Why is
julia> collect(partitions(1,2))
0-element Array{Any,1}
returned instead of
2-element Array{Any,1}:
[0,1]
[1,0]
and do I really have to
x = collect(partitions(n,m));
y = Array(Int64,length(x),length(x[1]));
for i in 1:length(x)
for j in 1:length(x[1])
y[i,j] = x[i][j];
end
end
to convert the result to a two-dimensional array?
From the wikipedia:
In number theory and combinatorics, a partition of a positive integer n, also called an integer partition, is a way of writing n as a sum of positive integers.
For array conversion, try:
julia> x = collect(partitions(5,3))
2-element Array{Any,1}:
[3,1,1]
[2,2,1]
or
julia> x = partitions(5,3)
Base.FixedPartitions(5,3)
then
julia> hcat(x...)
3x2 Array{Int64,2}:
3 2
1 2
1 1
Here's another approach to your problem that I think is a little simpler, using the Combinatorics.jl library:
multisets(n, k) = map(A -> [sum(A .== i) for i in 1:n],
with_replacement_combinations(1:n, k))
This allocates a bunch of memory, but I think your current approach does too. Maybe it would be useful to make a first-class version and add it to Combinatorics.jl.
Examples:
julia> multisets(2, 1)
2-element Array{Array{Int64,1},1}:
[1,0]
[0,1]
julia> multisets(3, 5)
21-element Array{Array{Int64,1},1}:
[5,0,0]
[4,1,0]
[4,0,1]
[3,2,0]
[3,1,1]
[3,0,2]
[2,3,0]
[2,2,1]
[2,1,2]
[2,0,3]
⋮
[1,2,2]
[1,1,3]
[1,0,4]
[0,5,0]
[0,4,1]
[0,3,2]
[0,2,3]
[0,1,4]
[0,0,5]
The argument order is backwards from yours to match mathematical convention. If you prefer the other way, that can easily be changed.
one robust solution can be achieved using lexicographic premutations generation algorithm, originally By Donald Knuth plus classic partitions(n).
that is lexicographic premutations generator:
function lpremutations{T}(a::T)
b=Vector{T}()
sort!(a)
n=length(a)
while(true)
push!(b,copy(a))
j=n-1
while(a[j]>=a[j+1])
j-=1
j==0 && return(b)
end
l=n
while(a[j]>=a[l])
l-=1
end
tmp=a[l]
a[l]=a[j]
a[j]=tmp
k=j+1
l=n
while(k<l)
tmp=a[k]
a[k]=a[l]
a[l]=tmp
k+=1
l-=1
end
end
end
The above algorithm will generates all possible unique
combinations of an array elements with repetition:
julia> lpremutations([2,2,0])
3-element Array{Array{Int64,1},1}:
[0,2,2]
[2,0,2]
[2,2,0]
Then we will generate all integer arrays that sum to n using partitions(n) (forget the length of desired arrays m), and resize them to the lenght m using resize_!
function resize_!(x,m)
[x;zeros(Int,m-length(x))]
end
And main function looks like:
function lpartitions(n,m)
result=[]
for i in partitions(n)
append!(result,lpremutations(resize_!(i, m)))
end
result
end
Check it
julia> lpartitions(3,4)
20-element Array{Any,1}:
[0,0,0,3]
[0,0,3,0]
[0,3,0,0]
[3,0,0,0]
[0,0,1,2]
[0,0,2,1]
[0,1,0,2]
[0,1,2,0]
[0,2,0,1]
[0,2,1,0]
[1,0,0,2]
[1,0,2,0]
[1,2,0,0]
[2,0,0,1]
[2,0,1,0]
[2,1,0,0]
[0,1,1,1]
[1,0,1,1]
[1,1,0,1]
[1,1,1,0]
The MATLAB script from http://www.mathworks.com/matlabcentral/fileexchange/28340-nsumk actually behaves the way I need, and is what I though that partitions() would do from the description given. The Julia version is
# k - sum, n - number of non-negative integers
function nsumk(k,n)
m = binomial(k+n-1,n-1);
d1 = zeros(Int16,m,1);
d2 = collect(combinations(collect((1:(k+n-1))),n-1));
d2 = convert(Array{Int16,2},hcat(d2...)');
d3 = ones(Int16,m,1)*(k+n);
dividers = [d1 d2 d3];
return diff(dividers,2)-1;
end
julia> nsumk(3,2)
4x2 Array{Int16,2}:
0 3
1 2
2 1
3 0
using daycaster's lovely hcat(x...) tidbit :)
I still wish there would be a more compact way of doing this.
