# Switch case on an array

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luca on 19 Sep 2019
Edited: Guillaume on 19 Sep 2019
Given the following code
TA = 0
TB = 0
TC = 0
x = [12 64 24];
s = num2str(x);
n = length(s)
for k=1:n
switch s(k)
case 12
TA = TA + 1;
case 24
TB = TB + 1;
case 64
TC = TC + 1;
end
end
I would like to create a switch case depending on the value on x.
Following the code, the final result should be
TA = 1
TB = 1
TC = 1
But I don't know how to write x in the right character array, and the code give me a wrong result.
Which is the most clever way to solve this problem?
luca on 19 Sep 2019
Edited: luca on 19 Sep 2019
The code has the following purpose:
I have a vector
x = [12 24 64 24 12 12]
and three quantity intially equal to 0
TA = 0;
TB = 0;
TC = 0;
I want to iterate on x's columns. Everytime I find 12 I want to increment of 1 TA, everytime i find 24 I want to increment by 1 TB and everytime I find 64 I want to increment by 1 TC.
In this case we should obtain
TA = 3
TB = 2
TC = 1
I hope I have been clearer now

Guillaume on 19 Sep 2019
Yes, it's a lot clearer. So, the transition to a char array is a complete red herring. It's totally unnecessary.
Using sequentially named variables is a bad idea. It will force you to write complicated code. If you have 10 different numbers that you want to count would you add another 7 cases to your switch? How about 100 numbers, 10000 numbers. Will you write all these cases?
The answer is to use array indexing, T(1) the count of your first, T(2) the count of your 2nd number, etc. Then you can use a loop. For example,one (inneficient) way to then implement your code would be:
%input
x = [12 24 64 24 12 12]
lookup = [12, 24, 64]; %values to compute the histogram of
%code
count = zeros(size(lookup))
for i = 1:numel(x)
[found, where] = ismember(x(i), lookup);
if found
count(where) = count(where) + 1;
end
end
But the whole thing can be achieve without loops, if or case. You want to compute an histogram, use the histcounts function:
x = [12 24 64 12 12];
values = unique(x)'; %get unique values of
count = histcounts(x, [values; Inf])'; %inf is right edge of last bin
table(values, count) %for pretty display
returns
ans =
3×2 table
values count
______ _____
12 3
24 1
64 1
##### 2 CommentsShow 1 older commentHide 1 older comment
Guillaume on 19 Sep 2019
Edited: Guillaume on 19 Sep 2019
You can but you shouldn't. Again, what if you want 1000 values, you'll be writing Txx = count(xx) a 1000 times.
Simply, continue using count(1) wherever you were going to use T1. Don't number variables, use indexing.

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