how to optimise the code with cell?

1 Ansicht (letzte 30 Tage)
Shubham Mohan Tatpalliwar
Shubham Mohan Tatpalliwar am 12 Feb. 2019
for this program i am calculating A by using for
A = cell(1,3);
for k = 1:3
A{k} = B*k + C*k;
end
And then every matrix is to be compared to get a Optimal matrix containg minimum of every every A matrix
Comparison = cat(3,A{1},A{2},A{3});
MinimumMatrix = min(Comparison./(Comparison~=0),[],3);
This is a example code
Nut in actual program i have to calculate
k=1000:10:5500
What function should be used in
Comparison = cat(3,A{1000},A{1010},.. A{5500});
to avoid writing the name of each variable for comparison matrix
  2 Kommentare
madhan ravi
madhan ravi am 12 Feb. 2019
cellfun(@(x)min(x),A,'un',0)
madhan ravi
madhan ravi am 12 Feb. 2019
oops I misinterpreted please upload cell A as .mat file

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Guillaume
Guillaume am 12 Feb. 2019
Why are you using cell arrays in the first place? It's just slower and has more overhead than matrices. You can just create your final matrix directly
stepcount = 3
A = zeros([size(B), stepcount]);
for k = 1:stepcount
A(:, :, k) = B*k + C*k;
end
If you really want to use an intermediate cell array, then you can use the expansion of cell arrays into comma-separated lists:
Comparison = cat(3,A{1},A{2},A{3});
%is equivalent to
Comparison = cat(3,A{:});
Comparison = cat(3,A{1000},A{1010},.. A{5500});
%is equivalent to
Comparison = cat(3,A{1000:5500});
  2 Kommentare
Shubham Mohan Tatpalliwar
Shubham Mohan Tatpalliwar am 12 Feb. 2019
Actually i was using eval
but after getting some error i switched to cell....
will try stepcount now
Shubham Mohan Tatpalliwar
Shubham Mohan Tatpalliwar am 12 Feb. 2019
It was really helpful
thank you

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Statistics and Machine Learning Toolbox finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by