find the position of en element if you have the increment

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How i will find mathematically what is the position of the element correspounding z=-2 it is 21 but I cant find it mathematically
p=[4.4 5.5 6.6 8];
h=[2 5 3 6];
x=[0:0.4:40];
z=[0:-0.1:-16]
vel=zeros(numel(z),numel(x));
vel(1:21,:)=4.4;
vel(22:71,:)=5.5;
vel(72:101,:)=6.6;
vel(102:161,:)=8;
  3 Comments
F.O
F.O on 11 Dec 2017
Edited: F.O on 11 Dec 2017
Hei Adam ,Actually ,sometimes I am very bad in asking and sorry for that. the thing is that every value in p corresspound to the values in h and I wanted to know how to get the position when z equals minus every value in h vector

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Accepted Answer

Stephen23
Stephen23 on 11 Dec 2017
Edited: Stephen23 on 11 Dec 2017
You need to compare the difference against a (carefully selected) tolerance:
>> z = 0:-0.1:-16; % do not use square brackets!
>> tol = 0.00001;
>> val = -4.4;
>> idx = find(abs(z-val)<=tol)
idx = 45
>>
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More Answers (1)

James Tursa
James Tursa on 11 Dec 2017
Edited: James Tursa on 11 Dec 2017
You need to be careful how you search for this element. Because of the way the colon operator works with floating point fractions, you cannot guarantee that values you think might obviously be in the vector are actually there exactly. E.g.,
>> z=[0:-0.1:-16];
>> find(z==-4.4)
ans =
45
>> num2strexact(-4.4)
ans =
-4.4000000000000003552713678800500929355621337890625
>> num2strexact(z(45))
ans =
-4.4000000000000003552713678800500929355621337890625
>> find(z==-4.1)
ans =
Empty matrix: 1-by-0
>> num2strexact(-4.1)
ans =
-4.0999999999999996447286321199499070644378662109375
>> num2strexact(z(42))
ans =
-4.10000000000000053290705182007513940334320068359375
So, for the -4.4 value the colon operator happen to pick the same "closest" IEEE double value that you get when you type the value at the command line. But this didn't happen for the -4.1 value. So to do these types of operations in your code, you need to account for this effect and either use tolerances or use some type of integer range that you know can be searched exactly.

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