In MATLAB, I can do the following operation:
[1 2 3; 4 5 6] + [-1 -2 -3]
This will return [0 0 0; 3 3 3], as it understands that I want to perform the addition of the vector with each row of the matrix.
How can I do this in Julia? If I try, it gives me this error:
ERROR: DimensionMismatch("dimensions must match: a has dims (Base.OneTo(2), Base.OneTo(3)), b has dims (Base.OneTo(1), Base.OneTo(3)), mismatch at 1")
Add a dot in front of + like this:
julia> [1 2 3; 4 5 6] .+ [-1 -2 -3]
2×3 Matrix{Int64}:
0 0 0
3 3 3
However, note that [-1 -2 -3] is not a vector in Julia. It is a matrix having one row:
julia> [-1 -2 -3]
1×3 Matrix{Int64}:
-1 -2 -3
In Julia a vector would be [1, 2, 3] and vectors are always treated as columnar:
julia> [1, 2, 3]
3-element Vector{Int64}:
1
2
3
The reason why:
[1 2 3; 4 5 6] + [-1 -2 -3]
fails is that it is an incorrect expression from a mathematical perspective. These two matrices: [1 2 3; 4 5 6] and [-1 -2 -3] are representing linear operators having different spaces they operate on and adding such linear operators is typically considered as not allowed in mathematics.
However, if matrices had the same shape you could add them as linear operators then have the same dimension of the from and to spaces so they can be added:
julia> [1 2; 3 4] + [5 6; 7 8]
2×2 Matrix{Int64}:
6 8
10 12
(such an operation is mathematically well defined)
You can read more about broadcasting here.
Related
Let's say I have a two-dimensional array
a = [1 2 3; 1 2 3]
2×3 Array{Int64,2}:
1 2 3
1 2 3
and I would like sum along a dimension, e.g. along dimension 1 yielding
[2, 4, 6]
or along dimension 2 yielding
[6, 6]
How is this done properly in Julia?
julia> sum(a; dims=1)
1×3 Array{Int64,2}:
2 4 6
julia> sum(a; dims=2)
2×1 Array{Int64,2}:
6
6
You can drop the dimension with vec.
What Jun Tian suggests is the standard way to do it. However, it is also worth to know a more general pattern:
julia> sum.(eachrow(a))
2-element Array{Int64,1}:
6
6
julia> sum.(eachcol(a))
3-element Array{Int64,1}:
2
4
6
In this case sum can be replaced by any collection aggregation function.
I'd like to know how can I operate with CartesianIndex. For example I have array
julia> A = rand(1:5, 10, 2)
10×2 Array{Int64,2}:
2 5
1 1
4 5
4 1
2 1
4 1
2 4
1 5
2 5
4 4
and I want to save all numbers which stay near (in pair) with number 1. I can use c=findall(x->x==1, A), but I will have a cartensian indexes of "1".
There is function x=getindex.(c, [1 2]) it makes an array which I can change, but I don't know how to convert it back to CartesianIndex. And I think that must be a better way to do this.
A[view(A.==1,:,[2,1])]
This literally returns "all numbers which stay in pair with number 1".
The order of returned numbers is columnar. If you want to return it by rows:
A'[view(A.==1,:,[2,1])']
Example:
julia> A = rand(1:5, 10, 2)
10×2 Array{Int64,2}:
1 4
3 3
1 3
3 3
5 1
1 5
2 1
3 3
1 3
2 3
julia> A'[view(A.==1,:,[2,1])']
6-element Array{Int64,1}:
4
3
5
5
2
3
If you rather want full rows than use filter!:
julia> filter!((x)->(1 in x), collect(eachrow(A)))
6-element Array{SubArray{Int64,1,Array{Int64,2},Tuple{Int64,Base.Slice{Base.OneTo{Int64}}},true},1}:
[1, 4]
[1, 3]
[5, 1]
[1, 5]
[2, 1]
[1, 3]
If I want to create 2D array with 1 row by 5 columns.
I could do this
julia> a = [1 2 3 4 5]
1×5 Array{Int64,2}:
1 2 3 4 5
But to create 2D array with 5 rows by 1 column. I have tried
julia> b = [1; 2; 3; 4; 5]
5-element Array{Int64,1}:
1
2
3
4
5
But I got back a 1D array which is NOT what I wanted
The only way to get it to work is
julia> b=reshape([1 2 3 4 5],5,1)
5×1 Array{Int64,2}:
1
2
3
4
5
Perhaps I am missing some crucial information here.
