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I'm new to Pytorch and having an issue with the gather() function:
I have a 3d tensor, x[i,j,k]:
x=tensor([[[1,2,3],
[4,5,6],
[7,8,9]],
[[10,11,12],
[13,14,15],
[16,17,18]]])
I have an index tensor:
index=tensor([[1,2,0]])
I want to use the values of index to iterate over x[j] and fetch the (complete) rows. I've tried gather() with all dims, squeezing, unsqueezing and it never seems to get the output I'm looking for, which would be:
output=tensor([[[4,5,6],
[7,8,9],
[1,2,3]],
[[13,14,15],
[16,17,18],
[10,11,12]]])
I've also tried repeating the values of index to get the same shape as x but it did not work.
I know I can do this with an if loop, but I'm pretty sure I can do it with gather() as well. Thanks for the help
Let us set up the two tensors x and index:
>>> x = torch.arange(1,19).view(2,3,3)
>>> index = torch.tensor([[1,2,0]])
What you are looking for is the torch.gather operation:
out[i][j][k] = x[i][index[i][j][k]][k]
In other to apply this function, you need to expand index to the same shape as out. Additionally, a transpose operation is required to flip your original index tensor.
>>> i = index.T.expand_as(x)
tensor([[[1, 1, 1],
[2, 2, 2],
[0, 0, 0]],
[[1, 1, 1],
[2, 2, 2],
[0, 0, 0]]])
If you compare with the pseudo code line above, you can see how every element of i represents the row of the original tensor x the operator will gather values from.
Applying the function gets us to the desired result:
x.gather(dim=1, index=index.T.expand_as(x))
tensor([[[ 4, 5, 6],
[ 7, 8, 9],
[ 1, 2, 3]],
[[13, 14, 15],
[16, 17, 18],
[10, 11, 12]]])
Let's say I have a vector a = [1, 0, 1, 2, 3, 4, 5, 0, 5, 6, 7, 8, 0, 9, 0] and I want to split it to smaller vectors based on a condition depending on value in that array. E.g. value being zero.
Thus I want to obtain vector of following vectors
[1, 0]
[1, 2, 3, 4, 5, 0]
[5, 6, 7, 8, 0]
[9, 0]
So far this was working for me as a naive solution, but it loses the type.
function split_by_λ(a::Vector, λ)
b = []
temp = []
for i in a
push!(temp, i)
if λ(i)
push!(b, temp)
temp = []
end
end
b
end
split_by_λ(a, isequal(0))
Then I tried to play with ranges, which feels a bit more idiomatic, and does not lose the type.
function split_by_λ(a::Vector, λ)
idx = findall(λ, a)
ranges = [(:)(i==1 ? 1 : idx[i-1]+1, idx[i]) for i in eachindex(idx)]
map(x->a[x], ranges)
end
split_by_λ(a, isequal(0))
but it still feels very cumbersome regarding it's a rather simple task.
Is there something I'm missing, some easier way?
Maybe someone has a shorter idea but here is mine:
julia> inds = vcat(0,findall(==(0),a),length(a))
julia> getindex.(Ref(a), (:).(inds[1:end-1].+1,inds[2:end]))
5-element Array{Array{Int64,1},1}:
[1, 0]
[1, 2, 3, 4, 5, 0]
[5, 6, 7, 8, 0]
[9, 0]
[]
Or if you want to avoid copying a
julia> view.(Ref(a), (:).(inds[1:end-1].+1,inds[2:end]))
5-element Array{SubArray{Int64,1,Array{Int64,1},Tuple{UnitRange{Int64}},true},1}:
[1, 0]
[1, 2, 3, 4, 5, 0]
[5, 6, 7, 8, 0]
[9, 0]
0-element view(::Array{Int64,1}, 16:15) with eltype Int64
Pretty much the same as Przemyslaw's answer, but maybe less cryptic dense:
function split_by(λ, a::Vector)
first, last = firstindex(a), lastindex(a)
splits = [first-1; findall(λ, a); last]
s1, s2 = #view(splits[1:end-1]), #view(splits[2:end])
return [view(a, i1+1:i2) for (i1, i2) in zip(s1, s2)]
end
Also, I changed the signature to the conventional one of "functions first", which allows you to use do-blocks. Additionally, this should work with offset indexing.