The the first mention of this approach seem to be https://au.mathworks.com/matlabcentral/newsreader/view_thread/52610, and as far as I can understand it is based on the "stars and bars" method https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)
Related
v = range(1e10, -1e10, step=-1e8) # velocities [cm/s]
deleteat!(v, findall(x->x==0,v))
I want to delete the value 0 from v. Following this tutorial, I tried deleteat! but I get the error
MethodError: no method matching deleteat!(::StepRangeLen{Float64, Base.TwicePrecision{Float64}, Base.TwicePrecision{Float64}, Int64}, ::Vector{Int64})
What am I missing here?
Notice the type that is returned by the function range.
typeof(range(1e10, -1e10, step=-1e8))
The above yields to
StepRangeLen{Float64, Base.TwicePrecision{Float64}, Base.TwicePrecision{Float64}, Int64}
Calling the help function for the function deleteat!.
? deleteat!()
deleteat!(a::Vector, inds)
Remove the items at the indices given by inds, and return the > modified a. Subsequent items are shifted to fill the resulting gap.
inds can be either an iterator or a collection of sorted and > unique integer indices, or a boolean vector of the same length as a with true indicating entries to delete.
We can convert the returned type of range using collect. Try the following code.
v = collect(range(1e10, -1e10, step=-1e8))
deleteat!(v,findall(x->x==0,v))
Notice that we can shorten x->x==0 to iszero which yields to
v = collect(range(1e10, -1e10, step=-1e8))
deleteat!(v,findall(iszero,v))
Use filter! or filter:
julia> filter!(!=(0), [1,0,2,0,4])
3-element Vector{Int64}:
1
2
4
In case of a range you can collect it or use:
julia> filter(!=(0), range(2, -2, step=-1))
4-element Vector{Int64}:
2
1
-1
-2
However for big ranges you might just not want to materialize them to save the memory footprint. In that case you could use:
(x for x in range(2, -2, step=-1) if x !== 0)
To see what is being generated you need to collect it:
julia> collect(x for x in range(2, -2, step=-1) if x !== 0)
4-element Vector{Int64}:
2
1
-1
-2
I'm quite new to Julia and I'm trying to convert a vector of tuples in array.
Here's an example
using Statistics
a = randn((10, 100))
q = (0.05, 0.95)
conf_intervals = [quantile(a[i,:], q) for i in 1:10]
and conf_intervals is a 10-element Vector{Tuple{Float64, Float64}}.
The expected result should be a 10×2 Matrix{Float64}
I tried splatting conf_intervals with [conf_intervals...] but the vector doesn't change.
Thank you very much
You can use a comprehension:
mat2x10 = [tup[k] for k in 1:2, tup in conf_intervals]
mat10x2 = [tup[k] for tup in conf_intervals, k in 1:2]
Or you can just re-interpret the same memory. This is more fragile -- it won't work for all vectors of tuples, e.g. Any[(i, i^2/2) for i in 1:10]. But for Vector{Tuple{Float64, Float64}}:
if VERSION >= v"1.6"
reinterpret(reshape, Float64, conf_intervals)
else
reshape(reinterpret(Float64, conf_intervals), 2, :)
end
mat2x10 == ans # true
You need to use collect to convert tuples to vectors, and then you can combine them:
julia> hcat(collect.(conf_intervals)...)
