I have an array of arrays, a
49455-element Array{Array{AbstractString,1},1}
the length varies, this is just one of many possibilities
I need to do a b = vcat(a...) giving me
195158-element Array{AbstractString,1}:
and convert it to a SharedArray to have all cores work on the strings in it (I'll convert to a Char matrix behind the curtians, but this is not important)
In a, every element is an array of some number of strings, which I do
map(x -> length(x), a)
49455-element Array{Int64,1}:
1
4
8
.
.
2
Is there a way I can easily resotre the array b to the same dimensions of a?
With the Iterators.jl package:
# `a` holds original. `b` holds flattened version. `newa` should == `a`
using Iterators # install using Pkg.add("Iterators")
lmap = map(length,a) # same length vector defined in OP
newa = [b[ib+1:ie] for (ib,ie) in partition([0;cumsum(lmap)],2,1)]
This is somewhat neat, and can also be used to produce a generator for the original vectors, but a for loop implementation should be just as fast and clear.
As a complement to Dan Getz's answer, we can also use zip instead of Iterators.jl's partition:
tails = cumsum(map(length,a))
heads = [1;tails+1][1:end-1]
newa = [b[i:j] for (i,j) in zip(heads,tails)]
Related
As I am new to Julia and I am trying to do some exercise
I have a vector
A = [[112.01507313113326, 60.7645449470438, 44.284185340771124, 16.4524736204982]
[101.46307715025503, 45.051658067785084, 29.896435433335395, 9.8679853915780]]
and I have B=[100,50, 50,100]
I wanted to get A with an entire row if any of the elements is greater than the value of B (in order)
when I use A[A.>B] I am getting only elements that are greater than the B value
Any help would be appreciated.
Assuming that A is (your code is incomplete):
A=[[112.01507313113326, 60.7645449470438, 44.284185340771124, 16.4524736204982] [101.46307715025503, 45.051658067785084, 29.896435433335395, 9.8679853915780]]
You could do something like:
julia> A[:,[any(col .> B) for col in eachcol(A)]]
4×2 Matrix{Float64}:
112.015 101.463
60.7645 45.0517
44.2842 29.8964
16.4525 9.86799
Since the OP states A is a vector, one can do an array comprehension,
[a for a in A if any(a.>B)]
or a direct indexing using broadcasting,
A[any.(A.>(B,))]
which both give a similar (same) vector:
2-element Vector{Vector{Float64}}:
[112.01507313113326, 60.7645449470438, 44.284185340771124, 16.4524736204982]
[101.46307715025503, 45.051658067785084, 29.896435433335395, 9.867985391578]
Surprisingly, the direct indexing with broadcasting is much faster at this short length.
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.
I'm mucking about with Julia and can't seem to get multidimensional array comprehensions to work. I'm using a nightly build of 0.20-pre for OSX; this could conceivably be a bug in the build. I suspect, however, it's a bug in the user.
Lets say I want to wind up with something like:
5x2 Array
1 6
2 7
3 8
4 9
5 10
And I don't want to just call reshape. From what I can tell, a multidimensional array should be generated something like: [(x, y) for x in 1:5, y in 6:10]. But this generates a 5x5 Array of tuples:
julia> [(x, y) for x in 1:5, y in 6:10]
5x5 Array{(Int64,Int64),2}:
(1,6) (1,7) (1,8) (1,9) (1,10)
(2,6) (2,7) (2,8) (2,9) (2,10)
(3,6) (3,7) (3,8) (3,9) (3,10)
(4,6) (4,7) (4,8) (4,9) (4,10)
(5,6) (5,7) (5,8) (5,9) (5,10)
Or, maybe I want to generate a set of values and a boolean code for each:
5x2 Array
1 false
2 false
3 false
4 false
5 false
Again, I can only seem to create an array of tuples with {(x, y) for x in 1:5, y=false}. If I remove the parens around x, y I get ERROR: syntax: missing separator in array expression. If I wrap x, y in something, I always get output of that kind -- Array, Array{Any}, or Tuple.
My guess: there's something I just don't get here. Anybody willing to help me understand what?
