I picked up Julia to do some numerical analysis stuff and was trying to implement a full pivot LU decomposition (as in, trying to get an LU decomposition that is as stable as possible). I thought that the best way of doing so was finding the maximum value for each column and then resorting the columns in descending order of their maximum values.
Is there a way of avoiding swapping every element of two columns and instead doing something like changing two references/pointers?
Following up on #longemen3000's answer, you can use views to swap columns. For example:
julia> A = reshape(1:12, 3, 4)
3×4 reshape(::UnitRange{Int64}, 3, 4) with eltype Int64:
1 4 7 10
2 5 8 11
3 6 9 12
julia> V = view(A, :, [3,2,4,1])
3×4 view(reshape(::UnitRange{Int64}, 3, 4), :, [3, 2, 4, 1]) with eltype Int64:
7 4 10 1
8 5 11 2
9 6 12 3
That said, whether this is a good strategy depends on access patterns. If you'll use elements of V once or a few times, this view strategy is a good one. In contrast, if you access elements of V many times, you may be better off making a copy or moving values in-place, since that's a price you pay once whereas here you pay an indirection cost every time you access a value.
Just for "completeness", in case you actually want to swap columns in-place,
function swapcols!(X::AbstractMatrix, i::Integer, j::Integer)
#inbounds for k = 1:size(X,1)
X[k,i], X[k,j] = X[k,j], X[k,i]
end
end
is simple and fast.
In fact, in an individual benchmark for small matrices this is even faster than the view approach mentioned in the other answers (views aren't always free):
julia> A = rand(1:10,4,4);
julia> #btime view($A, :, $([3,2,1,4]));
31.919 ns (3 allocations: 112 bytes)
julia> #btime swapcols!($A, 1,3);
8.107 ns (0 allocations: 0 bytes)
in julia there is the #view macro, that allows you to create an array that is just a reference to another array, for example:
A = [1 2;3 4]
Aview = #view A[:,1] #view of the first column
Aview[1,1] = 10
julia> A
2×2 Array{Int64,2}:
10 2
3 4
with that said, when working with concrete number types (Float64,Int64,etc), julia uses contiguous blocks of memory with the direct representation of the number type. that is, a julia array of numbers is not an array of pointers were each element of an array is a pointer to a value. if the values of an array can be represented by a concrete binary representation (an array of structs, for example) then an array of pointers is used.
I'm not a computer science expert, but i observed that is better to have your data tightly packed that using a lot of pointers when doing number crunching.
Another different case is Sparse Arrays. the basic julia representation of an sparse array is an array of indices and an array of values. here you can simply swap the indices instead of copying the values
Related
I would like a way to programmatically "deconstruct" a vector of variable-length vectors in Julia. I do not care about the resulting vector's order.
For example, suppose that my vector of vectors is A = [[1], [2,3], [4,5,6]]. I can deconstruct A by writing vcat(A[1], A[2], A[3]), which returns [1,2,3,4,5,6]. However, if the length of A is large, then this approach becomes cumbersome. Is there a better, more scalable way to obtain the same result?
Try Iterators.flatten:
julia> collect(Iterators.flatten(A))
6-element Vector{Int64}:
1
2
3
4
5
6
(This yields a lazy representation hence I collected this before showing the output)
While I would second Przemyslaw's answer for any situation where you can get away with using a lazy representation, maybe a more direct answer to your question is:
julia> vcat(A...)
6-element Vector{Int64}:
1
2
3
4
5
6
whenever you feel the need to type out all elements of a collection as function arguments, splatting ... is your friend.
Splatting can however negatively impact performance, so it is generally recommended to use reduce, which has a specialisation for vcat:
julia> reduce(vcat, A)
6-element Vector{Int64}:
1
2
3
4
5
6
In Julia, you can permanently append elements to an existing vector using append! or push!. For example:
julia> vec = [1,2,3]
3-element Vector{Int64}:
1
2
3
julia> push!(vec, 4,5)
5-element Vector{Int64}:
1
2
3
4
5
# or
julia> append!(vec, 4,5)
7-element Vector{Int64}:
1
2
3
4
5
But, what is the difference between append! and push!? According to the official doc it's recommended to:
"Use push! to add individual items to a collection which are not already themselves in another collection. The result
of the preceding example is equivalent to push!([1, 2, 3], 4, 5, 6)."
