The slicing is not complete!

Question

Lama Hamadeh 2021-12-16

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1612705-the-slicing-is-not-complete

评论： Voss 2021-12-17

Hi all,

I have the following 2D matrix and I want to slice it by taking just the elements on the first column that equal to one. The thing is tht the slicing is turns out t be 50X2 rather than 57x2.

a =[1.0000    0.0399;
0000    0.6043;
0000    0.4364;
0000    0.8750;
0000    0.8750;
0000    0.8750;
0000    0.8750;
0000    0.7557;
0000    0.8363;
0000    0.8123;
0000    0.5656;
0000    0.3738;
0000    0.0285;
0000    0.4316;
0000    0.3117;
0000    0.6250;
0000    0.6250;
0000    0.6250;
0000    0.6250;
0000    0.6250;
0000    0.9756;
0000    0.2671;
0000    0.5090;
0000    0.4370;
0000    0.3156;
0000    0.1238;
0000    0.7256;
0000    0.0171;
0000    0.2590;
0000    0.8757;
0000    0.1870;
0000    0.3750;
0000    0.3750;
0000    0.3750;
0000    0.3750;
0000    0.3750;
0000    0.3750;
0000    0.3750;
0000    0.9593;
0000    0.1816;
0000    0.0617;
0000    0.0656;
0000    0.2419;
0000    0.0057;
0000    0.0863;
0000    0.2919;
0000    0.0623;
0000    0.1250;
0000    0.1250;
0000    0.1250;
0000    0.1250;
0000    0.1250;
0000    0.1250;
0000    0.1250;
0000    0.1250;
0000    0.1250;
0000    0.9431]; %this is 57x2 matrix
 E = a(a(:,1)==1, :)  %this should be the same but it turns out to be 50x2!
    

Any help of why the slicing is throwing away 7 elements would be apprecited.

Thanks.

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Voss 2021-12-17

Can you show the value of E which is 50-by-2? I get E is 57-by-2 as expected when I run the code.

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Answer 1

Steven Lord 2021-12-16

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https://ww2.mathworks.cn/matlabcentral/answers/1612705-the-slicing-is-not-complete#answer_856850

在 MATLAB Online 中打开

Most likely some of the numbers that are displayed as 1 are stored as a value that is very close to but not down-to-the-last-bit equal to 1. Some of the elements of the following vector will be 0 and some will be very small but not zero.

% Using block comments so the rest of the lines in this answer can be
% executed
%{
difference = a(:, 1) - 1
%}

Compare using a tolerance.

x = 0:0.1:1;
x == 0.3 % none match
ans = 1×11 logical array
   0   0   0   0   0   0   0   0   0   0   0
abs(x - 0.3) < eps % element 4 matches
ans = 1×11 logical array
   0   0   0   1   0   0   0   0   0   0   0