I was wondering if anyone has a clear way to return a row vector from a slice of a matrix?
matrix[k,:]
actually returns a vector of size (n,), so it's automatically treated by Julia as a column vector, not a row vector. So far, I've been using the syntax
matrix[k,:]'
to represent the row vector of size (1,n), but somehow this seems clunky and not at all intuitive that this is a row vector and not a column vector. Most mathematicians I've talked to assume incorrectly that this isn't a row vector.
Is there a more Julianic way to get a row vector slice, that is more clear?
julia> m=rand(1:10,3,4)
3×4 Array{Int64,2}:
4 5 3 9
6 8 1 5
4 5 3 4
julia> m[[1],:]
1×4 Array{Int64,2}:
4 5 3 9
Note that it is almost always better to use views to avoid data copying:
julia> #view m[[1],:]
1×4 view(::Array{Int64,2}, [1], :) with eltype Int64:
4 5 3 9
Explanation:
Selection of Array elements can be done either by using scalars or by using iterables. Using a scalar causes the given dimension to be dropped. On the other hand using a collection does not drop the dimension. For better explanation consider yet another example:
julia> m[1:1,1:1]
1×1 Array{Int64,2}:
4
You can see that only one element has been selected but the dimensions have not been dropped. Hence you end-up having a Matrix (that is 2-dimensional Array) having only one row and one column.
EDIT (comment by Colin T Bowers - thank you!)
Deciding whether or not to use views is not trivia - mostly this is whether your code benefits from buffering. However, surprisingly often views are better.
julia> const vals = rand(200,200);
julia> using BenchmarkTools
julia> #btime sum(view(vals,1,:))
169.392 ns (1 allocation: 48 bytes)
95.08924081258299
julia> #btime sum(vals[1,:])
184.384 ns (1 allocation: 1.77 KiB)
95.08924081258299
Let us define our own summing function.
julia> function mysum(a::AbstractVector{A}) where A <: Number
v = zero(A)
#inbounds #simd for i in 1:length(a)
v += a[i]
end
v
end;
julia> #btime mysum(view(vals,1,:))
141.931 ns (0 allocations: 0 bytes)
95.08924081258299
julia> #btime mysum(vals[1,:])
174.934 ns (1 allocation: 1.77 KiB)
95.08924081258297
It can be clearly seen that when you are summing row once views are still better.
Last but not least summing by columns is of course several time faster and copying the data is expensive as hell:
julia> #btime sum(view(vals,:,1))
25.828 ns (1 allocation: 48 bytes)
96.04440265541243
julia> #btime mysum(view(vals,:,1))
13.927 ns (0 allocations: 0 bytes)
96.04440265541243
julia> #btime sum(vals[:,1])
167.745 ns (1 allocation: 1.77 KiB)
96.04440265541243
To create a vector from a matrix row, you can simply use:
Vector(m[i, :])
Related
I am trying to do a simple function to check the differences between factorial and Stirling's approximation:
using DataFrames
n = 24
df_res = DataFrame(i = BigInt[],
f = BigInt[],
st = BigInt[])
for i in 1:n
fac = factorial(big(i))
sterling = i^i*exp(-i)*sqrt(i)*sqrt(2*pi)
res = DataFrame(i = [i],
f = [fac],
st = [sterling])
df_res = [df_res;res]
end
first(df_res, 24)
The result for sterling when i= 16 and i= 24 is 0!. So, I checked power for both values and the result is 0:
julia> 16^16
0
julia> 24^24
0
I did the same code in R, and there are no issues. What am I doing wrong or what I don't know about Julia and I probably should?
It appears that Julia integers are either 32-bit or 64-bit, depending on your system, according to the Julia documentation for Integers and Floating-Point Numbers. Your exponentiation is overflowing your values, even if they're 64 bits.
Julia looks like it supports Arbitrary Precision Arithmetic, which you'll need to store the large resultant values.
According to the Overflow Section, writing big(n) makes n arbitrary precision.
While the question has been answered at the another post one more thing is worth saying.