So this is the main difference between these two functions! But, in the example above, I appended individual elements to an existing vector using append!. So why do they recommend using push! in these cases?
append!(v, x) will iterate x, and essentially push! the elements of x to v. push!(v, x) will take x as a whole and add it at the end of v. In your example there is no difference, since in Julia you can iterate a number (it behaves like an iterator with length 1). Here is a better example illustrating the difference:
julia> v = Any[]; # Intentionally using Any just to show the difference
julia> x = [1, 2, 3]; y = [4, 5, 6];
julia> push!(v, x, y);
julia> append!(v, x, y);
julia> v
8-element Vector{Any}:
[1, 2, 3]
[4, 5, 6]
1
2
3
4
5
6
In this example, when using push!, x and y becomes elements of v, but when using append! the elements of x and y become elements of v.
Since Julia is still in its early phase it'd be best if you follow the community standards, and one of the community standards, is your code making "sense" to other developers at first sight - I should know what your "intents" are immediately I read your code.
About append!, the doc says:
"For an ordered container collection, add the elements of each
collections to the end of it. !!! compat "Julia 1.6" Specifying
multiple collections to be appended requires at least Julia 1.6."
The append! method was added and requires Julia 1.6 to use for multiple collections; so in a sense it is the method that's going to be used in the future as Julia gets adopted a lot, Python uses it too, so adopters from there would likely use it too.
About push!, the doc says:
"Insert one or more items in collection. If collection is an ordered
container, the items are inserted at the end (in the given order). If
collection is ordered, use append! to add all the elements of another
collection to it."
The doc advises you use "append!" over "push!" when your collection is ordered. So as a Julia user, if I see append! on your code, I should know the collections its making changes on is in some way "ordered". That's just it. Otherwise, push! and append! does same things(something that might change in the future), but please follow community standards, it will help.
So use append! when you care about order and use push! when order doesn't matter in your collections. This way, any Julia user reading your code, will know your intents right away; but please don't mix them up
Let's say I have a vector V, and I want to either turn this vector into multiple m x n matrices, or get multiple m x n matrices from this Vector V.
For the most basic example: Turn V = collect(1:75) into 3 5x5 matrices.
As far as I am aware this can be done by first using reshape reshape(V, 5, :) and then looping through it. Is there a better way in Julia without using a loop?
If possible, a solution that can easily change between row-major and column-major results is preferrable.
TL:DR
m, n, n_matrices = 4, 2, 5
V = collect(1:m*n*n_matrices)
V = reshape(V, m, n, :)
V = permutedims(V, [2,1,3])
display(V)
From my limited knowledge about Julia:
When doing V = collect(1:m*n), you initialize a contiguous array in memory. From V you wish to create a container of m by n matrices. You can achieve this by doing reshape(V, m, n, :), then you can access the first matrix with V[:,:,1]. The "container" in this case is just another array (thus you have a three dimensional array), which in this case we interpret as "an array of matrices" (but you could also interpret it as a box). You can then transpose every matrix in your array by swapping the first two dimensions like this: permutedims(V, [2,1,3]).
How this works
From what I understand; n-dimensional arrays in Julia are contiguous arrays in memory when you don't do any "skipping" (e.g. V[1:2:end]). For example the 2 x 4 matrix A:
1 3 5 7
2 4 6 8
is in memory just 1 2 3 4 5 6 7 8. You simply interpret the data in a specific way, where the first two numbers makes up the first column, then the second two numbers makes the next column so on so forth. The reshape function simply specifies how you want to interpret the data in memory. So if we did reshape(A, 4, 2) we basically interpret the numbers in memory as "the first four values makes the first column, the second four values makes the second column", and we would get:
1 5
2 6
3 7
4 8
We are basically doing the same thing here, but with an extra dimension.
From my observations it also seems to be that permutedims in this case reallocates memory. Also, feel free to correct me if I am wrong.
Old answer:
I don't know much about Julia, but in Python using NumPy I would have done something like this:
reshape(V, :, m, n)
EDIT: As #BatWannaBe states, the result is technically one array (but three dimensional). You can always interpret a three dimensional array as a container of 2D arrays, which from my understanding is what you ask for.
I was delighted to learn that Julia allows a beautifully succinct way to form inner products:
julia> x = [1;0]; y = [0;1];
julia> x'y
1-element Array{Int64,1}:
0
This alternative to dot(x,y) is nice, but it can lead to surprises:
julia> #printf "Inner product = %f\n" x'y
Inner product = ERROR: type: non-boolean (Array{Bool,1}) used in boolean context
julia> #printf "Inner product = %f\n" dot(x,y)
Inner product = 0.000000
So while i'd like to write x'y, it seems best to avoid it, since otherwise I need to be conscious of pitfalls related to scalars versus 1-by-1 matrices.
