# Calling variables instead of values

18 ビュー (過去 30 日間)
Will 2011 年 10 月 31 日
I have this:
A=6; B=7; C=5;
x = [7 5 6]
I want to call x1 as:
x1 =
B C A
x = [2 3 1]
xc = {'A' 'B' 'C'}
x1 = xc(x)
This line of code only works for values that are within the number of columns of the matrix.

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### 採用された回答

Alex 2011 年 10 月 31 日
In the previous question you have a single set (x) that has several elements that you want to link to.
In the new problem, you do not have a single set, so you have a way to search each of the possible solutions. There is no easy way to do this. The two ways I know how to do this is with if/else tree (messy), or an enumerator and switch/case setup (not as messy).
if/else tree example
for i = 1:3
if(x(i) == A)
x1(i) = 'A';
elseif( x(i) == B)
...
...
end
end
The other option is using enumeration and switch/case. I've only used enumeration in older versions, so the following is a quick setup for 2009 A.
classdeff my_letters (Enumeration) < int32
proerties
A = 6;
B = 7;
C = 8;
end
methods (static)
function str = string(input)
switch int32(input)
case my_letter.A
str = 'A';
...
...
...
end
end
Now you can use a switch case as opposed to an if-else tree
Switch x(i)
case int32(my_letters.A)
X1(i) = my_letters.A.string();
...
...
end
The switch/Case is cleaner and faster code wise.
##### 2 件のコメント表示非表示 1 件の古いコメント
Andrei Bobrov 2011 年 10 月 31 日
S = {'A' 'B' 'C'}
C =[6 7 5]
x = [7 5 6]
[loc,loc] = ismember(x,C)
out = S(loc)

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### その他の回答 (1 件)

Alex 2011 年 10 月 31 日
My original comment about one-to-one, I realize, is not the best descriptor.
I've since edited the original post to reflect that and I'll try to explain the differences a little better.
In your original question, you have a base set (x) and you are looking to match values within that set. Since all the values are contained within a single set, this is easy.
In the new question, there are multiple sets (A, B, & C) that all have to be searched since each one could possibly contain the solution that you desire.
If there is a way to combine the multiple sets (like Will's above comment) into a single set (set S in Will's example), then the solution can be achieved much easier. Otherwise, the if/else tree or the switch/case ideas I mentioned are the best options, that I know of.

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