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I've had good read of the 'Sorting Functions' section of the Julia manual, and had a look at some of the similar questions that have already been asked on this board, but I don't think I've quite found the answer to my question. Apologies if I've missed something.
Essentially I have a vector of vectors, with the enclosed vectors containing integers. For the purposes of the example, each enclosed vector contains 3 integers, but it could be any number. I want to sort the enclosed vectors by the first element, then by the second element, then by the third element etc.
Let's start with the vector:
v = [[3, 6, 1], [2, 2, 6], [1, 5, 9], [2, 1, 8], [3, 7, 9],
[1, 1, 2], [2, 2, 2], [3, 6, 2], [1, 2, 5], [1, 5, 6],
[3, 7, 4], [2, 1, 4], [2, 2, 1], [3, 1, 2], [1, 2, 8]]
And continue with what I'm actually looking for:
v = [[1, 1, 2], [1, 2, 5], [1, 2, 8], [1, 5, 6], [1, 5, 9],
[2, 1, 4], [2, 1, 8], [2, 2, 1], [2, 2, 2], [2, 2, 6],
[3, 1, 2], [3, 6, 1], [3, 6, 2], [3, 7, 4], [3, 7, 9]]
So there should be no requirement for rocket science.
I can easily sort the vector by the first element of the enclosed vectors by one of two ways:
v = sort(v, lt = (x, y) -> isless(x[1], y[2]))
or:
v = sort(v, by = x -> x[1])
Both these methods produce the same answer:
v = [[1, 5, 9], [1, 1, 2], [1, 2, 5], [1, 5, 6], [1, 2, 8],
[2, 2, 6], [2, 1, 8], [2, 2, 2], [2, 1, 4], [2, 2, 1],
[3, 6, 1], [3, 7, 9], [3, 6, 2], [3, 7, 4], [3, 1, 2]]
So, as you can see, I have sorted by the first element of the enclosed vectors, but not by the subsequent elements.
So, to come back to the question in the title, is there a method of sorting by multiple elements using the sort() function?
I can actually get what I want using loops:
for i = 3:-1:1
v = sort(v, lt = (x, y) -> isless(x[i], y[i]))
end
or:
for i = 3:-1:1
v = sort(v, by = x -> x[i])
end
However, I don't want to re-invent the wheel, so if there's a way of doing it within the sort() function I'd love to learn about it.
You can use lexless function as lt keyword argument that does exactly what you want if I understood your question correctly:
julia> sort(v, lt=lexless)
15-element Array{Array{Int64,1},1}:
[1, 1, 2]
[1, 2, 5]
[1, 2, 8]
[1, 5, 6]
[1, 5, 9]
[2, 1, 4]
[2, 1, 8]
[2, 2, 1]
[2, 2, 2]
[2, 2, 6]
[3, 1, 2]
[3, 6, 1]
[3, 6, 2]
[3, 7, 4]
[3, 7, 9]
EDIT: I have just checked that this is a solution for Julia 0.6. In Julia 0.7 you can simply write:
julia> sort(v)
15-element Array{Array{Int64,1},1}:
[1, 1, 2]
[1, 2, 5]
[1, 2, 8]
[1, 5, 6]
[1, 5, 9]
[2, 1, 4]
[2, 1, 8]
[2, 2, 1]
[2, 2, 2]
[2, 2, 6]
[3, 1, 2]
[3, 6, 1]
[3, 6, 2]
[3, 7, 4]
[3, 7, 9]
Using combination method on Ruby,
[1, 2, 3, 4, 5, 6].combination(2).to_a
#=> [[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 3],
# [2, 4], [2, 5], [2, 6], [3, 4], [3, 5], [3, 6],
# [4, 5], [4, 6], [5, 6]]
we can get a 2-dimensional array having 15 (6C2) elements.
I would like to create a fair_combination method that returns an array like this:
arr = [[1, 2], [3, 5], [4, 6],
[3, 4], [5, 1], [6, 2],
[5, 6], [1, 3], [2, 4],
[2, 3], [4, 5], [6, 1],
[1, 4], [2, 5], [3, 6]]
So that every three sub-arrays (half of 6) contain all the given elements:
arr.each_slice(3).map { |a| a.flatten.sort }
#=> [[1, 2, 3, 4, 5, 6],
# [1, 2, 3, 4, 5, 6],
# [1, 2, 3, 4, 5, 6],
# [1, 2, 3, 4, 5, 6],
# [1, 2, 3, 4, 5, 6]]
This makes it kind of "fair", by using as different elements as possible as arrays go on.
To make it more general, what it needs to satisfy is as follows:
(1) As you follow the arrays from start and count how many times each number appears, at any point it should be as flat as possible;
(1..7).to_a.fair_combination(3)
#=> [[1, 2, 3], [4, 5, 6], [7, 1, 4], [2, 5, 3], [6, 7, 2], ...]