One could surely get rid of the intermediate allocations, but I think that gets ugly without yield:
function split_by(λ, a::Vector)
result = Vector{typeof(view(a, 1:0))}()
l = firstindex(a)
r = firstindex(a)
while r <= lastindex(a)
if λ(a[r])
push!(result, #view(a[l:r]))
l = r + 1
end
r += 1
end
push!(result, #view(a[l:end]))
return result
end
What is julian way to do yield (and yield from) as python do?
Edit: I will try to add small example in python.
Think 4x4 chess board. Find every N moves long path chess king could do. Don't waste memory -> make generator of every path.
if we sign every position with numbers:
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 16
point 0 has 3 neighbors (1, 4, 5). We could find table for every neighbors for every point:
NEIG = [[1, 4, 5], [0, 2, 4, 5, 6], [1, 3, 5, 6, 7], [2, 6, 7], [0, 1, 5, 8, 9], [0, 1, 2, 4, 6, 8, 9, 10], [1, 2, 3, 5, 7, 9, 10, 11], [2, 3, 6, 10, 11], [4, 5, 9, 12, 13], [4, 5, 6, 8, 10, 12, 13, 14], [5, 6, 7, 9, 11, 13, 14, 15], [6, 7, 10, 14, 15], [8, 9, 13], [8, 9, 10, 12, 14], [9, 10, 11, 13, 15], [10, 11, 14]]
Recursive function (generator) which enlarge given path from list of points or from generator of (generator of ...) points:
def enlarge(path):
if isinstance(path, list):
for i in NEIG[path[-1]]:
if i not in path:
yield path[:] + [i]
else:
for i in path:
yield from enlarge(i)
Function (generator) which give every path with given length
def paths(length):
steps = ([i] for i in range(16)) # first steps on every point on board
for _ in range(length-1):
nsteps = enlarge(steps)
steps = nsteps
yield from steps
We could see that there is 905776 paths with length 10:
sum(1 for i in paths(10))
Out[89]: 905776
In ipython we could timeit:
%timeit sum(1 for i in paths(10))
1.21 s ± 15.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
My julia implementation is ugly and much more complicated. And it seems to be slower.
Check out ResumableFunctions.jl
from the README
using ResumableFunctions
#resumable function fibonnaci(n::Int) :: Int
a = 0
b = 1
for i in 1:n-1
#yield a
a, b = b, a+b
end
a
end
for fib in fibonnaci(10)
println(fib)
end
You can use Julia Iterators.
struct Fibonacci
last::Int64
end
function Base.iterate(fibo::Fibonacci)
return 1, (1, 1, 2) # the first output, and the next iteration state
end
function Base.iterate(fibo::Fibonacci, state)
i, a, b = state
if i ≤ fibo.last
return a, (i + 1, b, a + b) # the output, and the next iteration state
end
end
You can then use it like this:
for i in Fibonacci(10)
print(i, " ")
end
Which outputs:
1 1 2 3 5 8 13 21 34 55 89
This can result in excellent performance, but it is often a bit verbose, and it's also tricky to decide on what iteration state to use, and how to find the next element given that state. In your chess example, I would avoid this approach, but in other cases it can come in really handy.
You can use Channels:
function fibo(n)
Channel() do ch
a, b = 0, 1
for _ in 1:n
a, b = b, a + b
put!(ch, a)
end
end
end
Use it like this:
for val in fibo(10)
print(val, " ")
end
Which outputs:
1 1 2 3 5 8 13 21 34 55
To get the yield from behavior, you can just use a for loop. For example, to get the Fibonacci sequence r times:
function repeat_fibo(r, n)
Channel() do ch
for _ in 1:r
for val in fibo(n)
put!(ch, val)
end
end
end
end
For more details, see the docs.
Note that the ResumableFunctions.jl library has some benchmarks showing that their solution is significantly faster than using Channels (see #gggg's answer). Perhaps Channel performance will improve in future Julia versions.
To get better Channel performance, you should set the channel's element type and the channel's size: for example, use Channel{Int64}(100) instead of Channel().
