Fast reorganization of cell arrays

26 vues (au cours des 30 derniers jours)
Christopher Smith
Christopher Smith le 15 Fév 2025
Commenté : Stephen23 le 6 Avr 2025 à 4:38
I have two cell arrays which store row indices to other arrays for processing. TB is a column cell array that stores the indices for all of the "targets" that interact with "sources" in the SB column cell array. The cell index of TB corresponds to the same cell index of SB. In other words, all of the indices in TB{1} interact with all of the indices in SB{1}. I would like to out of TB all of the unique indices, then find all of the indices in SB that interact with each unique index in TB. The tricky bit is that some of the indices in TB are repeated across cells. I have included a minimum working example that does what I want. It runs quickly enough for small arrays, but becomes slow when the TB and SB arrays/cells get large. Is there a way to make this faster and more compact? Thank you for any insight!
% build the two arrays
TB = cell(3,1); % "target" array
TB{1} = [1;2;3;4;5];
TB{2} = [4;5;6];
TB{3} = [7,8,9,10];
SB = cell(3,1); % "source" array
SB{1} = (1:1:15)';
SB{2} = (16:1:30)';
SB{3} = (31:1:45)';
% build the intermediate sorted array
NTB = length(TB);
out1 = []; % intermediate array
for cnt = 1:NTB
for cnt2 = 1:length(TB{cnt})
out1 = [out1;repmat(TB{cnt}(cnt2),size(SB{cnt},1),1),SB{cnt}];
end
end
% pull out of the intermediate array the desired arrays
TB_desired = unique(out1(:,1)); % all of the unique "targets"
NTB_desired = length(TB_desired);
SB_desired = cell(NTB_desired,1);
for cnt = 1:NTB_desired
logvar = TB_desired(cnt) == out1(:,1);
SB_desired{cnt} = out1(logvar,2); % all of the "sources" that interact with each unique "target"
end

Réponse acceptée

Stephen23
Stephen23 le 15 Fév 2025
Modifié(e) : Stephen23 le 15 Fév 2025
The timings might be similar for small arrays, but try testing on larger arrays!
tic
% Initialize target and source arrays
tgt = {[1;2;3;4;5],[4;5;6],[7,8,9,10]};
src = {(1:15).',(16:30).',(31:45).'};
% Preallocate the intermediate array for better performance
len = sum(cellfun(@numel,tgt) .* cellfun(@numel,src));
arr = zeros(len,2);
% Build the intermediate array more efficiently
idx = 1;
for ii = 1:numel(tgt)
val = tgt{ii}(:); % Ensure column vector
dat = src{ii};
% Calculate indices for current block:
num = numel(val) * numel(dat);
pos = idx:(idx+num-1);
% Create pairs using vectorized operations:
[tmp,raw] = ndgrid(val,dat);
arr(pos,:) = [tmp(:),raw(:)];
%
idx = idx + num;
end
% Create final mapping:
[tgx,~,ids] = unique(arr(:,1));
out = accumarray(ids,arr(:,2),[],@(x){x});
toc
Elapsed time is 0.014116 seconds.
SB_desired1 = out
SB_desired1 = 10x1 cell array
{15x1 double} {15x1 double} {15x1 double} {30x1 double} {30x1 double} {15x1 double} {15x1 double} {15x1 double} {15x1 double} {15x1 double}
Compared with your original code:
tic
% build the two arrays
TB = cell(3,1); % "target" array
TB{1} = [1;2;3;4;5];
TB{2} = [4;5;6];
TB{3} = [7,8,9,10];
SB = cell(3,1); % "source" array
SB{1} = (1:1:15)';
SB{2} = (16:1:30)';
SB{3} = (31:1:45)';
% build the intermediate sorted array
NTB = length(TB);
out1 = []; % intermediate array
for cnt = 1:NTB
for cnt2 = 1:length(TB{cnt})
out1 = [out1;repmat(TB{cnt}(cnt2),size(SB{cnt},1),1),SB{cnt}];
end
end
% pull out of the intermediate array the desired arrays
TB_desired = unique(out1(:,1)); % all of the unique "targets"
NTB_desired = length(TB_desired);
SB_desired0 = cell(NTB_desired,1);
for cnt = 1:NTB_desired
logvar = TB_desired(cnt) == out1(:,1);
SB_desired0{cnt} = out1(logvar,2); % all of the "sources" that interact with each unique "target"
end
toc
Elapsed time is 0.015339 seconds.
SB_desired0
SB_desired0 = 10x1 cell array
{15x1 double} {15x1 double} {15x1 double} {30x1 double} {30x1 double} {15x1 double} {15x1 double} {15x1 double} {15x1 double} {15x1 double}
isequal(SB_desired0,SB_desired1)
ans = logical
1
  3 commentaires
Christopher Smith
Christopher Smith le 5 Avr 2025 à 20:45
This answer has been super helpful, but the numbers of elements that need to be sorted are much larger than 1e3. This code is much faster than the original, but still is the bottleneck in the larger code that uses it. Is there a faster way to accomplish this task? I've thought about starting a new question, but figure I would post a comment to this answer first. Thanks!

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