But I'm new to Julia, and probably I'm not thinking in the right way. Do others use this succinct alternative to dot, and if so, when is it safe to do so?
There is a conceptual problem here. When you do
julia> x = [1;0]; y = [0;1];
julia> x'y
0
That is actually turned into a matrix * vector product with dimensions of 2x1 and 1 respectively, resulting in a 1x1 matrix. Other languages, such as MATLAB, don't distinguish between a 1x1 matrix and a scalar quantity, but Julia does for a variety of reasons. It is thus never safe to use it as alternative to the "true" inner product function dot, which is defined to return a scalar output.
Now, if you aren't a fan of the dots, you can consider sum(x.*y) of sum(x'y). Also keep in mind that column and row vectors are different: in fact, there is no such thing as a row vector in Julia, more that there is a 1xN matrix. So you get things like
julia> x = [ 1 2 3 ]
1x3 Array{Int64,2}:
1 2 3
julia> y = [ 3 2 1]
1x3 Array{Int64,2}:
3 2 1
julia> dot(x,y)
ERROR: `dot` has no method matching dot(::Array{Int64,2}, ::Array{Int64,2})
You might have used a 2d row vector where a 1d column vector was required.
Note the difference between 1d column vector [1,2,3] and 2d row vector [1 2 3].
You can convert to a column vector with the vec() function.
The error message suggestion is dot(vec(x),vec(y), but sum(x.*y) also works in this case and is shorter.
julia> sum(x.*y)
10
julia> dot(vec(x),vec(y))
10
Now, you can write x⋅y instead of dot(x,y).
To write the ⋅ symbol, type \cdot followed by the TAB key.
If the first argument is complex, it is conjugated.
Now, dot() and ⋅ also work for matrices.
Since version 1.0, you need
using LinearAlgebra
before you use the dot product function or operator.
In the following code I am using the Julia Optim package for finding an optimal matrix with respect to an objective function.
Unfortunately the provided optimize function only supports vectors, so I have to transform the matrix to a vector before passing it to the optimize function, and also transform it back when using it in the objective function.
function opt(A0,X)
I1(A) = sum(maximum(X*A,1))
function transform(A)
# reshape matrix to vector
return reshape(A,prod(size(A)))
end
function transformback(tA)
# reshape vector to matrix
return reshape(tA, size(A0))
end
obj(tA) = -I1(transformback(tA))
result = optimize(obj, transform(A0), method = :nelder_mead)
return transformback(result.minimum)
end
I think Julia is allocating new space for this every time and it feels slow, so what would be a more efficient way to tackle this problem?
So long as arrays contain elements that are considered immutable, which includes all primitives, then elements of an array are contained in 1 big contiguous blob of memory. So you can break dimension rules and simply treat a 2 dimensional array as a 1-dimensional array, which is what you want to do. So you don't need to reshape, but I don't think reshape is your problem
Arrays are column major and contiguous
Consider the following function
function enumerateArray(a)
for i = 1:*(size(a)...)
print(a[i])
end
end
This function multiplies all of the dimensions of a together and then loops from 1 to that number assuming a is one dimensional.
When you define a as the following
julia> a = [ 1 2; 3 4; 5 6]
3x2 Array{Int64,2}:
1 2
3 4
5 6
The result is
julia> enumerateArray(a)
135246
This illustrates a couple of things.
Yes it actually works
Matrices are stored in column-major format
reshape
So, the question is why doesn't reshape use that fact? Well it does. Here's the julia source for reshape in array.c
a = (jl_array_t*)allocobj((sizeof(jl_array_t) + sizeof(void*) + ndimwords*sizeof(size_t) + 15)&-16);
So yes a new array is created, but the only the new dimension information is created, it points back to the original data which is not copied. You can verify this simply like this:
b = reshape(a,6);
julia> size(b)
(6,)
julia> size(a)
(3,2)
julia> b[4]=100
100
julia> a
3x2 Array{Int64,2}:
1 100
3 4
5 6
So setting the 4th element of b sets the (1,2) element of a.
As for overall slowness
I1(A) = sum(maximum(X*A,1))
will create a new array.
You can use a couple of macros to track this down #profile and #time. Time will additionally record the amount of memory allocated and can be put in front of any expression.
For example
julia> A = rand(1000,1000);
julia> X = rand(1000,1000);
julia> #time sum(maximum(X*A,1))
elapsed time: 0.484229671 seconds (8008640 bytes allocated)
266274.8435928134
The statistics recorded by #profile are output using Profile.print()
Also, most methods in Optim actually allow you to supply Arrays, not just Vectors. You could generalize the nelder_mead function to do the same.