How do I use both Julia's dot notation to do elementwise operations AND ensure that the result is saved in an already existing array?
function myfun(x, y)
return x + y
end
a = myfun(1, 2) # Results in a == 3
a = myfun.([1 2], [3; 4]) # Results in a == [4 5; 5 6]
function myfun!(x, y, out)
out .= x + y
end
a = zeros(2, 2)
myfun!.([1 2], [3; 4], a) # Results in a DimensionMismatch error
Also, does #. a = myfun([1 2], [3; 4]) write the result to a in the same way as I am trying to achieve with myfun!()? That is, does that line write the result directly to a without saving storing the result anywhere else first?
Your code should be:
julia> function myfun!(x, y, out)
out .= x .+ y
end
myfun! (generic function with 1 method)
julia> myfun!([1 2], [3; 4], a)
2×2 Matrix{Float64}:
4.0 5.0
5.0 6.0
julia> a
2×2 Matrix{Float64}:
4.0 5.0
5.0 6.0
As for #. a = myfun([1 2], [3; 4]) - the answer is yes, it does not create temporary arrays and operates in-place.
This isn't commonly required, and there are usually better ways to achieve this, but it's possible to broadcast on an output argument by using Reference values that point inside the output array.
julia> a = zeros(2, 2)
2×2 Matrix{Float64}:
0.0 0.0
0.0 0.0
julia> function myfun!(out, x, y)
out[] = x + y
end
myfun! (generic function with 1 method)
julia> myfun!.((Ref(a, i) for i in LinearIndices(a)), [1 2], [3; 4])
2×2 Matrix{Int64}:
4 5
5 6
julia> a
2×2 Matrix{Float64}:
4.0 5.0
5.0 6.0
Edit: Changed out to be the first parameter as per the Style guide - thanks to #phipsgabler for the reminder.
Related
I have a list of vectors (vector of vectors) like the following:
A 2-element Vector{Vector{Int64}}: A= [347229118, 1954075737, 6542148346,347229123, 1954075753, 6542148341] [247492691, 247490813, -2796091443606465490, 247491615, 247492910, 247491620, -4267071114472318843, 747753505]
the goal is to have them all in just one vector. I did try collect, A[:], vec(A), flatten(A) but it still returns 2-element Vector{Vector{Int64}}
I don't know what command I should use. Is there anything
Assuming your input data is:
julia> x = [[1, 2], [3, 4], [5, 6]]
3-element Vector{Vector{Int64}}:
[1, 2]
[3, 4]
[5, 6]
here are some natural options you have.
Option 1: use Iterators.flatten:
julia> collect(Iterators.flatten(x))
6-element Vector{Int64}:
1
2
3
4
5
6
You can omit collect in which case you get a lazy iterator over the source data which is more memory efficient.
Option 2: use vcat:
julia> reduce(vcat, x)
6-element Vector{Int64}:
1
2
3
4
5
6
You could also write:
julia> vcat(x...)
6-element Vector{Int64}:
1
2
3
4
5
6
but splatting might get problematic if your x vector is very long. In which case I recommend you to use the reduce function as shown above.
is there any equivalent to scipy.sparse.linalg.spsolve in Julia? Here's the description of the function in Python.
In [59]: ?spsolve
Signature: spsolve(A, b, permc_spec=None, use_umfpack=True)
Docstring:
Solve the sparse linear system Ax=b, where b may be a vector or a matrix.
I couldn't find this in Julia's LinearAlgebra and SparseArrays. Is there anything I miss or any alternatives?
Thanks
EDIT
For example:
In [71]: A = sparse.csc_matrix([[3, 2, 0], [1, -1, 0], [0, 5, 1]], dtype=float)
In [72]: B = sparse.csc_matrix([[2, 0], [-1, 0], [2, 0]], dtype=float)
In [73]: spsolve(A, B).data
Out[73]: array([ 1., -3.])
In [74]: spsolve(A, B).toarray()
Out[74]:
array([[ 0., 0.],
[ 1., 0.],
[-3., 0.]])
