Increase size of array within cellfun
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I have a cell array with each cell containing a 4x4 double array. I want to add an additional value at position (5,5), and thus increase the size of each double array to 5x5. The double arrays will not always be 4x4, but whatever nxm is, I just need to append one additional element in the (n+1)x(m+1) position
I'd rather avoid looping over each cell, can this be done with cellfun? My code below uses '=' within cellfun which is not supported.
n = 4;
m = 4;
A = rand(n,m); B = rand(n,m); C = rand(n,m); D = rand(n,m);
ABCD = {A,B;C,D}; % 2x2 cell array
% ABCD = cellfun(@(x) x(end+1,end+1) = 0, ABCD) % code that doesn't work
Thanks!
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採用された回答
Voss
2023 年 9 月 19 日
cellfun performs a loop internally, so I think an explicit for loop in this case is fine.
However, if you want to use cellfun, here are a couple of ways:
n = 4;
m = 4;
A = rand(n,m); B = rand(n,m); C = rand(n,m); D = rand(n,m);
ABCD = {A,B;C,D}; % 2x2 cell array
Method 1: Use a helper function (here it's called extend_matrix_by_one and it's defined at the end):
ABCD_new = cellfun(@extend_matrix_by_one, ABCD, 'UniformOutput',false)
ABCD_new{1} % check the first cell
Method 2: Use subsasgn to do the assignment:
ABCD_new = cellfun(@(x) subsasgn(x,substruct('()',{size(x,1)+1,size(x,2)+1}),0), ABCD, 'UniformOutput',false)
ABCD_new{1} % check the first cell
% helper function definition
function x = extend_matrix_by_one(x)
x(end+1,end+1) = 0;
end
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その他の回答 (1 件)
Dyuman Joshi
2023 年 9 月 19 日
編集済み: Dyuman Joshi
2023 年 9 月 19 日
"I'd rather avoid looping over each cell, can this be done with cellfun?"
That's the thing - Cellfun is just a loop in disguise.
Looping is the way to go here.
%Bigger Cell array for testing
n = 800;
m = 1000;
A = rand(n,m); B = rand(n,m); C = rand(n,m); D = rand(n,m);
ABCD = {A,B;C,D;A,B;C,D;A,B;C,D;A,B;C,D;A,B;C,D;A,B;C,D;A,B;C,D;A,B;C,D};
%value to append
val = 2;
F1 = @() CELL(ABCD,n,m,val);
F2 = @() LOOP(ABCD,n,m,val);
isequal(F1(),F2())
fprintf('Time taken by cellfun = %f seconds', timeit(F1))
fprintf('Time taken by loop = %f seconds', timeit(F2))
%% If you want to use cellfun, this is one way to do it
function z = CELL(z,n,m,val)
z = cellfun(@(x) [x zeros(n,1);zeros(1,m) val], z, 'uni', 0);
end
%% Loop method
function z = LOOP(z,n,m,val)
for k=1:numel(z)
z{k}(n+1,m+1) = val;
end
end
1 件のコメント
Dyuman Joshi
2023 年 9 月 19 日
Let's compare the speed of all approaches in this thread -
%Bigger Cell array for testing
n = 800;
m = 1000;
A = rand(n,m); B = rand(n,m); C = rand(n,m); D = rand(n,m);
ABCD = {A,B,C,D;A,B,C,D;A,B,C,D;A,B,C,D;A,B,C,D;A,B,C,D};
%value to append
val = 2;
F1 = @() CELL(ABCD,n,m,val);
F2 = @() LOOP(ABCD,n,m,val);
F3 = @() SUBTITUTE(ABCD,n,m,val);
F4 = @() HELPER(ABCD,val);
%% Check if outputs are equal or not
isequal(F1(),F2(),F3(),F4())
fprintf('Time taken by cellfun = %f seconds', timeit(F1))
fprintf('Time taken by loop = %f seconds', timeit(F2))
fprintf('Time taken by substitute = %f seconds', timeit(F3))
fprintf('Time taken by helper = %f seconds', timeit(F4))
function z = CELL(z,n,m,val)
z = cellfun(@(x) [x zeros(n,1);zeros(1,m) val], z, 'uni', 0);
end
function z = LOOP(z,n,m,val)
for k=1:numel(z)
z{k}(n+1,m+1) = val;
end
end
function z = SUBTITUTE(z,n,m,val)
z = cellfun(@(x) subsasgn(x,substruct('()',{n+1,m+1}),val), z, 'UniformOutput',false);
end
function z = HELPER(z,val)
z = cellfun(@(x) extend_matrix_by_one(x,val), z, 'UniformOutput',false);
% helper function definition
function x = extend_matrix_by_one(x,val)
x(end+1,end+1) = val;
end
end
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