Julia equivalent of Python numpy "arange"

Question

In Python, I can create an array of evenly spaced values using

xi2 = np.arange(0, np.sqrt(6), 1e-3)

How do I write this in Julia? I tried,

xi2 = range(0,sqrt(6),step=1e-3)

but this returns 0.0:0.001:2.449

Answer 1

The result, 0.0:0.001:2.449 is indeed a range of evenly spaced values, and can be indexed, sliced, broadcasted on, and generally used just like any other AbstractArray. For example:

julia> xi2 = range(0,sqrt(6),step=1e-3)
0.0:0.001:2.449

julia> xi2[1]
0.0

julia> xi2[100]
0.099

julia> length(xi2)
2450

julia> isa(xi2, AbstractArray)
true

julia> sin.(xi2)
2450-element Vector{Float64}:
 0.0
 0.0009999998333333417
 0.0019999986666669333
 0.002999995500002025
 ⋮
 0.6408405168240852
 0.6400725224915994
 0.6393038880866445
 0.6385346143778549

If for any reason you want to turn xi2 into a full Array preemptively, you can do that with collect

julia> collect(xi2)
2450-element Vector{Float64}:
 0.0
 0.001
 0.002
 0.003
 ⋮
 2.446
 2.447
 2.448
 2.449

which will "materialize" the range, so to speak. This uses a lot more memory than the Range though, and is less often necessary than you might expect.