In Julia, with \ operator
julia> A = Float64.(sparse([3 2 0; 1 -1 0; 0 5 1]))
3×3 SparseMatrixCSC{Float64,Int64} with 6 stored entries:
[1, 1] = 3.0
[2, 1] = 1.0
[1, 2] = 2.0
[2, 2] = -1.0
[3, 2] = 5.0
[3, 3] = 1.0
julia> B = Float64.(sparse([2 0; -1 0; 2 0]))
3×2 SparseMatrixCSC{Float64,Int64} with 3 stored entries:
[1, 1] = 2.0
[2, 1] = -1.0
[3, 1] = 2.0
julia> A \ B
ERROR: MethodError: no method matching ldiv!(::SuiteSparse.UMFPACK.UmfpackLU{Float64,Int64}, ::SparseMatrixCSC{Float64,Int64})
Closest candidates are:
ldiv!(::Number, ::AbstractArray) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.3/LinearAlgebra/src/generic.jl:236
ldiv!(::SymTridiagonal, ::Union{AbstractArray{T,1}, AbstractArray{T,2}} where T; shift) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.3/LinearAlgebra/src/tridiag.jl:208
ldiv!(::LU{T,Tridiagonal{T,V}}, ::Union{AbstractArray{T,1}, AbstractArray{T,2}} where T) where {T, V} at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.3/LinearAlgebra/src/lu.jl:588
...
Stacktrace:
[1] \(::SuiteSparse.UMFPACK.UmfpackLU{Float64,Int64}, ::SparseMatrixCSC{Float64,Int64}) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.3/LinearAlgebra/src/factorization.jl:99
[2] \(::SparseMatrixCSC{Float64,Int64}, ::SparseMatrixCSC{Float64,Int64}) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.3/SparseArrays/src/linalg.jl:1430
[3] top-level scope at REPL[81]:1
Yes, it's the \ function.
julia> using SparseArrays, LinearAlgebra
julia> A = sprand(Float64, 20, 20, 0.01) + I # just adding the identity matrix so A is non-singular.
julia> typeof(A)
SparseMatrixCSC{Float64,Int64}
julia> v = rand(20);
julia> A \ v
20-element Array{Float64,1}:
0.5930744938331236
0.8726507741810358
0.6846427450637211
0.3135234897986168
0.8366321472466727
0.11338490488638651
0.3679058951515244
0.4931583108292607
0.3057947282994271
0.27481281228206955
0.888942874188458
0.905356044150361
0.17546911165214607
0.13636389619386557
0.9607381212005248
0.2518153541168824
0.6237205353883974
0.6588050295549153
0.14748809413104935
0.9806131247053784
Edit in response to question edit:
If you want v here to instead be a sparse matrix B, then we can proceed by using the QR decomposition of B (note that cases where B is truly sparse are rare:
function myspsolve(A, B)
qrB = qr(B)
Q, R = qrB.Q, qrB.R
R = [R; zeros(size(Q, 2) - size(R, 1), size(R, 2))]
A\Q * R
end
now:
julia> A = Float64.(sparse([3 2 0; 1 -1 0; 0 5 1]))
3×3 SparseMatrixCSC{Float64,Int64} with 6 stored entries:
[1, 1] = 3.0
[2, 1] = 1.0
[1, 2] = 2.0
[2, 2] = -1.0
[3, 2] = 5.0
[3, 3] = 1.0
julia> B = Float64.(sparse([2 0; -1 0; 2 0]))
3×2 SparseMatrixCSC{Float64,Int64} with 3 stored entries:
[1, 1] = 2.0
[2, 1] = -1.0
[3, 1] = 2.0
julia> mysolve(A, B)
3×2 Array{Float64,2}:
0.0 0.0
1.0 0.0
-3.0 0.0
and we can test to make sure we did it right:
julia> mysolve(A, B) ≈ A \ collect(B)
true
In Python, there is np.indices, which returns the indices of a grid:
Python 2.7.1
> import numpy as np
> x,y = np.indices((2,2))
> x
array([[0, 0],
[1, 1]])
> y
array([[0, 1],
[0, 1]])
What would be a similar function in Julia? Especially for multidimensional grids.
I tried eachindex, but it expects a grid as input, not just the dimensions. Also the output is a flat list of linear indices, instead of separated cartesian components.
