Compare 2 arrays using for loop and if statement

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Dua Fatima 2020-10-17

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/616693-compare-2-arrays-using-for-loop-and-if-statement

评论： Dua Fatima 2020-10-18

I want to compare enteries of 2 arrays and if they follow the condition, store the value at the same index in a separate array.

The following code is giving me an error. Please let me know what else can I use?

I want to return the 2D array M_filter_1

M =[1 2 3 ; 8 9 0];
e = 2;
M_filter_1=[[]];
n = size(M,2)
m = size(M,1)
A =[1 2 3 4 5 6 7 8 9 0];
for k = 1:1:size(A,2)
    for i= 1:1:m
        for j = 1:1:n
            if (M(i,j)> A(k) - e) & (M (i,j)< A(k)+ e)
                M_filter_1(i,j) = M(i,j);
            else 
               M_filter_1(i,j)= 0 ;
            end
        end
    end
end

16 个评论
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Walter Roberson 2020-10-17

在 MATLAB Online 中打开

if (M(i,j)> A(k) + e) & (M (i,j)< A(k)+ e)

So the M entry has to be strictly greater than A(k)+e, but it also has to be strictly less than A(k)+e .

There are no numbers that are both strictly greater than and strictly less than the same quantity.

The only way that the test could be satisfied, in theory, is if you rewrote it to

if ~(M(i,j) <= A(k) + e) & ~(M (i,j)>= A(k)+ e)

If you stick to real numbers, then P>Q is equivalent to ~(P<=Q) and so you would normally expect the two if statements to be the same. But double precision arithmetic allows for one quantity that is not a number, and which has strange comparison behaviours. The quantity that is not a number is known as NaN -- literally Not A Number. And for NaN, P>Q and P<=Q are both false -- but ~false is true, so ~(P<=Q) is true if P>Q or if one of the values is NaN, and likewise ~(P>=Q) is true if P<Q or one of the values is nan; when you & those two tests together, the combination cannot be satisfied if P and Q are finite or inf or +inf, but the combination can be satisfied of one of the quantities is NaN.

Testing for and of a condition and its opposite, has been used in real production code historically in order to detect NaN, in languages in which there is no isnan() test, so the kind of code you wrote is not entirely useless. It is just not appropriate for the circumstances, and would have to be changes along the lines I show if your intention was to test for NaN.

Walter Roberson 2020-10-18

The output I showed in https://www.mathworks.com/matlabcentral/answers/616693-compare-2-arrays-using-for-loop-and-if-statement#comment_1066638 with the [1 0 0; 0 0 0] output, was run on R2020b.

Dua Fatima 2020-10-18

Aright. Thankyou so much!

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回答（1 个）

Asad (Mehrzad) Khoddam 2020-10-18

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/616693-compare-2-arrays-using-for-loop-and-if-statement#answer_516823

在 MATLAB Online 中打开

M =[1 2 3 ; 8 9 0];
e = 2;
m = size(M,1);
n = size(M,2);
M_filter_1 = zeros(size(M));
A =[1  2 3 4 5 6 7 8 9 0];
for a = A
    for i= 1:1:m
        for j = 1:1:n
            if (M(i,j)> a - e) && (M (i,j)< a + e)
                M_filter_1(i,j) = M(i,j);
            end
        end
    end
end