The first 7 numbers make [1,2,...,7] and so do the following 7 numbers.
(2) Once number A comes in the same array with B, A does not want to be in the same array with B if possible.
(1..10).to_a.fair_combination(4)
#=> [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 1, 5], [2, 6, 9, 3], [4, 7, 10, 8], ...]
Is there any good algorithm that creates a "fair combination" like this ?
It's not guaranteed to give the best solution, but it gives a good enough one.
At each step, it chooses a minimal subpool which is the set of items of minimal height, for which there is still a combination to choose from (height is the number of times the items have been used before).
For instance, let the enumerator be
my_enum = FairPermuter.new('abcdef'.chars, 4).each
The first iteration may return
my_enum.next # => ['a', 'b', 'c', 'd']
At this point those letters have height 1, but there is not enough letters of height 0 to make a combination, so take just all of them for the next:
my_enum.next # => ['a', 'b', 'c', 'e'] for instance
Now the heights are 2 for a, b and c, 1 for d and e, and 0 for f, and still the optimal pool is the full initial set.
So this is not really optimized for combinations of large size. On the other side, if the size of the combination is at most half of the size of the initial set, then the algorithm is pretty decent.
class FairPermuter
def initialize(pool, size)
#pool = pool
#size = size
#all = Array(pool).combination(size)
#used = []
#counts = Hash.new(0)
#max_count = 0
end
def find_valid_combination
[*0..#max_count].each do |height|
candidates = #pool.select { |item| #counts[item] <= height }
next if candidates.size < #size
cand_comb = [*candidates.combination(#size)] - #used
comb = cand_comb.sample
return comb if comb
end
nil
end
def each
return enum_for(:each) unless block_given?
while combination = find_valid_combination
#used << combination
combination.each { |k| #counts[k] += 1 }
#max_count = #counts.values.max
yield combination
return if #used.size >= [*1..#pool.size].inject(1, :*)
end
end
end
Results for fair combinations of 4 over 6
[[1, 2, 4, 6], [3, 4, 5, 6], [1, 2, 3, 5],
[2, 4, 5, 6], [2, 3, 5, 6], [1, 3, 5, 6],
[1, 2, 3, 4], [1, 3, 4, 6], [1, 2, 4, 5],
[1, 2, 3, 6], [2, 3, 4, 6], [1, 2, 5, 6],
[1, 3, 4, 5], [1, 4, 5, 6], [2, 3, 4, 5]]
Results of fair combination of 2 over 6
[[4, 6], [1, 3], [2, 5],
[3, 5], [1, 4], [2, 6],
[4, 5], [3, 6], [1, 2],
[2, 3], [5, 6], [1, 6],
[3, 4], [1, 5], [2, 4]]
Results of fair combinations of 2 over 5
[[4, 5], [2, 3], [3, 5],
[1, 2], [1, 4], [1, 5],
[2, 4], [3, 4], [1, 3],
[2, 5]]
Time to get combinations of 5 over 12:
1.19 real 1.15 user 0.03 sys
Naïve implementation would be:
class Integer
# naïve factorial implementation; no checks
def !
(1..self).inject(:*)
end
end
class Range
# constant Proc instance for tests; not needed
C_N_R = -> (n, r) { n.! / ( r.! * (n - r).! ) }
def fair_combination(n)
to_a.permutation
.map { |a| a.each_slice(n).to_a }
.each_with_object([]) do |e, memo|
e.map!(&:sort)
memo << e if memo.all? { |me| (me & e).empty? }
end
end
end
▶ (1..6).fair_combination(2)
#⇒ [
# [[1, 2], [3, 4], [5, 6]],
# [[1, 3], [2, 5], [4, 6]],
# [[1, 4], [2, 6], [3, 5]],
# [[1, 5], [2, 4], [3, 6]],
# [[1, 6], [2, 3], [4, 5]]]
▶ (1..6).fair_combination(3)
#⇒ [
# [[1, 2, 3], [4, 5, 6]],
# [[1, 2, 4], [3, 5, 6]],
# [[1, 2, 5], [3, 4, 6]],
# [[1, 2, 6], [3, 4, 5]],
# [[1, 3, 4], [2, 5, 6]],
# [[1, 3, 5], [2, 4, 6]],
# [[1, 3, 6], [2, 4, 5]],
# [[1, 4, 5], [2, 3, 6]],
# [[1, 4, 6], [2, 3, 5]],
# [[1, 5, 6], [2, 3, 4]]]
▶ Range::C_N_R[6, 3]
#⇒ 20
Frankly, I do not understand how this function should behave for 10 and 4, but anyway this implementation is too memory consuming to work properly on big ranges (on my machine it gets stuck on ranges of size > 8.)
