Remove for loop and reduce time
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I want to reduce the running time of my code. I found that most of the time in my code is spent in this loop, as this loop is invoked multiple times in my code. Is there anyway to reduce time?? Code explanation: ac1 is an acceleration matrix having values ranging from -10 to 10. I am using discretize function to group different acceleration values into different bins, represented by edges_a. Now, I have to find how many values are there in each bin and store those values in the fourth column of acc_dis matrix.
Here is the code:
if true
edges_a=min(ac1):1.25:max(ac1);
grp_ac=discretize(ac1,edges_a);
for i=1:length(edges_a)
acc_dis(i,4)=sum(grp_ac==i);
end
end
5 commentaires
Réponse acceptée
Guillaume
le 29 Juin 2018
Points for using discretize instead of writing your own loop for that. The thing is, what you really want is the histogram and there's also a function for that: histcounts:
edges_a = min(ac1):1.25:max(ac1);
h = histcounts(ac1, edges_a);
acc_dis(1:numel(h), 4) = h;
Note: I don't know how many rows acc_dis has. Hopefully at least enough. If it already has the correct number of rows, that last line can be simplified to:
acc_dis(:, 4) = h;
Note 2: If you still need grp_ac, then:
[h, ~, grp_ac] = histcounts(ac1, edges_a);
Plus de réponses (1)
Nithin Banka
le 29 Juin 2018
Modifié(e) : Nithin Banka
le 29 Juin 2018
if true
edges_a=min(ac1):1.25:max(ac1);
grp_ac=discretize(ac1,edges_a);
len = length(edges_a);
auxArr = zeros(len, 1));
len1 = length(grp_ac);
for i=1:len1
if(grp_ac(i)<=len&grp_ac(i)>=1)
auxArr(grp_ac(i)) = auxArr(grp_ac(i)) + 1;
end
end
acc_dis(1:len, 4) = auxArr;
end
I think this should help.
3 commentaires
Nithin Banka
le 29 Juin 2018
The code you used runs through grp_ac in every iteration in the 'grp_ac==i' command. That will make it kind of for loop inside a for loop.
Voir également
Catégories
En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!