How to make a vector linearly spaced?

Hello everyone,

I have generated the timestamps for the charging power of a battery in the form of:

t=[7.2285 8.2285 16.4685 ...]

I have been trying to make this data linearly spaced, as in:

t2=[0.0000 0.0001 ... 7.2285 ... 8.2285 ... 16.4685]

I've tried the index method to find the index of the matching values as follows:

t3=0:0.0001:24;
for m=1:length(t3)
    ind=find(t(:,1)==t3(m,1),1);
end

It always returns a zero matrix.

I can then use the index of the matching values to find the corresponding values in another vector. I've been at it for the last three hours, but I can't seem to find a matching value between the two vectors.

Is there a way to simply add linearly increasing values in the vector?

Answer 1

[Stephen23]

It is quite likely that your two vectors do not actually have exactly the same values, due to floating point number precision. This has been explained in many forums and threads:

http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

http://www.mathworks.com/matlabcentral/answers/69-why-does-1-2-3-1-3-not-equal-zero

http://www.mathworks.com/matlabcentral/answers/57444-faq-why-is-0-3-0-2-0-1-not-equal-to-zero

https://www.mathworks.com/matlabcentral/answers/41536-why-is-2-24-10-22-4-not-equal-to-0

The solution is to compare the difference between the values against a tolerance value:

>> t = [7.2285 8.2285 16.4685];
>> vec = 0:0.0001:24;
>> tol = 0.00005;
>> idx = any(abs(bsxfun(@minus,vec,t(:)))<tol,1);
>> find(idx)
ans =
    72286    82286   164686
>> vec(idx)
ans =
    7.2285    8.2285   16.4685

[Guillaume]

I would use ismembertol instead of a bsxfun expression in this case.

As a bonus, by default, the tolerance of ismembertol is relative to the magnitude of the numbers.

[Hamb BB]

Unfortunately I have the 2013 version that doesn't have ismembertol.

I ended up using Stephen's method. I stored the find(idx) and then used the vector to simply pick the values from a corresponding vector of power values.

Answer 2

[Guillaume]

As Stephen's says, your t3 and t don't match because of floating point errors. You need to use a tolerance to perform the comparison instead of pure equality. I.e. instead of

a == b

use

abs(a-b) <= some_tolerance

However, since you want to find the location of elements of a vector within another vector, allowing some tolerance for equality, there is now (since 2015a) ismembertol. This will return all your indices in one go, no need for a loop:

t3 = 0:0.0001:24
[found, indices] = ismembertol(t, t3)  %use default relative tolerance of 1e-12
assert(all(found), 'some elements of t were not found in t3 within the given tolerance')

Answer 3

[Thorsten]

I cannot reproduce the error. This works fine for me:

t=[7.2285 8.2285 16.4685];
t2 = 0:0.0001:t(end);
[f idx] = ismember(t, t2)