1×0 empty double row vector using find

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Nadhynee
Nadhynee 2023-6-1
编辑： VBBV 2023-6-2
Hi, I have problem with this code:
clc; clear; close all
x=[0 0.1 0.2 0.3 0.4 0.5];
y=[1 7 4 3 5 2];
h=0.1;
n=(max(x)-min(x))/h
suma=0;
for i=2:n
aux=h*(i-1)
[row,col] = find(x==aux)
suma=suma+y(col);
end
when I run the for cicle and aux is equal to 0.3, the result of find is "1×0 empty double row vector", but there is a 0.3 in x. I'm really confused about this, someone can help me, please?
Thanks in advance.
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回答（2 个）

James Tursa
James Tursa 2023-6-1
编辑：James Tursa 2023-6-1
Ran in:
Welcome to the world of floating point arithmetic. For your specific example, they are not equal. E.g.,
x=[0 0.1 0.2 0.3 0.4 0.5];
h=0.1;
i = 4;
aux=h*(i-1);
[row,col] = find(x==aux)
row = 1×0 empty double row vector col = 1×0 empty double row vector
fprintf('%20.18f\n',x(4))
0.299999999999999989
fprintf('%20.18f\n',aux)
0.300000000000000044
isequal(0.3,3*0.1)
ans = logical
0
You can see that these numbers are close but not exactly equal. They differ by one least significant bit in the floating point bit pattern:
num2hex(0.3)
ans = '3fd3333333333333'
num2hex(3*0.1)
ans = '3fd3333333333334'
To understand why you get this difference between 0.3 and 3*0.1, see this link:
https://www.mathworks.com/matlabcentral/answers/703427-why-is-0-3-0-1-0-2-0-1-0-2-0-3?s_tid=srchtitle
It is usually bad practice to test for exact equality when floating point arithmetic is involved. Your code needs to be written to account for these small differences.
VBBV
VBBV 2023-6-1
Ran in:
clc; clear; close all
x=[0 0.1 0.2 0.3 0.4 0.5];
y=[1 7 4 3 5 2];
h=0.1;
n=(max(x)-min(x))/h
n = 5
suma=0;
for i=2:n
aux=h*(i-1)
[row,col] = find((x==round(aux,1)))
suma=suma+y(col);
end
aux = 0.1000
row = 1
col = 2
aux = 0.2000
row = 1
col = 3
aux = 0.3000
row = 1
col = 4
aux = 0.4000
row = 1
col = 5
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显示 1更早的评论
John D'Errico
John D'Errico 2023-6-1
Ran in:
Be careful with rounding decimals. Even then, you do not get exactly what you think you get. For example,
x = 0.3;
sprintf('%0.55f',x)
ans = '0.2999999999999999888977697537484345957636833190917968750'
y = round(x,1);
sprintf('%0.55f',y)
ans = '0.2999999999999999888977697537484345957636833190917968750'
0.3 is not exactly representable in floating point arithmetic, and rounding will not change that fact.
VBBV
VBBV 2023-6-1
编辑：VBBV 2023-6-2
Ran in:
round function will output the same what we expect based on the number of input decimal precision given to the function. However, sprintf is different thing, which again displays outputs based on the input precision specified in the function
x = 0.3;
sprintf('%0.55f',x)
ans = '0.2999999999999999888977697537484345957636833190917968750'
y = round(x,1)
y = 0.3000
% round while displaying
sprintf('%0.9f',y)
ans = '0.300000000'
0.3 is not exactly representable in floating point arithmetic, and rounding will not change that fact.
Then what is the purpose of round function ?

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