I am new in Julia. I want to do a program where I can use group Actions.
For example: Take a vector $(a,b,c,d)$ in $\mathbb{R}^{4}$, and consider the action of elements of $S_{4}$ on this, such as the cycle $(1,2,3,4)$. I would like to have a program which computes:
$$(1,2,3,4)(a,b,c,d) = (d,a,b,c)$$
It would be great if this would be possible for any permutation. Do you have any ideas which packages must I download? and How must I write it?
thank you for your help.
You might like the package Permutations, https://github.com/scheinerman/Permutations.jl.
julia> using Permutations
julia> p = Permutation([2,3,4,1]) # an element of S_4
(1,2,3,4) # printed in cycle notation
julia> p(1) # apply it to one element
2
julia> two_row(p) # alternative way to print this
2×4 Matrix{Int64}:
1 2 3 4
2 3 4 1
julia> inv(p)
(1,4,3,2)
julia> input = [:a, :b, :c, :d]; # an array of any 4 things
julia> [input[p(i)] for i in eachindex(input)] # permute those?
4-element Vector{Symbol}:
:b
:c
:d
:a
julia> ans == input[p.(eachindex(input))] # another way to write the same
true
julia> input[inv(p).(eachindex(input))] # with inv(p)
4-element Vector{Symbol}:
:d
:a
:b
:c
julia> Permutation([2,3,1,4,6,5])
(1,2,3)(4)(5,6) # it really prints the cycles
Related
Suppose we have
A = [1 2; 3 4]
In numpy, the following syntax will produce
A[[1,2],[1,2]] = [1,4]
But, in julia, the following produce a permutation which output
A[[1,2],[1,2]] = [1 2; 3 4]
Is there a concise way to achieve the same thing as numpy without using for loops?
To get what you want I would use CartesianIndex like this:
julia> A[CartesianIndex.([(1,1), (2,2)])]
2-element Vector{Int64}:
1
4
or
julia> A[[CartesianIndex(1,1), CartesianIndex(2,2)]]
2-element Vector{Int64}:
1
4
Like Bogumil said, you probably want to use CartesianIndex. But if you want to get your result from supplying the vectors of indices for each dimensions, as in your Python [1,2],[1,2] example, you need to zip these indices first:
julia> A[CartesianIndex.(zip([1,2], [1,2]))]
2-element Vector{Int64}:
1
4
How does this work? zip traverses both vectors of indices at the same time (like a zipper) and returns an iterator over the tuples of indices:
julia> zip([1,2],[1,2]) # is a lazy iterator
zip([1, 2], [1, 2])
julia> collect(zip([1,2],[1,2])) # collect to show all the tuples
2-element Vector{Tuple{Int64, Int64}}:
(1, 1)
(2, 2)
and then CartesianIndex turns them into cartesian indices, which can then be used to get the corresponding values in A:
julia> CartesianIndex.(zip([1,2],[1,2]))
2-element Vector{CartesianIndex{2}}:
CartesianIndex(1, 1)
CartesianIndex(2, 2)
I want to evaluate a set of vectors (or tuples) in a function $f$ but Julia says me that is imposible.
For example: If I have an array of tuples p=[(1,1), (1,-1), (-1,1), (-1,-1)] and a function f(x,y)=x+y. I would like to calculate f(p[1]) = f(1,1)= 2. But Julia says me that the types are incompatible.
Can you help me please?
You have to splat a tuple like this:
julia> p=[(1,1), (1,-1), (-1,1), (-1,-1)]
4-element Array{Tuple{Int64,Int64},1}:
(1, 1)
(1, -1)
(-1, 1)
(-1, -1)
julia> f(x,y)=x+y
f (generic function with 1 method)
julia> f(p[1]...)
2
you could also define a higher order function splat that would conveniently wrap any function and perform splatting. It is useful as then you can e.g. broadcast such function:
julia> splat(f) = x -> f(x...)
splat (generic function with 1 method)
julia> splat(f)(p[1])
2
julia> splat(f).(p)
4-element Array{Int64,1}:
2
0
0
-2
Alternatively you can define your function f like this:
julia> f((x,y),)=x+y
f (generic function with 1 method)
julia> f(p[1])
2
and now you do not have to do splatting.
Just use the ... operator to unpack the tuple as parameters:
julia> f(p[1]...)
2
In addition to other answers, if your task allows you, you can just define
julia> f(x) = f(x...)
and use it as
julia> f.(p)
4-element Vector{Int64}:
2
0
0
-2
Let's assume I have two vectors x = [1, 2] and y = [3, 4]. How to best combine them to get a matrix m = [1 2; 3 4] in Julia Programming language? Thanks in advance for your support.
Note that in vcat(x', y') the operation x' is adjoint so it should not be used if you are working with complex numbers or vector elements that do not have adjoint defined (e.g. strings). Therefore then permutedims should be used but it will be slower as it allocates. A third way to do it is (admittedly it is more cumbersome to type):
julia> [reshape(x, 1, :); reshape(y, 1, :)]
2×2 Array{Int64,2}:
1 2
3 4
It is non allocating like [x'; y'] but does not do a recursive adjoint.
EDIT:
Note for Cameron:
julia> x = repeat(string.('a':'z'), 10^6);
julia> #btime $x';
1.199 ns (0 allocations: 0 bytes)
julia> #btime reshape($x, 1, :);
36.455 ns (2 allocations: 96 bytes)
so reshape allocates but only minimally (it needs to create an array object, while x' creates an immutable struct which does not require allocation).
