This is an extremely basic question and I have to be missing something, but when trying to normalize matrix A below so that rows sum to one, some (small) differences remain: A = rand(100,4) A = A./sum(A, dims = 2) sum(sum(A, dims =2) .== 1) # fewer than 100! Why does this happen? Is there a wa…

The result is due to floating point inaccuracies. The following holds: A = rand(100,4) A = A./sum(A, dims = 2) @assert sum(sum(A, dims =2) .≈ 1) == 100 Alternate name for ≈ is isapprox.

Yes, but I need exact sums–I’m thinking of columns as probabilities.

Almost all scientific computing code works with inexact probabilities since the standard level of inexactness offered by floating point is usually not an issue in practice. What level of inexactness creates problems for you?

Well, that eventually results in some distance measures between probability distributions being negative. But now that I know the reason, I will just force them ex-post to be zero.

As @johnmyleswhite mentions with Float types we should always expect these tiny inaccuracies that we force into [0,1] when dealing with probabilities. However, if you need to guarantee calculations without forcing the outcome an option is to use Rational numbers to declare probabilities. A = rand(…

The big limitation here is the rational numbers overflow in ways potentially more surprising than floating point numbers: julia> x = 1//10 1//10 julia> typeof(x) Rational{Int64} julia> for i in 1:1_000 x *= 1//10 println((i, x)) end (1, 1//100) (2, 1//1000) (3, 1/…

I think this is relevant here: [image] PSA: floating-point arithmetic Offtopic Sometimes people are surprised by the results of floating-point calculations such as julia> 5/6 0.8333333333333334 # shouldn't the last digit be 3? julia> 2.…

Normalizing a vector to sum to one does not work?

General Usage

Sukera October 11, 2021, 12:41pm 8

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Normalizing a vector to sum to one does not work?

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