MATLAB: Generalizing an IF and FPRINTF expression

fprintfifMATLAB

Hello,

I got a part of code looking like this:

A = ... % can be a scalar or a vector
B = ... % can be a scalar or a vector if A is scalar
if any(B-A) <= 0 && length(B-A) == length(B)
    fprintf('B > %g or A < %g',A,B(1,find(B-A<=0,1)))
    return
elseif any(B-A) <= 0 && length(B-A) == length(A)
    fprintf('B > %g or A < %g',A(1,find(B-A<=0,1)),B)
    return
end

A and B can't both be vectors at the same time. They can both be scalars at the same time

Is there a way to have only one FPRINTF ?

It would need the code not to acces out of (1,1) for the scalar variable.

It's to have less lines in the final code. I would use it elsewhere too. It may be a bit perfectionism but would be usefull.

Thank you in advance for answering.

EDIT: actually the question is "How to return 1 if the index exceeds matrix dimensions" ?

Best Answer

Since your actual question is much easier to understand, I'll try to answer that one. Below you will find two strategies that ensure you will either get a specific index, or index 1. I removed some of the logic you had, and I also edited the find in the second strategy.

I would also urge you to have a look at the doc for any, as any(B-A) <= 0 is equivalent to ~any(B-A)

%example A and B
A=4;
B=[5 2 1 9];
ind_A=find(B-A<=0,1);
ind_B=find(B-A<=0,1);
if ind_A>size(A,2),ind_A=1;end
if ind_B>size(B,2),ind_B=1;end
fprintf('B > %g or A < %g\n',A(1,ind_A),B(1,ind_B))
%or:
ind=find(B<=A,1);%ind=find(B-A<=0,1);
fprintf('B > %g or A < %g\n',...
    A(1,min(ind,end)),B(1,min(ind,end)))

The trick in the second one is that within an indexing context, the end keyword will also work in another function call.

Related Solutions

MATLAB: Find the number of outputs of a value.

@Zach Adams, first, more feedback. Please take some time to digest that. Second, I've improved your code without changing the algorithm and I've shown you how to determine the number of outputs.

Feedback on code

1) What is H and where does it come from? If H is supposed to be a variable, doesn't it cause an error? If H is a script or a function without any inputs, that needs improved for 2 reasons. a) "H" is a really bad function/file name. b) calling a script from a function defeats the advantages of functions - reproducibility and self-containment. If H is a function that merely asks for input you could move that code into his function. If H is a function that is used by other function, it's OK to keep it as it is but the name should really change. Functions should generally contain verbs or describe what it does. Example: getUserInput().

2) I see you have a variable named "max". You should avoid naming variables after common functions for two reasons. a) if you have a variable named max you will not be able to use the max() function anywhere in the same workspace and b) other people reading your code, including your future-self, will get confused because they are used to the "max" being a function. You can rename it to something like "maxC" if you'd like.

3) Use smart-indentation by selecting the entire code (ctrl + a) and then ctrl+i. I've corrected the code in your comment so that it has smart indentation. This makes it MUCH easier to read.

4) Your final for-loop is actually a conditional statement. Instead of "for hypotenuse > A &&..." use "if hypotenuse > A &&..."

5) 'M' is a very bad function name (same reasons as in point #1 above).

Cleaner version of your code

See inline comments.

function [A, B, hypotenuse] = findPythagTriplets(C)
%This calculates the pythagorean triples given the length of the hypotenuse.
% H;  % H isn't even used anywhere
disp(1:1:C);
A = []; 
B = []; 
hypotenuse = [];
for maxC = 1:1:C
    for m = round(sqrt(maxC/2)):round(sqrt(maxC))
       % m == sqrt(maxC/2) && m == sqrt(maxC);   %This line doesn't do anything at all
        if m == floor(m)
            n = sqrt(maxC - m^2);
        end
        A(end+1) = m^2 -n^2; %* The A(end+1) is generally bad practice but I didn't want to make major changes to your code
        B(end+1) = 2*m*n;    % this too
        hypotenuse(end+1) = m^2 + n^2; %this too
        if A(end) > B(end)
            temp_A=A(end);
            A(end) = B(end);
            B(end) = temp_A;
        end
        if hypotenuse(end) > A(end) && hypotenuse(end) > B(end) && hypotenuse(end)^2 == B(end)^2+A(end)^2 &&  ...
                A(end) >= 0 && A(end) == round(A(end)) &&  B(end) >= 0 && B(end) == round(B(end))
            
            fprintf('Side A:')
            disp(A(end))
            fprintf('Side B:')
            disp(B(end))
            fprintf('Hypotenuse:')
            disp(hypotenuse(end))
        else
            A(end) = []; 
            B(end) = []; 
            hypotenuse(end) = []; 
        end
    end
end

Determine number of values in the output

To determine the number of triangles, just call the function with the output and count the number of values in the output variables.

[A, B, hypotenuse] = findPythagTriplets(10)
numberOfTriangles = numel(A)
% numberOfTriangles =
%      2

MATLAB: (simple) problem for function that doesn’t return value

It works fine for me. Here is test2.m:

function test2
 clc;    % Clear the command window.
 p62(10,200, 300); % Call the function
function p62(a,b,c)
 fprintf('a = %.1f\n',a)
 fprintf('b = %.1f\n',b)
 fprintf('c = %.1f\n',c)

I put both functions into the same file, but you don't have to. And here is what it reports to the command window:

a = 10.0
b = 200.0
c = 300.0

What did you do differently?

Best Answer

Related Solutions

MATLAB: Find the number of outputs of a value.

MATLAB: (simple) problem for function that doesn’t return value

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