I have a structure called (level3) and inside it some fields (e.g x0, vx). each field is an array with some values as shown in the screenshot.
I would like to go through these fields values and multply them as follows:
x(1,1)*vx(1,1)
x(2,1)*vx(2,1)
x(3,1)*vx(3,1)
.... and so on.
so I got at the end 10 arrays of results of x
I tried to do it like this but I just got one array with the last value.
fields = fieldnames(level3);
for k=1:10
x = [x0. * vx];
end
Thanks in advance

 Réponse acceptée

Mohammed Hammad
Mohammed Hammad le 25 Jan 2019
Modifié(e) : Mohammed Hammad le 26 Jan 2019

0 votes

I got the answer after many tries as follows:
for n = 1:10
x{n} = level3(n).x0. * (level3(n).vx)
end

2 commentaires

Stephen23
Stephen23 le 26 Jan 2019
Modifié(e) : Stephen23 le 26 Jan 2019
Those square brackets are superfluous. Get rid of them.
Note that your inner loop serves no purpose, because you do not use its loop index anywhere and none of its iterations depend on previous iterations, so the results of every of its iteration will get discarded except for the last one.
Mohammed Hammad
Mohammed Hammad le 26 Jan 2019
Thank you for your notes. yeah I removed the inner loop and I got the desired results.

Connectez-vous pour commenter.

Plus de réponses (1)

Bob Thompson
Bob Thompson le 24 Jan 2019

1 vote

fields = fieldnames(level3);
for k=1:10
x = [x0. * vx];
end
If I understand this correctly you're trying to capture the array of multiplications (x0 .* vx) for all of the different elements of the structure. I believe that all you're missing then is either an index on x, or to concatonate the results of x with the previous values.
fields = fieldnames(level3)
x = [];
for k = 1:10
x = [x;[x0.*vx]];
end

1 commentaire

Mohammed Hammad
Mohammed Hammad le 25 Jan 2019
It works , thank you. but the answer is Double(50*1) and I want it 10 arrays, each array has different number of values. the length must be the same as x0 and vx arrays length

Connectez-vous pour commenter.

Catégories

En savoir plus sur Loops and Conditional Statements dans Centre d'aide et File Exchange

Produits

Version

R2016a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by