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Adding element within vector without overwriting the existing value

19 vues (au cours des 30 derniers jours)
Alice
Alice le 16 Fév 2015
Modifié(e) : Thorsten le 18 Fév 2015
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

Réponses (4)

Geoff Hayes
Geoff Hayes le 16 Fév 2015
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'
  2 commentaires
Alice
Alice le 16 Fév 2015
Is it possible to use this to insert a new variable at multiple positions without repeating the previous sections of the list?
Geoff Hayes
Geoff Hayes le 16 Fév 2015
Alice - please clarify what you mean by without repeating the previous sections of the list. What are the previous sections?

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Thorsten
Thorsten le 16 Fév 2015
X = {'asdf' 'foo' 'bar'};
n = 3; name = 'foooo';
X = {X{1:n-1} name X{n:end}};
  2 commentaires
Alice
Alice le 16 Fév 2015
How would you implement this when n is a vector? So the resulting list would not have duplicated information from above?
Thorsten
Thorsten le 18 Fév 2015
Modifié(e) : Thorsten le 18 Fév 2015
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

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Stephen23
Stephen23 le 16 Fév 2015
Modifié(e) : Stephen23 le 17 Fév 2015
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.

Ajay Pherwani
Ajay Pherwani le 16 Fév 2015
expitemorder is a 8x1 cell array/vector , to update any element : - expitemorder{ any_index,1}='value';
  1 commentaire
Thorsten
Thorsten le 18 Fév 2015
No, this overwrites the value at any_index , but Alice asked for without overwriting the existing values.

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