Here's a Julia implementation of your chess problem using Channels:
NEIG = [[1, 4, 5], [0, 2, 4, 5, 6], [1, 3, 5, 6, 7], [2, 6, 7], [0, 1, 5, 8, 9],
[0, 1, 2, 4, 6, 8, 9, 10], [1, 2, 3, 5, 7, 9, 10, 11], [2, 3, 6, 10, 11],
[4, 5, 9, 12, 13], [4, 5, 6, 8, 10, 12, 13, 14],
[5, 6, 7, 9, 11, 13, 14, 15], [6, 7, 10, 14, 15], [8, 9, 13],
[8, 9, 10, 12, 14], [9, 10, 11, 13, 15], [10, 11, 14]]
function paths(start, length)
Channel{Vector{Int64}}(100) do ch
if length == 1
put!(ch, [start])
else
for path in paths(start, length - 1)
for next_step in NEIG[path[end] + 1]
next_step in path || put!(ch, [path; next_step])
end
end
end
end
end
function paths(length)
Channel{Vector{Int64}}(100) do ch
for start in 0:15
for path in paths(start, length)
put!(ch, path)
end
end
end
end
You can count all the paths of length 10 much like in Python:
sum(1 for _ in paths(10))
You can also time it:
#time sum(1 for _ in paths(10))
On my machine this takes about 4 seconds. There are probably ways to optimize this further, but this does show that Channel performance still has some room for improvement.
Using Elliptic curve factorization (in Python) I am able to find the PRIME factors of a 50 digit number in ~0.5 sec. Is there any way I can convert the prime factors to the factors of the number?
What I have realized by testing on small digits (496 and 28), multiplying the prime factors together in a specific order. Then multiplying those numbers almost gives me the factors, but it is not very flexible because I have only gotten the formula of what I need to multiply together from small list of prime factors (1,2,3,5).
Here's my version, which computes the element-wise products of the powerset of the set of prime factors, keeping only those products that are unique:
def divisors(n, fs=[]):
if fs == []: fs = factors(n)
divs = [1]
for f in fs:
temp = divs[:]
for d in divs:
if f * d not in temp:
temp.append(f*d)
divs = temp
return sorted(divs)
And here it is in action:
>>> factors(496)
[2, 2, 2, 2, 31]
>>> divisors(496)
[1, 2, 4, 8, 16, 31, 62, 124, 248, 496]
>>> factors(28)
[2, 2, 7]
>>> divisors(28)
[1, 2, 4, 7, 14, 28]
If the number is factored into powers of primes, like p^a q^b r^c then the possible factors of the number are all numbers of the form p^x q^y r^z for 0 ≤ x ≤ a, 0 ≤ y ≤ b, and 0 ≤ r ≤ z.
Since you can have different numbers of prime factors, this is a little programming problem. Have fun.
Maybe you're looking for the product of each (unique) set in the powerset. So for 18=2*3*3 you want the product for each set in {{},{2},{3},{3},{2,3},{2,3},{3,3},{2,3,3}}, giving you {1, 2, 3, 3, 6, 9, 18}.
You can use sympy module like this:
import sympy
sympy.ntheory.factorint(4960) #Factorization.
sympy.ntheory.divisors(4960) #List all divisors.
I'm almost a complete programming beginner and I've started to go through a Swift ebook from Apple.
The things I read are pretty clear, but once you start to experiment things get tricky :).
I'm stuck with the experiment in the Control Flow section. Here is the initial code:
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
largest = number
}
}
}
largest
And here is the task:
Add another variable to keep track of which kind of number was the
largest, as well as what that largest number was.
As I understand, they want me to add up all the values in each number kind (get a total sum for Prime, Fibonacci and Square) and then compare the result to show the largest result.
But I can't figure out the syntax.
Can someone share any advice on how to tackle this experiment?
Maybe I'm not understanding the problem?
They're just asking you to keep track of which number category the largest number belongs to:
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
var largestkind = ""
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
largest = number
largestkind = kind
}
}
}
largest
largestkind
Alternately you can use closure to make the tasks simpler.
The for loop calculate the sum of each series.
The final reduce finds the series tuple that contains maximum number.
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var sums = Array<(String, Int)>()
for (kind, numbers) in interestingNumbers {
sums = sums + [(kind, numbers.reduce(0, +))]
}
let maxSeries = sums.reduce(("", Int.min), { $0.1 > $1.1 ? $0 : $1 })
println(sums)
println(maxSeries)
Here it from playground using Xcode 8.3 and Swift 3.0
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
let largest = interestingNumbers.map{$0.value}.flatMap{$0}.max()
print(largest)
Optional(25)