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# What's wrong with the code?

1 回表示 (過去 30 日間)
George Ansari 2017 年 8 月 8 日

The code supposed to delete an element from B cell if corresponding P = 0. Please help me to fix it. Thanks, George.
L = [1 10 15];
L_cell = num2cell(L);
P_max = [100;150;200];
Pg = [90 150 150];
B = [1;2;3];
for k = 1:length(L)
P_d{k} = P_max*L_cell{k};
P{k}=P_d{k} - Pg';
B_cell{k} = num2cell(B);
if (P{k}(:) == 0)
B_cell{k}(:) = [];
end
end

### 回答 (1 件)

Adam 2017 年 8 月 8 日

Simple use of the debugger or even just the command line would help you solve this.
>> P{k}(:)
ans =
1410
2100
2850
This cannot be tested by equality against 0 in an if statement (or rather not in the way you would want it to be).
It isn't obvious what you want in this case, but
doc all
doc any
will solve the problem in 2 different ways depending on what logic you want, e.g.
if ( all( P{k}(:) == 0 ) )
or
if ( any( P{k}(:) == 0 ) )
##### 2 件のコメントなしを表示なしを非表示
George Ansari 2017 年 8 月 8 日
Thank you Adam, I read documentation on 'all' and 'any', but don't think that any of these fit my needs.
What I'm trying to implement is the following. Assume there are 3 power generating units, each has its values of P_max, Pg and B. As it can be seen from the code 'P' is also individual to each unit. I calculate it for different 'L'(which is general to all units). When generator's P = 0, there is no power output from the generator, and therefore the corresponding B value must be deleted from the B vector (in other words B vector is reduced to 2 elements). I hope I was clear. George.
Adam 2017 年 8 月 8 日
The point is though that P{k} contains 3 elements, not just one. So P == 0 does not make sense in this context.

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