    ______    _____    _____    _____
       4      true     false    1
       6      true     false    2
       8      false    true     nan
      14      false    true     nan
      90      true     false    5

[Fidele Adanvo]

Thanks a lot. But I think that this solution will make the simulation more delayed. Against another way to do it.

I will share them here in case someone needs it.

A=[4, 6 ,8 ,14, 90 ]  ; A1=[4, 6 ,90] ; A2=[8, 14]  
Lbel=1:numel(A) % is a vector from 1 to the size of A containing the position of each element
Lbel(A(1))=0; Lbel(A(4))=0; Lbel(find(Lbel==0))=[]; 
A1(1)=0; A2(1)=0; A(A1(1))=0; A(A2(1))=0;  
 A(find(A==0))=[]; A1(find(A1==0))=[];A2(find(A2==0))=[];
 [~,A1]=ismember(A1,Lbel); [~,A2]=ismember(A2,Lbel)
%ismember  it does not work in all versions of matlab

[Sindar]

Some notes

• you don't need to use "find" when indexing, the logical vector will work fine
• Using A(i) as an index doesn't do what I think you want

A=[4, 6 ,8 ,14, 90 ]  ; 
A1=[4, 6 ,90] ; 
A2=[8, 14]  
Lbel=1:numel(A) % is a vector from 1 to the size of A containing the position of each element
% this sets the 4th (A(1)) element of Lbel to zero
Lbel(A(1))=0; 
% this sets the 14th (A(4)) element of Lbel to zero, 
% as well as any new elements needed to reach this point
Lbel(A(4))=0; 
% this removes all zeroes, including the extra ones created by the last line
Lbel(Lbel==0)=[]; 

% set 1st element of A1, A2 to zero
A1(1)=0; 
A2(1)=0; 
% set the 0th element of A to zero... this should throw an error
A(A1(1))=0; 
A(A2(1))=0;  


A(A==0)=[];
A1(A1==0)=[];
A2(A2==0)=[];
[~,A1]=ismember(A1,Lbel);
[~,A2]=ismember(A2,Lbel);

[Fidele Adanvo]

Thanks