Julia is one of very few languages that allows you to define your very own primitive types - so you can be still with fast fixed precision numbers yet handle huge values. There is a package BitIntegers for that:
BitIntegers.#define_integers 512
Now you can do:
julia> Int512(2)^500
3273390607896141870013189696827599152216642046043064789483291368096133796404674554883270092325904157150886684127560071009217256545885393053328527589376
Usually you will get better performance for for even big fixed point arithmetic numbers. For an example:
julia> #btime Int512(2)^500;
174.687 ns (0 allocations: 0 bytes)
julia> #btime big(2)^500;
259.643 ns (9 allocations: 248 bytes)
There is a simple solution to your problem that does not involve using BigInt or any specialized number types, and which is much faster. Simply tweak your mathematical expression slightly.
foo(i) = i^i*exp(-i)*sqrt(i)*sqrt(2*pi) # this is your function
bar(i) = (i / exp(1))^i * sqrt(i) * sqrt(2*pi) # here's a better way
OK, let's test it:
1.7.2> foo(16)
0.0 # oops. not what we wanted
1.7.2> foo(big(16)) # works
2.081411441522312838373895982304611417026205959453251524254923609974529540404514e+13
1.7.2> bar(16) # also works
2.0814114415223137e13
Let's try timing it:
1.7.2> using BenchmarkTools
1.7.2> #btime foo(n) setup=(n=16)
18.136 ns (0 allocations: 0 bytes)
0.0
1.7.2> #btime foo(n) setup=(n=big(16))
4.457 μs (25 allocations: 1.00 KiB) # horribly slow
2.081411441522312838373895982304611417026205959453251524254923609974529540404514e+13
1.7.2> #btime bar(n) setup=(n=16)
99.682 ns (0 allocations: 0 bytes) # pretty fast
2.0814114415223137e13
Edit: It seems like
baz(i) = float(i)^i * exp(-i) * sqrt(i) * sqrt(2*pi)
might be an even better solution, since the numerical values are closer to the original.
I have a concrete Array and want to efficiently construct a similar array of the same dimensions filled with ones. What would be the recommended approach?
Here's a random array to work with:
julia> A = rand(0:1, 10, 5)
10×5 Matrix{Int64}:
or A = rand(0:1., 10, 5) (with a dot on 0. and/or 1.) for a random matrix of floats.
Two approaches are very natural. I could do this:
julia> zero(A) .+ 1
5×10 Matrix{Int64}:
Or I could do it this way:
julia> repeat(ones(size(A)[2])', outer = size(A)[1])
5×10 Matrix{Float64}:
The first approach is more elegant. The second approach feels more clunky and prone to error (accidentally exchanging [1] and [2]), but at the same time it doesn't involve the addition operation and so possibly involves fewer allocations (or maybe not because the compiler is super smart Edit: quick benchmark below suggests the compiler is super smart).
And of course there may be another, better approach.
using BenchmarkTools
A = rand(0:1, 1000, 1000)
#btime zero(A) .+ 1
## 1.609 ms (6 allocations: 15.26 MiB)
#btime repeat(ones(size(A)[2])', outer = size(A)[1])
## 3.032 ms (10 allocations: 7.64 MiB)
Edit 2: Follow-up onBogumił's answer
The following method for a unit-array J, defined for convenience, is efficient:
function J(A::AbstractArray{T,N}) where {T,N}
ones(T, size(A))
end
J(A)
#btime J(A)
## 789.929 μs (2 allocations: 7.63 MiB)
What about:
ones(Int, size(A))
or
fill(1, size(A))
Let's say I have a vector
v = Any[1,2,3,4]
And I would like to recompute its eltype in such a way that
typeof(v) = Vector{Int}
Is it possible to accomplish this without having to manually concatenate each of the elements in v?
You can't "retype" the existing v, just create a copy of it with the more concrete type1.
Conversion
Assuming you already (statically) know the result type, you have multiple options. Most readable (and IMO, idiomatic) would be
Vector{Int}(v)
which is almost equivalent to
convert(Vector{Int}, v)
except that the latter does not copy if the input types is already the target type. Alternatively:
convert.(Int, v)
which surely copies as well.
What to convert to
If you don't know what the "common type" would be, there are multiple options how to get one that matches. In general, typejoin can be used to find a least upper bound:
mapreduce(typeof, typejoin, v; init=Union{})
The result will most likely be abstract, e.g. Real for an array of Ints and Float64s. So, for numeric types, you might be better off with promote_type:
mapreduce(typeof, promote_type, v; init=Union{}) # or init=Number
This at least gives you Float64 for mixed Ints and Float64s.