To adjust this to more robust solution one needs to get rid of permutation there in favor of “smart concatenate permuted arrays.”
Hope this is good for starters.
I was studying how to list out all divisors of a number and came across this solution by Marc-Andre here. In his solution, there is one part of the code which does something like this:
array.product(*arrays_of_array) # the asterisk seems to have done sth.
I tried it in irb to try play around but I couldn't make sense of the outputs. I tried:
a=[0,1,2]
b=[3,4]
c=[[5,6],[7,8]]
I understand that array.product(other_array) is a method to list all combinations of the two arrays into one. With this knowledge, I tested out several experiments
a.product(b) => [[0, 3], [0, 4], [1, 3], [1, 4], [2, 3], [2, 4]] / 6 elements
a.product(*b) => TypeError: no implicit conversion of Fixnum into Array
a.product(c) => [[0, [5, 6]], [0, [7, 8]], [1, [5, 6]], [1, [7, 8]], [2, [5, 6]], [2, [7, 8]]] / 6 elements
a.product(*c) => [[0, 5, 7], [0, 5, 8], [0, 6, 7], [0, 6, 8], [1, 5, 7], [1, 5, 8], [1, 6, 7], [1, 6, 8], [2, 5, 7], [2, 5, 8], [2, 6, 7], [2, 6, 8]]
From observation, It seems the asterisk (*) has to be applied to a multi-dimensional array? (i.e. matrix?). Without the asterisk, the product returns 6 elements and the combinations only one level. While with the asterisk, the combination will go 1 level deeper and returns 12 elements, and combine until there is no array within the combinations. Where can I find more examples to study this behaviour of the asterisk?
Edit:
I tried to introduce one more variable
d=[[[9,0],[1,2]],[[3,4],[5,6]]]
a.product(*d) => [[0, [9, 0], [3, 4]], [0, [9, 0], [5, 6]], [0, [1, 2], [3, 4]], [0, [1, 2], [5, 6]], [1, [9, 0], [3, 4]], [1, [9, 0], [5, 6]], [1, [1, 2], [3, 4]], [1, [1, 2], [5, 6]], [2, [9, 0], [3, 4]], [2, [9, 0], [5, 6]], [2, [1, 2], [3, 4]], [2, [1, 2], [5, 6]]]
So the asterisk sign only makes it go one level deeper.
In the context of finding the list of divisors. Can anyone explain what the code exactly does?
require 'prime'
def factors_of(number)
primes, powers = number.prime_division.transpose
exponents = powers.map{|i| (0..i).to_a}
divisors = exponents.shift.product(*exponents).map do |powers|
primes.zip(powers).map{|prime, power| prime ** power}.inject(:*)
end
divisors.sort.map{|div| [div, number / div]}
end
p factors_of(4800) # => [[1, 4800], [2, 2400], ..., [4800, 1]]
*(splat) is used to expand collections.
In your example, with b = [3,4],
a.product(*b)
is equivalent to
a.product(3, 4)
which generates an error because Array#product expects an Array as argument, not two integers.
I am trying to find sum of each array.
(1..9).to_a.combination(3).to_a.each{ |item| item.inject{:+}}
But my code gives the followings.
[[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 6], [1, 2, 7], [1, 2, 8], [1, 2, 9],
[1, 3, 4], [1, 3, 5], [1, 3, 6], [1, 3, 7], [1, 3, 8], ...
What I am expecting is something like this.
[6, 7, 8, 9,...]
How can I find sum of each array?
You are very close, a little change on your code may help:
(1..9).to_a.combination(3).map { |a| a.inject(:+) }
(1..9).to_a.combination(3).to_a.map { |item| item.inject(:+) }
I found another way with #reduce.
(1..9).to_a.combination(3).map { |item| item.reduce(:+) }
I have the following multi-dimensional array in Ruby:
[[1,2], [3], [4,5,6]]
I need to have the following output:
[[1,3,4], [1,3,5], [1,3,6], [2,3,4], [2,3,5], [2,3,6]]
I have tried creating a recursive function, but I'm not having much luck.
Are there any Ruby functions that would help with this? Or is the only option to do it recursively?
Thanks
Yup, Array#product does just that (Cartesian product):
a = [[1,2], [3], [4,5,6]]
head, *rest = a # head = [1,2], rest = [[3], [4,5,6]]
head.product(*rest)
#=> [[1, 3, 4], [1, 3, 5], [1, 3, 6], [2, 3, 4], [2, 3, 5], [2, 3, 6]]
Another variant:
a.inject(&:product).map(&:flatten)
#=> [[1, 3, 4], [1, 3, 5], [1, 3, 6], [2, 3, 4], [2, 3, 5], [2, 3, 6]]