I am trying to switch from Mathematica to IJulia for data exploration, and I was wondering if there's a n analogue for the following Mathematica one-liner:
ListPlot[Import["/tmp/output.tsv"], Joined -> True]
output.tsv is a tab delineated list of (X,Y) pairs
Here's a lame attempt:
In [1]: using Gadfly; plot(readdlm("/tmp/output.tsv", '\t', Float64))
no method plot(Array{Float64,2},)
at In[1]:1
Gadfly will accept arrays, but you need to specify the x and y values. Also, you need to pass an aesthetic.
julia> a = [1 2 3; 4 5 6]
2x3 Array{Int64,2}:
1 2 3
4 5 6
julia> plot(a)
ERROR: no method plot(Array{Int64,2})
julia> plot(x=a[1,:], y=a[2,:], Geom.line)
Here is a screenshot from REPL (not IJulia):
You'd probably have to read it into a dataframe (DataFrames.readtable), since that's what Gadfly operates on. Other plotting packages such as Winston operate on raw data, but since you are reading structure data anyway, the DataFrames approach is probably best.
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
Looking to find the numeric placement of letters in a random letter vector using a function equivalent to foo.
myletters = ["a","c","b","d","z"]
foo(myletters)
# [1,3,2,4,26]
Edit: If you're looking for the numeric distance from 'a', here's one solution:
julia> Int.(first.(["a","c","b","d","z"])) - Int('a') + 1
5-element Array{Int64,1}:
1
3
2
4
26
It will gracefully handle unicode (those simply are later code points and thus will have larger values) and longer strings (by only looking at the first character). Capitals, numbers, and some symbols will appear as negative numbers since their code points come before a.
Previous answer: I think you're looking for sortperm. It gives you a vector of indices that, if you index back into the original array with it, will put it in sorted order.
julia> sortperm(["a","c","b","d"])
4-element Array{Int64,1}:
1
3
2
4
I came up with the somewhat convoluted solution:
[reshape((1:26)[myletters[i] .== string.('a':'z')],1)[1] for i=1:length(myletters)]
Or using map
map(x -> reshape((1:26)[x .== string.('a':'z')],1)[1], myletters)
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.