It can be obtained using the CartesianIndices function:
julia> inds = CartesianIndices((2,2))
2×2 CartesianIndices{2,Tuple{Base.OneTo{Int64},Base.OneTo{Int64}}}:
CartesianIndex(1, 1) CartesianIndex(1, 2)
CartesianIndex(2, 1) CartesianIndex(2, 2)
julia> Tuple.(inds)
2×2 Array{Tuple{Int64,Int64},2}:
(1, 1) (1, 2)
(2, 1) (2, 2)
julia> getindex.(inds, 1)
2×2 Array{Int64,2}:
1 1
2 2
julia> getindex.(inds, 2)
2×2 Array{Int64,2}:
1 2
1 2
Given a vector z = [1, 2, 3], I want to create a vector of vectors with all circular permutations of z (i.e. zp = [[1,2,3], [3,1,2], [2,3,1]]).
I can print all elements of zp with
for i in 1:length(z)
push!(z, shift!(z)) |> println
end
How can I store the resulting permutations? Note that
zp = Vector(length(z))
for i in 1:length(z)
push!(z, shift!(z))
push!(zp, z)
end
doesn't work as it stores the same vector z 3 times in zp.
One way would just be to copy the vector before pushing it:
z = [1, 2, 3];
zp = Vector();
for i in 1:length(z)
push!(z, shift!(z))
push!(zp, copy(z))
end
gives me
julia> zp
3-element Array{Any,1}:
[2,3,1]
[3,1,2]
[1,2,3]
But I tend to prefer avoiding mutating operations when I can. So I'd instead write this as
julia> zp = [circshift(z, i) for i=1:length(z)]
3-element Array{Array{Int64,1},1}:
[3,1,2]
[2,3,1]
[1,2,3]
This seems to execute pretty quick on my machine (faster than a comprehension):
julia> z=[1,2,3]
3-element Array{Int64,1}:
1
2
3
julia> zp=Vector{typeof(z)}(length(z))
3-element Array{Array{Int64,1},1}:
#undef
#undef
#undef
julia> for i=1:length(z)
zp[i]=circshift(z,i-1)
end
julia> zp
3-element Array{Array{Int64,1},1}:
[1,2,3]
[3,1,2]
[2,3,1]
julia>
I have an empty matrix initially:
m = Matrix(0, 3)
and a row that I want to add:
v = [2,3]
I try to do this:
[m v]
But I get an error
ERROR: ArgumentError: number of rows of each array must match
What's the proper way to do this?
That is because your matrix sizes don't match. Specifically v does not contain enough columns to match m. And its transposed
So this doesnt work
m = Matrix(0, 3)
v = [2,3]
m = cat(1, m, v) # or a = [m; v]
>> ERROR: DimensionMismatch("mismatch in dimension 2 (expected 3 got 1)")
whereas this does
m = Matrix(0, 3)
v = [2 3 4]
m = cat(1, m, v) # or m = [m; v]
>> 1x3 Array{Any,2}:
>> 2 3 4
and if you run it again it creates another row
m = cat(1, m, v) # or m = [m; v]
>> 2x3 Array{Any,2}:
>> 2 3 4
>> 2 3 4
Use the vcat (concatenate vertically) function:
help?> vcat
search: vcat hvcat VecOrMat DenseVecOrMat StridedVecOrMat AbstractVecOrMat levicivita is_valid_char #vectorize_2arg
vcat(A...)
Concatenate along dimension 1
Notice you have to transpose the vector v, ie. v', else you get a DimensionMismatch error:
julia> v = zeros(3)
3-element Array{Float64,1}:
0.0
0.0
0.0
julia> m = ones(3, 3)
3x3 Array{Float64,2}:
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
julia> vcat(m, v') # '
4x3 Array{Float64,2}:
1.0 1.0 1.0
1.0 1.0 1.0
1.0 1.0 1.0
0.0 0.0 0.0
julia> v' # '
1x3 Array{Float64,2}:
0.0 0.0 0.0
julia> vcat(m, v)
ERROR: DimensionMismatch("mismatch in dimension 2 (expected 3 got 1)")
in cat_t at abstractarray.jl:850
in vcat at abstractarray.jl:887
Note: the comments; # ' are there just to make syntax highlighting work well.
Isn't that Matrix creates a two-dimensional array in Julia? If you try with m =[0, 3], which creates a one-dimensional Vector for you, you can append it by [m; v].
I think using [m v] is create a two-dimensional array as well, from the Julia Document