Manipulating arrays within a cell array

8 vues (au cours des 30 derniers jours)
mathman
mathman le 4 Jan 2018
Modifié(e) : mathman le 7 Jan 2018
I have a cell array, A_C which has 401 cells (1x401) each with arrays of varying sizes, but same dimensionality. I'm trying to divide each element within an array by the number of elements within that array. And then replace the array with the new one.
So for example, the array A(1,2) = 1x101, I want to divide each element within this array by 101, B = A(1,2)/101 and then I want to replace the array at A(1,2) by B.
  1 commentaire
mathman
mathman le 4 Jan 2018
I think I solved it, I'll leave the thread up incase someone wants to do something similar.

Connectez-vous pour commenter.

Réponse acceptée

Jan
Jan le 4 Jan 2018
Modifié(e) : Jan le 4 Jan 2018
% Create some test data: (your code)
dxs = 0:5e-06:0.002;
nCell = length(dxs);
xs = cell(1, nCell); % Pre-allocate!!!
for k = 1:nCell
xs{k} = 0:dxs(k):5e-04;
end
% Divide each cell by its length:
B = cell(size(xs)); % Pre-allocate!!!
for k = 1:nCell
B{k} = xs{k} / length(xs{k});
end
Simple with a loop, isn't it? I find "nCell" nicer than "num_cells_dxs" and avoid "l" as variable, because it is confused with 1 easily.
I assume the cellfun is nicer, but slower:
B = cellfun(@(c) c / length(c), xs, 'UniformOutput', false)

Plus de réponses (1)

mathman
mathman le 4 Jan 2018
for n = 1:l
xs_current1{n} = xs_current1{1,n}/num_point_sources(1,n);
end

Catégories

En savoir plus sur Resizing and Reshaping Matrices dans Help Center et File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by