Info

Diese Frage ist geschlossen. Öffnen Sie sie erneut, um sie zu bearbeiten oder zu beantworten.

improve performance of this code

1 Ansicht (letzte 30 Tage)
Salvatore Mazzarino
Salvatore Mazzarino am 10 Nov. 2012
Geschlossen: MATLAB Answer Bot am 20 Aug. 2021
suppose to have a matrix called A
3 0 3 3
0 0 4 0
and have this function
while(1)
if all(A(:,i) == 0)
i = i + 1;
else
i = i + 1
break;
end
end
How can I improve the performance of this code?
  2 Kommentare
Azzi Abdelmalek
Azzi Abdelmalek am 10 Nov. 2012
What do you want to find?
Salvatore Mazzarino
Salvatore Mazzarino am 10 Nov. 2012
I want to find the index that rappresents the column where there are at least one element

Antworten (3)

Jan
Jan am 10 Nov. 2012
Bearbeitet: Jan am 10 Nov. 2012
I cannot imagine that you can feel the speedup for such a tiny problem.
k = 1; % Avoid using "i" as variable
while(1)
if any(A(:, k)) % This saves the creation of the temp logical vector
k = k + 1;
break;
else
k = k + 1
end
end
Note that you function will crash, when A contains zeros only. More secure:
index = NaN;
for k = 1:size(A, 2)
if any(A:, k)
index = k;
break;
end
end
Now index is a NaN, if no column matches the condition.
Shorter:
k = find(any(A, 1), 1);
For large A this must be slower, if a matching column appears early.
  1 Kommentar
Salvatore Mazzarino
Salvatore Mazzarino am 10 Nov. 2012
i variable is important in my code. it's incremented during the execution of my code. it cannot be fixed

Matt J
Matt J am 10 Nov. 2012
Bearbeitet: Matt J am 10 Nov. 2012
i = find(any(A,1) ,1)

Azzi Abdelmalek
Azzi Abdelmalek am 10 Nov. 2012
Bearbeitet: Azzi Abdelmalek am 10 Nov. 2012
out= min(find(any(A)))
%or
ii= 1;
while(1)
if any(A(:,ii))
break;
else
ii= ii+1
end
end

Diese Frage ist geschlossen.

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by