Adding element within vector without overwriting the existing value

Is there a way of inserting values at points within an existing vector/array without overwriting the existing value?

So I have a list of values and I have a vector with positions in this list, is there a way of inserting the new values at these positions?

For example, is there a way of taking this list

expitemorder =

    'Strong_Alt_Prime2.png'
    'Strong_Alt_Target2.png'
    'Strong_Alt_Prime1.png'
    'Strong_Alt_Target1.png'
    'Strong_Alt_Prime3.png'
    'Strong_Alt_Target3.png'
    'Strong_Alt_Prime4.png'
    'Strong_Alt_Target7.png'

and inserting new/additional values without overwriting the existing values to get something like this

expitemorder =

    'Strong_Alt_Prime2.png'
    'Strong_Alt_Target2.png'
    'NEW VALUE'
    'Strong_Alt_Prime1.png'
    'Strong_Alt_Target1.png'
    'Strong_Alt_Prime3.png'
    'Strong_Alt_Target3.png'
    'NEW VALUE'
    'Strong_Alt_Prime4.png'
    'Strong_Alt_Target7.png'

So a way of adding new rows at predetermined positions within the list? Not just vertically concatenating the two.

Thanks

Answer 1

[Geoff Hayes]

Alice - short of defining a special class to handle this sort of insertion into a list (i.e a linked list), I think that concatenation is one of the ways to go about creating your new list. So if

 expitemorder = {...

    'Strong_Alt_Prime2.png'
    'Strong_Alt_Target2.png'
    'Strong_Alt_Prime1.png'
    'Strong_Alt_Target1.png'
    'Strong_Alt_Prime3.png'
    'Strong_Alt_Target3.png'
    'Strong_Alt_Prime4.png'
    'Strong_Alt_Target7.png'};

and you wish to insert a new string into the third position, then the code to do this would be

 expitemorder = [expitemorder(1:2) ; 'NEW_VALUE' ; expitemorder(3:end)]

which produces

 expitemorder = 

    'Strong_Alt_Prime2.png'
    'Strong_Alt_Target2.png'
    'NEW_VALUE'
    'Strong_Alt_Prime1.png'
    'Strong_Alt_Target1.png'
    'Strong_Alt_Prime3.png'
    'Strong_Alt_Target3.png'
    'Strong_Alt_Prime4.png'
    'Strong_Alt_Target7.png'

[Alice]

Is it possible to use this to insert a new variable at multiple positions without repeating the previous sections of the list?

[Geoff Hayes]

Alice - please clarify what you mean by without repeating the previous sections of the list. What are the previous sections?

Answer 2

[Thorsten]

 X = {'asdf' 'foo' 'bar'};
 n = 3; name = 'foooo'; 
 X = {X{1:n-1} name X{n:end}};

[Alice]

How would you implement this when n is a vector? So the resulting list would not have duplicated information from above?

[Thorsten]

There is probably a more elegant way w/o a for-loop, but this works:

X = {'asdf' 'foo' 'bar' 'yuck'};
pos = [2 4]; name = {'hi' 'this is new'};
for i = 1:numel(pos)
  n = pos(i);
  X = {X{1:n-1} name{i} X{n:end}};
end

Answer 3

[Stephen23]

This is easy using some MATLAB indexing and mat2cell, and just three lines of code:

X = [2,6]; % insert new element AFTER these elements.
A = {'Strong_Alt_Prime2.png',...
    'Strong_Alt_Target2.png',...
    'Strong_Alt_Prime1.png',...
    'Strong_Alt_Target1.png',...
    'Strong_Alt_Prime3.png',...
    'Strong_Alt_Target3.png',...
    'Strong_Alt_Prime4.png',...
    'Strong_Alt_Target7.png'};

B = mat2cell(A,1,[X(1),diff(X),numel(A)-X(end)]);
B(2,:) = {'NEW VALUE'}; % new element
B = [B{1:end-1}];

When you run this code, the value of B will be:

B = {'Strong_Alt_Prime2.png'
     'Strong_Alt_Target2.png'
     'NEW VALUE'
     'Strong_Alt_Prime1.png'
     'Strong_Alt_Target1.png'
     'Strong_Alt_Prime3.png'
     'Strong_Alt_Target3.png'
     'NEW VALUE'
     'Strong_Alt_Prime4.png'
     'Strong_Alt_Target7.png'}

Of course you can define X to be any positions for the new element, given the following conditions:

• X cannot be empty.
• all values in X are 0<=X<=numel(A).
• the values are sorted in order.
• zero places a new element before the first element.
• accepts multiple new elements in the same location.

Answer 4

[Ajay Pherwani]

expitemorder is a 8x1 cell array/vector , to update any element : - expitemorder{ any_index,1}='value';

[Thorsten]

No, this overwrites the value at any_index , but Alice asked for without overwriting the existing values.