Hi Everyone, I need your help on my problem.

2 次查看（过去 30 天）
Engdaw Chane
Engdaw Chane 2018-3-7
评论： Walter Roberson 2018-3-7
I have a 3D matrix (abc), and I need to do if and elseif conditions to this matrix. However, the only the first expression(aa=abc-5) is applied to all elements. %
%
ab=[10,2,2;4,12,6;7,5,9];
abc=repmat(ab,1,1,3);
if 1<abc<=5
aa=abc-5;
elseif 6<abc<=10
aa=abc+10;
end
How can I make the elseif expression work?
Thank you.
Chane

请先登录，再回答此问题。

采纳的回答

Jan
Jan 2018-3-7
编辑：Jan 2018-3-7
1 < abc <= 5
This will not do, what you expect. Matlab processes this expression from left to right:
  1. 1 < abc. This is either TRUE or FALSE, which is interpreted as 1 or 0
  2. 1 <= 5 or 0 <= 5. This is TRUE in both cases.
You want:
1 < abc && abc <= 5
Then your if and elseif will work.
For productive code using logical indexing as suggested by Birdman is usually more efficient and nicer. But understanding why a<b<c fails is essential.
  3 个评论
显示 1更早的评论
Jan
Jan 2018-3-7
If your Hx20_50 is a vector, you cannot use && but need &. But then the condition of the if command is a vector also. Because if requires a scalar condition, a all() is inserted automatically. This is most likely not, what you want.
Either use a loop:
t20_50 = zeros(size(Hx20_50)); % Pre-allocate!
for k = 1:numel(Hx20_50)
if 0<Hx20_50(k) && Hx20_50(k)<=0.5
t20_50(k) = sqrt(log(1 ./ (Hx20_50(k) .^ 2)));
elseif 0.5<Hx20_50(k) && Hx20_50(k)<=1
t20_50(k) = sqrt(log(1 ./ (1-Hx20_50(k)) .^ 2));
end
end
or use the logical indexing shown by Walter.
Engdaw Chane
Engdaw Chane 2018-3-7
@Jan This was what I have been looking for. The others also work but I wasn't getting the dimensions I needed because of a reason in my code. Thank you.

请先登录，再进行评论。

更多回答（1 个）

Birdman
Birdman 2018-3-7
编辑：Birdman 2018-3-7
Actually, you do not need a ifelse statement. Simple logical indexing will give you what you want.
ab=[10,2,2;4,12,6;7,5,9];
abc=repmat(ab,1,1,3);
abc(abc>1 & abc<=5)=abc(abc>1 & abc<=5)-5;
abc(abc>6 & abc<=10)=abc(abc>6 & abc<=10)+10
  6 个评论
显示 4更早的评论
Engdaw Chane
Engdaw Chane 2018-3-7
编辑：Engdaw Chane 2018-3-7
@Walter Hx20_50 is probability of elements, and the values are just between 0 and 1. just the dimensions are missing. Thank you.
Walter Roberson
Walter Roberson 2018-3-7
If the inputs are certain to be in that range then you can simplify to
mask = Hx20_50 <= 0.5;
t20_50(mask) = sqrt(log(1./(Hx20_50(mask).^2)));
t20_50(~mask) = sqrt(log(1./(1-Hx20_50(~mask)).^2));

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Programming 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by