But all of this is not really recommended, since it might be fragile, surprising, and is certainly not type stable.
1For certain combinations of types, with compatible binary form, reinterpret will work and return a view with a different type, but this is only possible for bits types, which Any is not. For converting Any[1,2,3] to Int[1,2,3] copying is fundamentally necessary because the two arrays have different layouts in memory: the former is an array of pointers to individually allocated integers objects, whereas the latter stores the Int values inline in contiguous memory.
If you don't know the output type, then consider using a comprehension
foo(v) = [x for x in v]
This is considerably faster on my computer than identity.(v):
julia> v = Any[1,2,3,4];
julia> #btime foo($v);
153.018 ns (2 allocations: 128 bytes)
julia> #btime identity.($v);
293.908 ns (5 allocations: 208 bytes)
julia> #btime foo(v) setup=(v=Any[rand(0:9) for _ in 1:1000]);
1.331 μs (2 allocations: 7.95 KiB)
julia> #btime identity.(v) setup=(v=Any[rand(0:9) for _ in 1:1000]);
25.498 μs (494 allocations: 15.67 KiB)
This is a quick and dirty trick that usually solves the problem
julia> v = Any[1,2,3,4]
4-element Array{Any,1}:
1
2
3
4
julia> identity.(v)
4-element Array{Int64,1}:
1
2
3
4
I have two versions of code that seem to do the same thing:
sum = 0
for x in 1:100
sum += x
end
sum = 0
for x in collect(1:100)
sum += x
end
Is there a practical difference between the two approaches?
In Julia, 1:100 returns a particular struct called UnitRange that looks like this:
julia> dump(1:100)
UnitRange{Int64}
start: Int64 1
stop: Int64 100
This is a very compact struct to represent ranges with step 1 and arbitrary (finite) size. UnitRange is subtype of AbstractRange, a type to represent ranges with arbitrary step, subtype of AbstractVector.
The instances of UnitRange dynamically compute their elements whenever the you use getindex (or the syntactic sugar vector[index]). For example, with #less (1:100)[3] you can see this method:
function getindex(v::UnitRange{T}, i::Integer) where {T<:OverflowSafe}
#_inline_meta
val = v.start + (i - 1)
#boundscheck _in_unit_range(v, val, i) || throw_boundserror(v, i)
val % T
end
This is returning the i-th element of the vector by adding i - 1 to the first element (start) of the range. Some functions have optimised methods with UnitRange, or more generally with AbstractRange. For instance, with #less sum(1:100) you can see the following
function sum(r::AbstractRange{<:Real})
l = length(r)
# note that a little care is required to avoid overflow in l*(l-1)/2
return l * first(r) + (iseven(l) ? (step(r) * (l-1)) * (l>>1)
: (step(r) * l) * ((l-1)>>1))
end
This method uses the formula for the sum of an arithmetic progression, which is extremely efficient as it's evaluated in a time independent of the size of the vector.
On the other hand, collect(1:100) returns a plain Vector with one hundred elements 1, 2, 3, ..., 100. The main difference with UnitRange (or other types of AbstractRange) is that getindex(vector::Vector, i) (or vector[i], with vector::Vector) doesn't do any computation but simply accesses the i-th element of the vector. The downside of a Vector over a UnitRange is that generally speaking there aren't efficient methods when working with them as the elements of this container are completely arbitrary, while UnitRange represents a set of numbers with peculiar properties (sorted, constant step, etc...).
If you compare the performance of methods for which UnitRange has super-efficient implementations, this type will win hands down (note the use of interpolation of variables with $(...) when using macros from BenchmarkTools):
julia> using BenchmarkTools
julia> #btime sum($(1:1000_000))
0.012 ns (0 allocations: 0 bytes)
500000500000
julia> #btime sum($(collect(1:1000_000)))
229.979 μs (0 allocations: 0 bytes)
500000500000
Remember that UnitRange comes with the cost of dynamically computing the elements every time you access them with getindex. Consider for example this function:
function test(vec)
sum = zero(eltype(vec))
for idx in eachindex(vec)
sum += vec[idx]
end
return sum
end
Let's benchmark it with a UnitRange and a plain Vector:
julia> #btime test($(1:1000_000))
812.673 μs (0 allocations: 0 bytes)
500000500000
julia> #btime test($(collect(1:1000_000)))
522.828 μs (0 allocations: 0 bytes)
500000500000
In this case the function calling the plain array is faster than the one with a UnitRange because it doesn't have to dynamically compute 1 million elements.
