Basic 'for loop' problem: it stops after several iterations

Question

Kuzan 2022-11-21

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评论： Steven Lord 2022-11-21

Hi all

Let me show you something interesting, so here is my code with a simple for loop:

clc;
c = zeros(11);
for a = 0:0.1:1
    b = 10*a + 1;
    c (1, b) = a;
end

It should work perfectly, but I got the errors, and the iterations stoped at the 7th when the a value was 0.6:

" Index in position 2 is invalid. Array indices must be positive integers or

logical values.

Error in delete (line 8)

c (1, b) = a;

"

Someone knows the reason? it is confusing. Thanks for your time.

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Jan 2022-11-21

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Welcome to the world of numerical maths.

Your problem is very near to the frequently asked question: Why is 0.1 + 0.2 ~= 0.3? Another variant concerning the same effect:

any(0:0.1:1 == 0.3)
ans = logical
   0

The most decimal number do not have an exact binary representation (and vice versa). Because computers use a binary representation of numbers usually (See: IEEE754), rounding effects are usual. This must be considered, when a result of a calculation with numbers having fractional parts or huge numbers > 2^53 is used for indexing or a comparison of floating point values.

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Answer 1

KSSV 2022-11-21

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I don't know why you are doing this, but the elow should work.

c = zeros(11);
for a = 0:0.1:1
    b = 10*a + 1;
    c (1, fix(b)) = a;
end

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Answer 2

Image Analyst 2022-11-21

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编辑：Image Analyst 2022-11-21

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Because b is not an perfect integer. It has a small fractional part. This should have been taught in your linear algebra class or your numerical analysis class (it was for me). Google truncation error or rounding error.

clc;
c = zeros(11);
format long g
for a = 0:0.1:1
    b = 10*a + 1;
    fprintf('%.25f\n', b)
    c (1, b) = a;
end
1.0000000000000000000000000
2.0000000000000000000000000
3.0000000000000000000000000
4.0000000000000000000000000
5.0000000000000000000000000
6.0000000000000000000000000
7.0000000000000008881784197
Index in position 2 is invalid. Array indices must be positive integers or logical values.

See the FAQ

https://matlab.fandom.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_(or_similar)_not_equal_to_zero?

There are several ways to fix it, for example you could round the values

clc;
c = zeros(11);
format long g
for a = 0:0.1:1
    b = round(10*a + 1);
    c (1, b) = a;
end

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Steven Lord 2022-11-21

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Another approach would be to iterate over a vector with integer entries. Use those values directly as indices and transform those values when you need to use them as data.

c = zeros(1, 11);
for a = 0:10
    c(1, a+1) = a/10;
end
format longg
disp(c.')
                         0
                       0.1
                       0.2
                       0.3
                       0.4
                       0.5
                       0.6
                       0.7
                       0.8
                       0.9
                         1

Or for this particular example, avoid the for loop entirely.

c2 = 0:0.1:1;
disp(c2.')
                         0
                       0.1
                       0.2
                       0.3
                       0.4
                       0.5
                       0.6
                       0.7
                       0.8
                       0.9
                         1

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