Also I think it was a design decision to allocate. As for isbitsunion types actually reshape returns a struct so it does not allocate (similarly like for ranges):
julia> #btime reshape($x, 1, :)
12.211 ns (0 allocations: 0 bytes)
1×2 reshape(::Array{Union{Missing, Int64},1}, 1, 2) with eltype Union{Missing, Int64}:
1 missing
Two ways I know of:
julia> x = [1,2];
julia> y = [3,4];
julia> vcat(x', y')
2×2 Array{Int64,2}:
1 2
3 4
julia> permutedims(hcat(x, y))
2×2 Array{Int64,2}:
1 2
3 4
One more option - this one works both with numbers and other objects as Strings:
julia> rotl90([y x])
2×2 Array{Int64,2}:
1 2
3 4
What about
vcat(transpose(x), transpose(y))
or
[transpose(x); transpose(y)]
I'd like to be able to splat an array of tuples into a function in a vectorized fashion. For example, if I have the following function,
function foo(x, y)
x + y
end
and the following array of tuples,
args_array = [(1, 2), (3, 4), (5, 6)]
then I could use a list comprehension to obtain the desired result:
julia> [foo(args...) for args in args_array]
3-element Array{Int64,1}:
3
7
11
However, I would like to be able to use the dot vectorization notation for this operation:
julia> foo.(args_array...)
ERROR: MethodError: no method matching foo(::Int64, ::Int64, ::Int64)
But as you can see, that particular syntax doesn't work. Is there a vectorized way to do this?
foo.(args_array...) doesn't work because it's doing:
foo.((1, 2), (3, 4), (5, 6))
# which is roughly equivalent to
[foo(1,3,5), foo(2,4,6)]
In other words, it's taking each element of args_array as a separate argument and then broadcasting foo over those arguments. You want to broadcast foo over the elements directly. The trouble is that running:
foo.(args_array)
# is roughly equivalent to:
[foo((1,2)), foo((3,4)), foo((5,6))]
In other words, the broadcast syntax is just passing each tuple as a single argument to foo. We can fix that with a simple intermediate function:
julia> bar(args) = foo(args...);
julia> bar.(args_array)
3-element Array{Int64,1}:
3
7
11
Now that's doing what you want! You don't even need to construct the second argument if you don't want to. This is exactly equivalent:
julia> (args->foo(args...)).(args_array)
3-element Array{Int64,1}:
3
7
11
And in fact you can generalize this quite easily:
julia> splat(f) = args -> f(args...);
julia> (splat(foo)).(args_array)
3-element Array{Int64,1}:
3
7
11
You could zip the args_array, which effectively transposes the array of tuples:
julia> collect(zip(args_array...))
2-element Array{Tuple{Int64,Int64,Int64},1}:
(1, 3, 5)
(2, 4, 6)
Then you can broadcast foo over the transposed array (actually an iterator) of tuples:
julia> foo.(zip(args_array...)...)
(3, 7, 11)
However, this returns a tuple instead of an array. If you need the return value to be an array, you could use any of the following somewhat cryptic solutions:
julia> foo.(collect.(zip(args_array...))...)
3-element Array{Int64,1}:
3
7
11
julia> collect(foo.(zip(args_array...)...))
3-element Array{Int64,1}:
3
7
11
julia> [foo.(zip(args_array...)...)...]
3-element Array{Int64,1}:
3
7
11
How about
[foo(x,y) for (x,y) in args_array]
How do you do simply select a subset of an array based on a condition? I know Julia doesn't use vectorization, but there must be a simple way of doing the following without an ugly looking multi-line for loop
julia> map([1,2,3,4]) do x
return (x%2==0)?x:nothing
end
4-element Array{Any,1}:
nothing
2
nothing
4
Desired output:
[2, 4]
Observed output:
[nothing, 2, nothing, 4]
You are looking for filter
http://docs.julialang.org/en/release-0.4/stdlib/collections/#Base.filter
Here is example an
filter(x->x%2==0,[1,2,3,5]) #anwers with [2]
There are element-wise operators (beginning with a "."):
julia> [1,2,3,4] % 2 .== 0
4-element BitArray{1}:
false
true
false
true
julia> x = [1,2,3,4]
4-element Array{Int64,1}:
1
2
3
4
julia> x % 2 .== 0
4-element BitArray{1}:
false
true
false
true
julia> x[x % 2 .== 0]
2-element Array{Int64,1}:
2
4
julia> x .% 2
4-element Array{Int64,1}:
1
0
1
0
You can use the find() function (or the .== syntax) to accomplish this. E.g.:
julia> x = collect(1:4)
4-element Array{Int64,1}:
1
2
3
4
julia> y = x[find(x%2.==0)]
2-element Array{Int64,1}:
2
4
julia> y = x[x%2.==0] ## more concise and slightly quicker
2-element Array{Int64,1}:
2
4
Note the .== syntax for the element-wise operation. Also, note that find() returns the indices that match the criteria. In this case, the indices matching the criteria are the same as the array elements that match the criteria. For the more general case though, we want to put the find() function in brackets to denote that we are using it to select indices from the original array x.
Update: Good point #Lutfullah Tomak about the filter() function. I believe though that find() can be quicker and more memory efficient. (though I understand that anonymous functions are supposed to get better in version 0.5 so perhaps this might change?) At least in my trial, I got:
x = collect(1:100000000);
#time y1 = filter(x->x%2==0,x);
# 9.526485 seconds (100.00 M allocations: 1.554 GB, 2.76% gc time)
#time y2 = x[find(x%2.==0)];
# 3.187476 seconds (48.85 k allocations: 1.504 GB, 4.89% gc time)
#time y3 = x[x%2.==0];
# 2.570451 seconds (57.98 k allocations: 1.131 GB, 4.17% gc time)
Update2: Good points in comments to this post that x[x%2.==0] is faster than x[find(x%2.==0)].
Another updated version:
v[v .% 2 .== 0]
Probably, for the newer versions of Julia, one needs to add broadcasting dot before both % and ==