Hi!
I've an array that is A but I'd like to create another array, B, that containes in order the position of this values.
For example: if A is [1000,700,500,350,200,100,50,15], B must be [8,7,6,5,4,3,2,1].
I tried to do:
a = 8; % a is number of values in array
A = [1000,700,500,350,200,100,50,15];
for i = 1:a
B = A(i)
end
but B is not an array. How can I do?

4 commentaires

Dennis
Dennis le 7 Mai 2018
Modifié(e) : Dennis le 7 Mai 2018
So B is just descending from length of A?
B=fliplr([1:length(A)])
You need to change B to B(i) in your code to get a vector. Right now you are overwriting B in every iteration. But then B would have the same values as A.
Stephen23
Stephen23 le 7 Mai 2018
Modifié(e) : Stephen23 le 7 Mai 2018
" I'd like to create another array, B, that containes in order the position of this values"
Then use sort:
[~,B] = sort(A)
EDIT: see also earlier question, with an impossibly complex approach to this problem:
sc
sc le 7 Mai 2018
'So B is just descending from length of A?'
No, because for example: if: A = [1000,500,700,350,100,200,50,15]; B must be B = [8,6,7,5,3,4,2,1].
In practice I should associate the values of A to the values of B, but values however may not be descendants.
It's hard to explain: I would like to create the ascending order of A's values, but insert them in order of appearance in B.
Is it possible?
Jan
Jan le 11 Mai 2018
@sc: Please stop closing your question, when they got accepted answers already. After closing the questions are deleted, and this would remove the invested time of the ones, who help you. This is counter-productive in a public forum.

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 Réponse acceptée

Jan
Jan le 7 Mai 2018

0 votes

Maybe:
B = numel(A):-1:1
if the array A is sorted already. Otehrwise:
[~, B] = sort(A, 'ascend')

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sc
le 7 Mai 2018

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Jan
le 11 Mai 2018

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