I don't think a comprehension is appropriate for what you're trying to do. The reason can be found in the Array Comprehension section of the Julia Manual:
A = [ F(x,y,...) for x=rx, y=ry, ... ]
The meaning of this form is that F(x,y,...) is evaluated with the variables x, y, etc. taking on each value in their given list of values. Values can be specified as any iterable object, but will commonly be ranges like 1:n or 2:(n-1), or explicit arrays of values like [1.2, 3.4, 5.7]. The result is an N-d dense array with dimensions that are the concatenation of the dimensions of the variable ranges rx, ry, etc. and each F(x,y,...) evaluation returns a scalar.
A caveat here is that if you set one of the variables to a >1 dimensional Array, it seems to get flattened first; so the statement that the "the result is... an array with dimensions that are the concatenation of the dimensions of the variable ranges rx, ry, etc" is not really accurate, since if rx is 2x2 and ry is 3, then you will not get a 2x2x3 result but rather a 4x3. But the result you're getting should make sense in light of the above: you are returning a tuple, so that's what goes in the Array cell. There is no automatic expansion of the returned tuple into the row of an Array.
If you want to get a 5x2 Array from a comprhension, you'll need to make sure x has a length of 5 and y has a length of 2. Then each cell would contain the result of the function evaluated with each possible pairing of elements from x and y as arguments. The thing is that the values in the cells of your example Arrays don't really require evaluating a function of two arguments. Rather what you're trying to do is just to stick two predetermined columns together into a 2D array. For that, use hcat or a literal:
hcat(1:5, 6:10)
[ 1:5 5:10 ]
hcat(1:5, falses(5))
[ 1:5 falses(5) ]
If you wanted to create a 2D Array where column 2 contained the result of a function evaluated on column 1, you could do this with a comprehension like so:
f(x) = x + 5
[ y ? f(x) : x for x=1:5, y=(false,true) ]
But this is a little confusing and it seems more intuitive to me to just do
x = 1:5
hcat( x, map(f,x) )
I think you are just reading the list comprehension wrong
julia> [x+5y for x in 1:5, y in 0:1]
5x2 Array{Int64,2}:
1 6
2 7
3 8
4 9
5 10
When you use them in multiple dimensions you get two variables and need a function for the cell values based on the coordinates
For your second question I think that you should reconsider your requirements. Julia uses typed arrays for performance and storing different types in different columns is possible. To get an untyped array you can use {} instead of [], but I think the better solution is to have an array of tuples (Int, Bool) or even better just use two arrays (one for the ints and one for the bool).
julia> [(i,false) for i in 1:5]
5-element Array{(Int64,Bool),1}:
(1,false)
(2,false)
(3,false)
(4,false)
(5,false)
I kind of like the answer #fawr gave for the efficiency of the datatypes while retaining mutability, but this quickly gets you what you asked for (working off of Shawn's answer):
hcat(1:5,6:10)
hcat({i for i=1:5},falses(5))
The cell-array comprehension in the second part forces the datatype to be Any instead of IntXX
This also works:
hcat(1:5,{i for i in falses(5)})
I haven't found another way to explicitly convert an array to type Any besides the comprehension.
Your intuition was to write [(x, y) for x in 1:5, y in 6:10], but what you need is to wrap the ranges in zip, like this:
[i for i in zip(1:5, 6:10)]
Which gives you something very close to what you need, namely:
5-element Array{(Int64,Int64),1}:
(1,6)
(2,7)
(3,8)
(4,9)
(5,10)
To get exactly what you're looking for, you'll need:
hcat([[i...] for i in zip(1:5, 6:10)]...)'
This gives you:
5x2 Array{Int64,2}:
1 6
2 7
3 8
4 9
5 10
This is another (albeit convoluted) way:
x1 = 1
x2 = 5
y1 = 6
y2 = 10
x = [x for x in x1:x2, y in y1:y2]
y = [y for x in x1:x2, y in y1:y2]
xy = cat(2,x[:],y[:])
As #ivarne noted
[{x,false} for x in 1:5]
would work and give you something mutable
I found a way to produce numerical multidimensional arrays via vcat and the splat operator:
R = [ [x y] for x in 1:3, y in 4:6 ] # make the list of rows
A = vcat(R...) # make n-dim. array from the row list
Then R will be a 3x3 Array{Array{Int64,2},2} while A is a 9x2 Array{Int64,2}, as you want.
For the second case (a set of values and a Boolean code for each), one can do something like
R = [[x y > 5] for x in 1:3, y in 4:6] # condition is y > 5
A = vcat(R...)
where A will be a 9x2 Array{Int64,2}, where true/false is denote by 1/0.
I have tested those in Julia 0.4.7.