less than or equal and greater than of equal operations

Question

ektor 2018-8-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/413578-less-than-or-equal-and-greater-than-of-equal-operations

编辑： jonas 2018-8-4

采纳的回答： Star Strider

在 MATLAB Online 中打开

Dear all,

I have this vector

and I want to make the following tranformation:

The values of DIFF<-0.02 to be tranformed to 1

The values of DIFF>=-0.02 & DIFF<-0.01 to be tranformed to 2

The values of DIFF>=-0.01 & DIFF<=0 to be tranformed to 3

The values of DIFF>0 & DIFF<=0.01 to be tranformed to 4

The values of DIFF>0.01 & DIFF<=0.02 to be tranformed to 5

The values of DIFF>0.02 to be tranformed to 6

So I constructed the following code

 c1= -0.02;
 c2= -0.01;
 c3= 0;
 c4= 0.01;
 c5= 0.02;
kk=size(DIFF,1);
 vy=zeros(kk,1); 
 f1=find( DIFF<c1);
f2=find(DIFF>=c1 & DIFF<c2);
f3=find(DIFF>=c2 & DIFF<=c3);
f4=find(DIFF>c3 & DIFF<=c4);
f5=find(DIFF>c4 & DIFF<=c5);
f6=find(DIFF>c5);
 vy(f1)=1;
vy(f2)=2;
vy(f3)=3;
vy(f4)=4;
vy(f5)=5;
vy(f6)=6;

And I finally obtain the following matrix

 DATA_NEW=[vy DIFF]
0000         0
0000    0.0500
0000   -0.0200
0000    0.0100
0000   -0.0400
0000         0
0000         0
0000         0
0000    0.0500
0000         0
0000    0.0200
0000         0
0000    0.0100
0000         0
0000   -0.0100
0000         0
0000   -0.1500
0000         0
0000   -0.0100
0000         0

But as you can see from the third element of DATA_NEW, the value of 1 is assigned to -0.02. But according to my categorization, the value of 2 should have been assigned to -0.02.

Similarly, the value of 6 is assigned to 0.05, but it should be the value of 5.

How could I rectify this problem given the fact that my real DIFF vector contains thousands of elements.

Thanks in advance

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Answer 1

Star Strider 2018-8-4

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The vector you posted may not be showing the full precision of the values in it. If you calculated them (or imported them without looking at them), they may not be what they seem. See Why is 0.3 - 0.2 - 0.1 (or similar) not equal to zero? (link) for an explanation.

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ektor 2018-8-4

编辑：ektor 2018-8-4

在 MATLAB Online 中打开

the displayed vector is of full precision. I also used

format shortG

I use Matlab R2017a.

Star Strider 2018-8-4

在 MATLAB Online 中打开

With the data you posted, I got (in R2018a):

DATA_NEW =
            3            0
            6         0.05
            2        -0.02
            4         0.01
            1        -0.04
            3            0
            3            0
            3            0
            6         0.05
            3            0
            5         0.02
            3            0
            4         0.01
            3            0
            3        -0.01
            3            0
            1        -0.15
            3            0
            3        -0.01
            3            0

If you use:

format long g

you will likely see the problem.

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Answer 2

jonas 2018-8-4

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https://ww2.mathworks.cn/matlabcentral/answers/413578-less-than-or-equal-and-greater-than-of-equal-operations#answer_331614

编辑：jonas 2018-8-4

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There is no problem with your code. In fact, if you copy your code as written in the question, you get the following results.

ans
0000         0
0000    0.0500
0000   -0.0200
0000    0.0100
0000   -0.0400
0000         0
0000         0
0000         0
0000    0.0500
0000         0
0000    0.0200
0000         0
0000    0.0100
0000         0
0000   -0.0100
0000         0
0000   -0.1500
0000         0
0000   -0.0100
0000         0

I am guessing that the value displayed as -0.0200 is actually less than -0.02.

Try writing

DIFF(3)-0.02

The answer will be less than zero, which implies that DIFF(3) is less than -0.02.

As a sidenote, you can substitute the find command with logical indexing. Instead of

f1=find(DIFF<c1);
vy(f1)=1;

you can write

vy(DIFF<c1)=1

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ektor 2018-8-4

编辑：ektor 2018-8-4

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I set

DIFF(3)==-0.02

but again the same problem. you can try yourself to verify this.

I use Matlab R2017a

jonas 2018-8-4

编辑：jonas 2018-8-4

在 MATLAB Online 中打开

What do you mean, again the same problem? I am quite sure that

DIFF(3)==-0.02

returns 0 in your case, because the number displayed as -0.02 is actuall less than -0.02 (e.g. -0.02000000000001), otherwise you would not be asking this question. It is not a problem, the code works as intended.

I can not verify this, because the numbers you posted are not the same as the ones you have in your workspace. Instead, you can try to clear your workspace and replace it with the code you gave us in your question. I am confident that you will get the (according to you) correct result.

Note that you do not SET anything when you use two equal signs, you TEST whether DIFF(3) is equal to -0.02. Change it to single equal sign and you will get the "correct" results.

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