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Uniform spacing and the problem of round-off error
The vector [3 4 5 6 7 8 9] is uniformly spaced with a step size of 1. So is [3 2 1 0 -1 -2] but with a step size of -1.
The vector [1 2 4 8] is not uniformly spaced.
A vector v with uniform spacing has the same finite interval or step size between consecutive elements of the vector. But sometimes round-off error poses a problem in calculating uniformity.
Take, for example, the vector produced by
format shortg
v = linspace(1,9,7)
v = 1x7
1 2.3333 3.6667 5 6.3333 7.6667 9
Linspace produces linearly spaced vectors but the intervals between elements of v, computed by diff(v), are not identical.
dv = diff(v)
dv = 1x6
1.3333 1.3333 1.3333 1.3333 1.3333 1.3333
dv == dv(1)
ans = 1×6 logical array
1 0 0 1 0 1
diff(dv)
ans = 1x5
4.4409e-16 0 -4.4409e-16 8.8818e-16 -8.8818e-16
Some extra steps are therefore necessary to set a tolerance that ignores error introduced by floating point arithmetic.
New in R2022b: isuniform
Determining uniformity of a vector became a whole lot easier in MATLAB R2022b with the new isuniform function.
[tf,step] = isuniform(v)
isuniform returns a logical scalar indicating whether vector v is uniformly spaced within a round-off tolerance and returns the step size (or NaN if v is not uniform).
Let's look at the results for our vector v,
[tf,step] = isuniform(v)
tf = logical
1
step =
1.3333
How about non-uniformly spaced vector?
[tf,step] = isuniform(logspace(1,5,4))
tf = logical
0
step =
NaN
Give it a shot in MATLAB R2022b
  • What happens when all elements of v are equal?
  • Can you produce a vector with uniform spacing without using colons or linspace?
  • What additional steps would be needed to use isuniform with circular data?
References
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