2×10 Matrix{Float64}:
-1.59757 -2.10057 -1.4437 -1.32868 -1.10686 -1.41256 -1.5696 -1.67288 -1.51947 -1.72257
1.24604 1.61692 1.77684 1.3599 1.90853 1.30831 1.10667 1.58356 1.56811 1.70685
If you need to transpose the result, add an apostrophe ' end the end of the command
I know there is a function that does this, for example:
A = [1,2,0,0,4,0]
find(A)
3-element Array{Int64,1}:
1
2
5
I am trying to do it on my own way, however, I am stuck here
for i=1:endof(A)
if A[i] != 0
[]
end
end
Thanks in advance.
Here's one alternative:
function myfind(c)
a = similar(c, Int)
count = 1
#inbounds for i in eachindex(c)
a[count] = i
count += (c[i] != zero(eltype(c)))
end
return resize!(a, count-1)
end
It actually outperformed find for all the cases I tested, though for the very small example vector you posted, the difference was negligible. There is perhaps some performance advantage to avoiding the branch and dynamically growing the index array.
I have notice that the question is really confusion (because is poorly formulated, sorry about that). Therefore, there are two possible answers: one is [1,2,4]which is an array with the non-zero elements; the other is [1,2,5] which is an array of the indices of the non-zero elements.
Let´s begin with the first option
A = [1,2,0,0,4,0]
B = []
for i=1:endof(A)
if A[i] != 0
push!(B,i)
end
end
println(B)
The output is Any[1,2,5]
However, the type is not the one I wanted. Using typeof(B) it shows Array{Any,1} so I added this code:
B = Array{Int64}(B)
println(B)
typeof(B)
And the result is the desired
[1,2,5]
Array{Int64,1}
To improve its efficiency, following with the recommendations in the comments, I have specified the type of B with the eltype() function before the loop as follows:
A1 = [1,2,0,0,4,0] #The default type is Array{Int64,1}
B1 = eltype(A1)[] #Define the type of the 0 element array B with the type of A
#B1 = eltype(typeof(A))[] this is also valid
for i=1:endof(A1)
if A1[i] != 0
push!(B1::Array{Int64,1},i::Int64)
end
end
println(B1)
typeof(B1)
Then, the output is again the desired
[1,2,5]
Array{Int64,1}
The simplest way of doing this is using the function find(). However, since I´m a beginner, I wanted to do it in another way. However, there is another alternative provided by #DNF that outperform find() for the cases he has tested it (see below answers).
The second option, which creates an output matrix with the non-zero elements has been provided by other users (#Harrison Grodin and #P i) in this discussion.
Thanks all of you for the help!
You have a few options here.
Using the strategy you started with, you can use push! inside the for loop.
julia> B = Int[]
0-element Array{Int64,1}
julia> for element = A
if element != 0
push!(B, element)
end
end
julia> B
3-element Array{Int64,1}:
1
2
4
You can also opt to use short-circuit evaluation.
julia> for element = A
element != 0 && push!(B, element)
end
julia> for element = A
element == 0 || push!(B, element)
end
Both filter and list comprehensions are valid, as well!
julia> B = [element for element = A if element != 0]
3-element Array{Int64,1}:
1
2
4
julia> filter(n -> n != 0, A)
3-element Array{Int64,1}:
1
2
4
Edit: Thanks to the OP's comment, I have realized that the desired result is the indices of the nonzero elements, not the elements themselves. This can be achieved simply with the following. :)
julia> find(A)
3-element Array{Int64,1}:
1
2
5
Just because I don't see a simple solution posted here, this is approach I took to the same problem. I think it is the simplest and most elegant solution. Hope this closes the thread!
julia> A = [1,2,0,0,4,0];
julia> findall(!iszero,A)
3-element Vector{Int64}:
1
2
5
I am trying to port some code and now I've hit a sticky bit. The original code is in C++. I need to port a union that has two 32 bit ints (in an array) and a double.
So far I have:
I1 = UInt32(56) # arbitrary integer values for example
I2 = UInt32(1045195987)
# do transforms on I1 and I2 as per the code I'm porting
A = bits(I1)
B = bits(I2)
return parse(Float64, string(A,B))
Is this the way to do it? The string operation seems expensive. Any advice appreciated.
