Histogram calculation in julia-lang

Question

refer to julia-lang documentations :

hist(v[, n]) → e, counts

Compute the histogram of v, optionally using approximately n bins. The return values are a range e, which correspond to the edges of the bins, and counts containing the number of elements of v in each bin. Note: Julia does not ignore NaN values in the computation.

I choose a sample range of data

testdata=0:1:10;

then use hist function to calculate histogram for 1 to 5 bins

hist(testdata,1) # => (-10.0:10.0:10.0,[1,10])
hist(testdata,2) # => (-5.0:5.0:10.0,[1,5,5])
hist(testdata,3) # => (-5.0:5.0:10.0,[1,5,5])
hist(testdata,4) # => (-5.0:5.0:10.0,[1,5,5])
hist(testdata,5) # => (-2.0:2.0:10.0,[1,2,2,2,2,2])

as you see when I want 1 bin it calculates 2 bins, and when I want 2 bins it calculates 3.

why does this happen?

Lyda

Answer 1

As the person who wrote the underlying function: the aim is to get bin widths that are "nice" in terms of a base-10 counting system (i.e. 10^k, 2×10^k, 5×10^k). If you want more control you can also specify the exact bin edges.

Piccolo

Thanks for your response, I think some lack of documentation exists there. – Lyda

Is there an easy way to plot the results of the hist function? – Software

Yes - using StatsBase; h = fit(Histogram, randn(1000)); using StatPlots; plot(h) – Trussell

This function seems to be deprecated in Base module. Is there any alternative? – Explore

Answer 2

The key word in the doc is approximate. You can check what hist is actually doing for yourself in Julia's base module here.

When you do hist(test,3), you're actually calling

hist(v::AbstractVector, n::Integer) = hist(v,histrange(v,n))

That is, in a first step the n argument is converted into a FloatRange by the histrange function, the code of which can be found here. As you can see, the calculation of these steps is not entirely straightforward, so you should play around with this function a bit to figure out how it is constructing the range that forms the basis of the histogram.

Mentor

Answer 3

In new versions of Julia, the hist function is not present.

To calculate a histogram, one should use StatsBase.Histogram and StatsBase.fit, e.g.:

    using StatsBase
    h = fit(Histogram, rand(100))
    print(h)

Output:

Histogram{Int64, 1, Tuple{StepRangeLen{Float64, Base.TwicePrecision{Float64}, Base.TwicePrecision{Float64}, Int64}}}
edges:
  0.0:0.2:1.0
weights: [21, 22, 17, 16, 24]
closed: left
isdensity: false

Champollion