Of course, in these toy examples it'd be more sensible to iterate over all elements of vec rather than its indices, but in real world cases a situation like these may be more sensible. This last example, however, shows that a UnitRange is not necessarily more efficient than a plain array, especially if you need to dynamically compute all of its elements. UnitRanges are more efficient when you can take advantage of specialised methods (like sum) for which the operation can be performed in constant time.
As a file remark, note that if you originally have a UnitRange it's not necessarily a good idea to convert it to a plain Vector to get good performance, especially if you're going to use it only once or very few times, as the conversion to Vector involves itself the dynamic computation of all elements of the range and the allocation of the necessary memory:
julia> #btime collect($(1:1000_000));
422.435 μs (2 allocations: 7.63 MiB)
julia> #btime test(collect($(1:1000_000)))
882.866 μs (2 allocations: 7.63 MiB)
500000500000
In Julia I want to find the column index of a matrix for the maximum value in each row, with the result being a Vector{Int}. Here is how I am doing it currently (Samples has 7 columns and 10,000 rows):
mxindices = [ i[2] for i in findmax(Samples, dims = 2)[2]][:,1]
This works but feels rather clumsy and verbose. Wondered if there was a better way.
Even simpler: Julia has an argmax function and Julia 1.1+ has an eachrow iterator. Thus:
map(argmax, eachrow(x))
Simple, readable, and fast — it matches the performance of Colin's f3 and f4 in my quick tests.
UPDATE: For the sake of completeness, I've added Matt B.'s excellent solution to the test-suite (and I also forced the transpose in f4 to generate a new matrix rather than a lazy view).
Here are some different approaches (yours is the base-case f0):
f0(x) = [ i[2] for i in findmax(x, dims = 2)[2]][:,1]
f1(x) = getindex.(argmax(x, dims=2), 2)
f2(x) = [ argmax(vec(x[n,:])) for n = 1:size(x,1) ]
f3(x) = [ argmax(vec(view(x, n, :))) for n = 1:size(x,1) ]
f4(x) = begin ; xt = Matrix{Float64}(transpose(x)) ; [ argmax(view(xt, :, k)) for k = 1:size(xt,2) ] ; end
f5(x) = map(argmax, eachrow(x))
Using BenchmarkTools we can examine the efficiency of each (I've set x = rand(100, 200)):
julia> #btime f0($x);
76.846 μs (13 allocations: 4.64 KiB)
julia> #btime f1($x);
76.594 μs (11 allocations: 3.75 KiB)
julia> #btime f2($x);
53.433 μs (103 allocations: 177.48 KiB)
julia> #btime f3($x);
43.477 μs (3 allocations: 944 bytes)
julia> #btime f4($x);
73.435 μs (6 allocations: 157.27 KiB)
julia> #btime f5($x);
43.900 μs (4 allocations: 960 bytes)
So Matt's approach is the fairly obvious winner, as it appears to just be a syntactically cleaner version of my f3 (the two probably compile to something very similar, but I think it would be overkill to check that).
I was hoping f4 might have an edge, despite the temporary created via instantiating the transpose, since it could operate on the columns of a matrix rather than the rows (Julia is a column-major language, so operations on columns will always be faster since the elements are synchronous in memory). But it doesn't appear to be enough to overcome the disadvantage of the temporary.
Note, if it is ever the case that you want the full CartesianIndex, that is, both the row and column index of the maximum in each row, then obviously the appropriate solution is just argmax(x, dims=2).
Mapslices function is also a great option for this problem:
julia> Samples = rand(10000, 7);
julia> res = mapslices(row -> findmax(row)[2], Samples, dims=[2])[:,1];
julia> res[1:10]
10-element Array{Int64,1}:
3
1
3
5
4
4
1
4
5
3
Although this is a lot slower than what Colin suggested above, it might be more readable for some people. This is essentially exactly the same code as you started out with, but uses mapslices instead of list comprehensions.