equivalent of numpy.c_ in julia - arrays

Question

Hi I am going through the book https://nnfs.io/ but using JuliaLang (it's a self-challenge to get to know the language better and use it more often.. rather than doing the same old same in Python..)
I have come across a part of the book in which they have custom wrote some function and I need to recreate it in JuliaLang...
source: https://cs231n.github.io/neural-networks-case-study/
python
N = 100 # number of points per class
D = 2 # dimensionality
K = 3 # number of classes
X = np.zeros((N*K,D)) # data matrix (each row = single example)
y = np.zeros(N*K, dtype='uint8') # class labels
for j in range(K):
ix = range(N*j,N*(j+1))
r = np.linspace(0.0,1,N) # radius
t = np.linspace(j*4,(j+1)*4,N) + np.random.randn(N)*0.2 # theta
X[ix] = np.c_[r*np.sin(t), r*np.cos(t)]
y[ix] = j
# lets visualize the data:
plt.scatter(X[:, 0], X[:, 1], c=y, s=40, cmap=plt.cm.Spectral)
plt.show()
my julia version so far....
N = 100 # Number of points per class
D = 2 # Dimensionality
K = 3 # Number of classes
X = zeros((N*K, D))
y = zeros(UInt8, N*K)
# See https://docs.julialang.org/en/v1/base/math/#Base.range
for j in range(0,length=K)
ix = range(N*(j), length = N+1)
radius = LinRange(0.0, 1, N)
theta = LinRange(j*4, (j+1)*4, N) + randn(N)*0.2
X[ix] = ????????
end
notice the ??????? area because I am now trying to decipher if Julia has an equivalent for this numpy function
https://numpy.org/doc/stable/reference/generated/numpy.c_.html
Any help is appreciated.. or just tell me if I need to write something myself

Answer 1

This is a special object to provide nice syntax for column concatanation. In Julia this is just built into the language hence you can do:
julia> a=[1,2,3];
julia> b=[4,5,6];
julia> [a b]
3×2 Matrix{Int64}:
1 4
2 5
3 6
For your case the Julian equivalent of np.c_[r*np.sin(t), r*np.cos(t)] should be:
[r .* sin.(t) r .* cos.(t)]
To understand Python's motivation you can also have a look at :
numpy.r_ is not a function. What is it?

Answer 2

The equivalent of numpy.c_ would seem to be horizontal concatenation, which you can do with either the hcat function or with (e.g.) simply [a b]. Fixing a few other issues with the translation so far, we end up with
N = 100 # Number of points per class
D = 2 # Dimensionality
K = 3 # Number of classes
X = zeros(N*K, D)
y = zeros(UInt8, N*K)
for j in range(0,length=K)
ix = (N*j+1):(N*(j+1))
radius = LinRange(0.0, 1, N)
theta = LinRange(j*4, (j+1)*4, N) + randn(N)*0.2
X[ix,:] .= [radius.*sin.(theta) radius.*cos.(theta)]
y[ix] .= j
end
# visualize the data:
using Plots
scatter(X[:,1], X[:,2], zcolor=y, framestyle=:box)