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Naime
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choose 2 index separately inside cell

Naime
さんによって質問されました 2019 年 2 月 22 日
最新アクティビティ Naime
さんによって 編集されました 2019 年 3 月 1 日
index={[1,2,5,9,10,13,17,18,21],[4,5,7,12,13,15,18,20,21],[3,4,6,11,12,14,18,19,20],[8,16,20,22]};
amount={[10,11,12,10,11,10,12,13,20],[11,15,12,13,11,10,15,13,12],[13,16,16,17,11,10,15,13,12],[4,5,6,10]};
change_amount=cellfun(@(m) m/2, index, 'UniformOutput', false);
changed_num=2;
p=cellfun(@(m) randperm(size(m,2)), index, 'UniformOutput', false);
changed_mes=cellfun(@(m) m(1:changed_num), p, 'UniformOutput', false);
I do not want repeated index in changed_mes(does not have any intersection). if changed_mes become {[3,9],[9,3],[3,2],[1,4]} as [3,9] and [9,3] is same, need to do randperm again. so I wrote this code
finally i can change amount of 2 mes in each array
result=cellfun(@(m,u) m(u)=0, change_amount,p 'UniformOutput', false);
but does not give correct result

  1 件のコメント

Jan
2019 年 2 月 22 日
You have posted some code, which does not do, what you want. It is not easy to find out, what you consider as "correct result". It would be much easier if you mention, what you want to achieve actually. What is the wanted result?
The excessive use of cellfun makes it much harder to understand and debug the code. Maybe some simple loops are ways better. Optimize the code with some smart cellfun calls only after you have an already working solution and if you can prove, that it is really faster.

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1 件の回答

Jan
回答者: Jan
2019 年 2 月 22 日

index = {[1,2,5,9,10,13,17,18,21],[4,5,7,12,13,15,18,20,21],[3,4,6,11,12,14,18,19,20],[8,16,20,22]};
amount = {[10,11,12,10,11,10,12,13,20],[11,15,12,13,11,10,15,13,12],[13,16,16,17,11,10,15,13,12],[4,5,6,10]};
num = 2;
result = cell(size(amount));
for k = 1:numel(result)
m = amount{k} / 2;
... ??? I cannot follow the rest of the code
... You want a set of [num] randomly selected indices to be set to 0, correctly?
end
Sorry, I tried to recreate the code by a simple loop, but failed to understand the logic.

  2 件のコメント

Naime
2019 年 2 月 22 日
Simple code is here
index=[1,2,5,9,10,13,17,18,21];
amount=[10,11,12,10,11,10,12,13,20];
changed_num=2;
p = randperm(size(index,2));
changed_mes_index=p(1:changed_num);
p(1:changed_num)=[];
b=amount/2;
b(p)=0;
result=amount+b;
but index and amount is cell.
index={[1,2,5,9,10,13,17,18,21],[4,5,7,12,13,15,18,20,21],[3,4,6,11,12,14,18,19,20],[8,16,20,22]};
amount={[10,11,12,10,11,10,12,13,20],[11,15,12,13,11,10,15,13,12],[13,16,16,17,11,10,15,13,12],[4,5,6,10]};
the other problem is I want to get result that has unique changed_mes_index.
for example 5 is common in first and second index, so if in first index choose 5 and 6 as changed_mes_index, in second one should be anything expect 5.
example of acceptable changed_mes={[1,9],[4,15],[3,12],[16,20]}
do not acceptable changed_mes={[13,17],[5,13],[20,12],[8,20]} as13 and 20 is repeated
Jan
2019 年 2 月 25 日
A clear description of what the code should do would be a good idea, preferrably as comments. A samll simplification:
index = [1,2,5,9,10,13,17,18,21];
amount = [10,11,12,10,11,10,12,13,20];
num = 2;
p = randperm(numel(index));
changed_mes_index = p(1:num);
b = amount / 2;
b(p(num+1:end)) = 0;
result = amount + b;
I have no idea, how this problem can be modified to work with cells.
You go forward with mentioning "the other problem". What exactly is the first problem?
If you want the elements in the cell array changed_mes to be unique, create it accordingly:
num = 2; % a [1x2] vector per cell element
numC = 4; % Number of cell elements
maxV = 20; % The maximum value
V = (1:maxV);
V = V(randperm(maxV, num * numC));
changed_mes = num2cell(reshape(V, [], 2), 2);

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