I also come from mostly C/C++ programming, and this is what I do to handle the problem:
First, create an immutable type with two UInt32 elements:
immutable MyType
a::UInt32
b::UInt32
end
Then you can convert a vector of Float64 to that type with reinterpret.
For example:
julia> x = [1.5, 2.3]
2-element Array{Float64,1}:
1.5
2.3
julia> immutable MyType ; a::UInt32 ; b::UInt32 ; end
julia> y = reinterpret(MyType, x)
2-element Array{MyType,1}:
MyType(0x00000000,0x3ff80000)
MyType(0x66666666,0x40026666)
julia> x[1]
1.5
julia> y[1]
MyType(0x00000000,0x3ff80000)
julia> y[1].a
0x00000000
julia> y[1].b
0x3ff80000
Note: the two vectors still point to the same memory, so you can even update elements, using either type.
julia> x[1] = 10e91
1.0e92
julia> y[1].a
0xbf284e24
julia> y[1].b
0x53088ba3
julia> y[1] = MyType(1,2)
MyType(0x00000001,0x00000002)
julia> x[1]
4.2439915824e-314
In python I can do nested list comprehensions, for instance I can flatten the following array thus:
a = [[1,2,3],[4,5,6]]
[i for arr in a for i in arr]
to get [1,2,3,4,5,6]
If I try this syntax in Julia I get:
julia> a
([1,2,3],[4,5,6],[7,8,9])
julia> [i for arr in a for i in arr]
ERROR: syntax: expected ]
Are nested list comprehensions in Julia possible?
This feature has been added in julia v0.5:
julia> a = ([1,2,3],[4,5,6],[7,8,9])
([1,2,3],[4,5,6],[7,8,9])
julia> [i for arr in a for i in arr]
9-element Array{Int64,1}:
1
2
3
4
5
6
7
8
9
List comprehensions work a bit differently in Julia:
> [(x,y) for x=1:2, y=3:4]
2x2 Array{(Int64,Int64),2}:
(1,3) (1,4)
(2,3) (2,4)
If a=[[1 2],[3 4],[5 6]] was a multidimensional array, vec would flatten it:
> vec(a)
6-element Array{Int64,1}:
1
2
3
4
5
6
Since a contains tuples, this is a bit more complicated in Julia. This works, but likely isn't the best way to handle it:
function flatten(x, y)
state = start(x)
if state==false
push!(y, x)
else
while !done(x, state)
(item, state) = next(x, state)
flatten(item, y)
end
end
y
end
flatten(x)=flatten(x,Array(Any, 0))
Then, we can run:
> flatten([(1,2),(3,4)])
4-element Array{Any,1}:
1
2
3
4
You can get some mileage out of using the splat operator with the array constructor here (transposing to save space)
julia> a = ([1,2,3],[4,5,6],[7,8,9])
([1,2,3],[4,5,6],[7,8,9])
julia> [a...]'
1x9 Array{Int64,2}:
1 2 3 4 5 6 7 8 9
Any reason why you're using a tuple of vectors? It's much simpler with arrays, as Ben has already shown with vec. But you can also use comprehensions pretty simply in either case:
julia> a = ([1,2,3],[4,5,6],[7,8,9]);
julia> [i for i in hcat(a...)]
9-element Array{Any,1}:
1
2
⋮
The expression hcat(a...) "splats" your tuple and concatenates it into an array. But remember that, unlike Python, Julia uses column-major array semantics. You have three column vectors in your tuple; is that what you intend? (If they were row vectors — delimited by spaces — you could just use [a...] to do the concatenation). Arrays are iterated through all elements, regardless of their dimensionality.
Don't have enough reputation for comment so posting a modification #ben-hammer. Thanks for the example of flatten(), it was helpful to me.
But it did break if the tuples/arrays contained strings. Since strings are iterables the function would further break them down to characters. I had to insert condition to check for ASCIIString to fix that. The code is below
function flatten(x, y)
state = start(x)
if state==false
push!(y, x)
else
if typeof(x) <: String
push!(y, x)
else
while (!done(x, state))
(item, state) = next(x, state)
flatten(item, y)
end
end
end
y
end
flatten(x)=flatten(x,Array(Any, 0))