You could also do a = [1 2 3 4 5]'.
On a side note, for Julia versions > 0.6 the type of a wouldn't be Array{Int64, 2} but a LinearAlgebra.Adjoint{Int64,Array{Int64,2}} as conjugate transpose is lazy in this case. One can get <= 0.6 behavior by a = copy([1 2 3 4 5]').
AFAIK there is no syntactic sugar for it.
I usually write:
hcat([1, 2, 3, 4, 5])
which is short and I find it easy to remember.
If you use reshape you can replace one dimension with : which means you do not have to count (it is useful e.g. when you get an input vector as a variable):
reshape([1 2 3 4 5], :, 1)
Finally you could use:
permutedims([1 2 3 4 5])
What is the best way to subtract a vector of length N from a matrix of size (N, K) in Julia?
Of course, for loop or repmat should work but they do not seem to be the most efficient.
Can I use broadcast somehow?
julia> [1 2 3; 4 5 6; 7 8 9] .- [1; 2; 3]
3×3 Array{Int64,2}:
0 1 2
2 3 4
4 5 6
(obviously, subtracting horizontal vectors is also broadcasted)
julia> [1 2 3; 4 5 6; 7 8 9] .- [1 2 3]
3×3 Array{Int64,2}:
0 0 0
3 3 3
6 6 6
Also, note that the broadcasted call .- in the top example is essentially equivalent to
julia> (-).([1 2 3; 4 5 6; 7 8 9], [1; 2; 3])
3×3 Array{Int64,2}:
0 1 2
2 3 4
4 5 6
as of julia 0.6, unifying the f.(args) syntax / under-the-hood implementation for broadcasting functions with that for broadcasting operators.
(i.e. .- is no longer a separately defined operator, which happens to be a 'broadcasted' version of - ).
If the two Int arrays are, a = [1;2;3] and b = [4;5;6], how do we concatenate the two arrays in both the dimensions? The expected outputs are,
julia> out1
6-element Array{Int64,1}:
1
2
3
4
5
6
julia> out2
3x2 Array{Int64,2}:
1 4
2 5
3 6
Use the vcat and hcat functions:
julia> a, b = [1;2;3], [4;5;6]
([1,2,3],[4,5,6])
help?> vcat
Base.vcat(A...)
Concatenate along dimension 1
julia> vcat(a, b)
6-element Array{Int64,1}:
1
2
3
4
5
6
help?> hcat
Base.hcat(A...)
Concatenate along dimension 2
julia> hcat(a, b)
3x2 Array{Int64,2}:
1 4
2 5
3 6
Square brackets can be used for concatenation:
julia> a, b = [1;2;3], [4;5;6]
([1,2,3],[4,5,6])
julia> [a; b]
6-element Array{Int64,1}:
1
2
3
4
5
6
julia> [a b]
3×2 Array{Int64,2}:
1 4
2 5
3 6
You can use the cat function to concatenate any number of arrays along any dimension. The first input is the dimension over which to perform the concatenation; the remaining inputs are all of the arrays you wish to concatenate together
a = [1;2;3]
b = [4;5;6]
## Concatenate 2 arrays along the first dimension
cat(1,a,b)
6-element Array{Int64,1}:
1
2
3
4
5
6
## Concatenate 2 arrays along the second dimension
cat(2,a,b)
3x2 Array{Int64,2}:
1 4
2 5
3 6
## Concatenate 2 arrays along the third dimension
cat(3,a,b)
3x1x2 Array{Int64,3}:
[:, :, 1] =
1
2
3
[:, :, 2] =
4
5
6
when encountered Array{Array,1}, the grammer is a little bit different, like this:
julia> a=[[1,2],[3,4]]
2-element Array{Array{Int64,1},1}:
[1, 2]
[3, 4]
julia> vcat(a)
2-element Array{Array{Int64,1},1}:
[1, 2]
[3, 4]
julia> hcat(a)
2×1 Array{Array{Int64,1},2}:
[1, 2]
[3, 4]
julia> vcat(a...)
4-element Array{Int64,1}:
1
2
3
4
julia> hcat(a...)
2×2 Array{Int64,2}:
1 3
2 4
ref:
... combines many arguments into one argument in function definitions
In the context of function definitions, the ... operator is used to combine many different arguments into a single argument. This use of ... for combining many different arguments into a single argument is called slurping
Functional way to concatanate 2 arrays is to use reduce function.
a = rand(10, 1)
b = rand(10, 1)
c = reduce(hcat, [ a, b])