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Is there a function, to partition a collection (Enumerable) by some non-boolean predicate into more than 2 partitions? (my math knowledge got rusty - but I think, this was called "factorize" - or at least in German "Faktorisierung")
This follows my own yesterday question (Ruby: how to split a collection by predicate into two collections in ruby?)
and I wonder, is there another standard ruby method I just missed.
Here´s an example, showing the principle:
def factorize(eee)
h={}; h.default_proc= proc{|h,k| h[k]=[] }
eee.each{|e| k=yield e; h[k]<<e }
h.values
end
factorize( 1..10 ){|n| n%3 } #=> [[1, 4, 7, 10], [2, 5, 8], [3, 6, 9]]
You can use group_by: https://apidock.com/ruby/v2_5_5/Enumerable/group_by
Example:
(1..10).group_by { |n| n % 3 }.values
=> [[1, 4, 7, 10], [2, 5, 8], [3, 6, 9]]
While the Enumerable#group_by solution is clearly the answer, it may be instructive to streamline your original attempt using #each_with_object.
def factorize(eee)
eee.each_with_object({}) do |e, hsh|
k = yield e
hsh[k] ||= []
hsh[k] << e
end.values
end
factorize(1..10) { |x| x % 3 }
# => [[1, 4, 7, 10], [2, 5, 8], [3, 6, 9]]
I have this array:
ary = [[1, 6, 7], [1, 4, 9], [1, 8, 3]]
I want to sort it by the first odd number, or the last number if they are all even, in each subarray.
Since the first element in each array is the same object 1 for this particular ary, I can solve this like this:
ary2 = ary.sort_by { |a, b, c| b.odd? ? b : c }
But when I try a more general one:
arr2 = ary.sort_by { |a, b, c| a.odd? ? a : b.odd? ? b : c }
ary2 comes back unsorted.
I tried removing the ternary operators like this:
ary2 = ary.sort_by do |a, b, c|
if a.odd?
a
elsif b.odd?
b
else
c
end
end
with the same effect (i.e., none).
Is there some reason that elsif can't be used in blocks passed to the sort_by method?
Edit: Axiac pointed out the problem with my logic. It looks like conditional logic has to deal with all of the possible permutations of odd and even values. This works:
arr2 = arr.sort_by do |a, b, c|
if a.odd?
if b.odd?
if c.odd?
[a, b, c]
else
[a, b]
end
elsif c.odd?
[a, c]
else
a
end
elsif b.odd?
if c.odd?
[b, c]
else
b
end
else
c
end
end
Maybe there's a more succinct and less brittle way to do it, but it's probably a good idea to do it this way instead:
arr2 = arr.sort_by do |sub_arr|
temp = sub_arr.select do |e|
e.odd?
end
temp.empty? ? Array(sub_arr.last) : temp
end
I'll see myself out.
Regarding your original question, just as axiac points out in the comment, the result of the sorting should be exactly the same as the input array because they are all sorted by the first odd element in each subarray, which is 1, and the sort method is stable in MRI.
Regarding your question after the edit, my answer would be:
ary.sort_by{|a| a[0...-1].select(&:odd?) << a.last}
# => [[1, 8, 3], [1, 6, 7], [1, 4, 9]]
I am pretty confident that this is what you wrote after the edit that you wanted, but I am not sure if this is what you wanted since the sorting mechanism looks strange to me.
I find the statement of the question ambiguous. I will give an answer that is consist with one interpretation. If that is not what you want, please clarify hte question.
def my_sort(arr)
arr.sort_by {|a| a.any?(&:odd?) ? a.map {|e| e.odd? ? e : Float::INFINITY} : [a.last]}
end
my_sort [[1, 6, 7], [1, 4, 9], [1, 2, 3]]
#=> [[1, ∞, 7], [1, ∞, 9], [1, ∞, 3]] (sort_by)
#=> [[1, 2, 3], [1, 6, 7], [1, 4, 9]]
my_sort [[3, 6, 7], [4, 1, 9], [5, 8, 1]]
#=> [[3, ∞, 7], [∞, 1, 9], [5, ∞, 1]] (sort_by)
#=> [[3, 6, 7], [5, 8, 1], [4, 1, 9]]
my_sort [[2, 6, 8], [4, 1, 4], [8, 6, 2]]
#=> [[8], [∞, 1, ∞], [2]] (sort_by)
#=> [[8, 6, 2], [2, 6, 8], [4, 1, 4]]
my_sort [[8, 6, 2], [5, 1, 1], [6, 8, 4]]
#=> [[2], [5, 1, 1], [4] (sort_by)
#=> [[8, 6, 2], [6, 8, 4], [5, 1, 1]]
For each example I've shown the arrays used by sort_by to produce the sort shown on the following line.
I'm trying to remove pairs of the smallest and largest elements from an Array and store them in a second one. Is there a better way to do this or a Ruby method I don't know about that could accomplish something like this?
Here's my code:
nums = [1, 2, 3, 4, 5, 6]
pairs = []; for n in nums
pairs << [n, nums.last]
nums.delete nums.last
nums.delete n
end
Current result:
nums
#=> [2, 4]
pairs
#=> [[1, 6], [3, 5]]
Expected result:
nums
#=> []
pairs
#=> [[1, 6], [2, 5], [3, 4]]
Assuming nums is sorted and can be modified, I like this way because it has a mechanical feel about it:
pairs = (nums.size/2).times.map { [nums.shift, nums.pop] }
#=> [[1, 6], [2, 5], [3, 4]]
nums
#=> []
I see #Drenmi has the same idea of using shift and pop.
If you don't want to modify nums, you could of course operate on a copy.
Enumerating over an Array while deleting it's content is generally not advisible. Here's an alternative solution:
nums = *(1..6)
#=> [1, 2, 3, 4, 5, 6]
pairs = []
#=> []
until nums.size < 2 do
pairs << [nums.shift, nums.pop]
end
pairs
#=> [[1, 6], [2, 5], [3, 4]]
I have an Array of Arrays that I want to sort by longest length to shortest. I achieved this easily enough with a sort_by
> a = [ [1, 2, 9],
[4, 5, 6, 7],
[1, 2, 3] ]
> a.sort_by(&:length).reverse # or a.sort_by {|e| e.length}.reverse
=> [[4, 5, 6, 7], [1, 2, 3], [1, 2, 9]]
What I want, however is to have a sort of tie-breaker for lists of equal length. If two lists' lengths are equal, the list whose last entry is greater should come first. So in the above, [1, 2, 9] and [1, 2, 3] should be switched.
I don't care abouth the case where two lists have both equal length and equal last element, they can be in whatever order if that occurs. I don't know if/how I can acheive this with ruby built-in sorting.
You can still do this with sort_by, you just need to realize that Ruby arrays compare element-by-element:
ary <=> other_ary → -1, 0, +1 or nil
[...]
Each object in each array is compared (using the <=> operator).
Arrays are compared in an “element-wise” manner; the first two elements that are not equal will determine the return value for the whole comparison.
That means that you can use arrays as the sort_by key, then throw in a bit of integer negation to reverse the sort order and you get:
a.sort_by { |e| [-e.length, -e.last] }
That will give you the [[4, 5, 6, 7], [1, 2, 9], [1, 2, 3]] that you're looking for.
If you're not using numbers so the "negation to reverse the order" trick won't work, then use Shaunak's sort approach.
There you go :
a = [ [1, 2, 9],[4, 5, 6, 7],[1, 2, 3] ]
a.sort { |a, b| (b.count <=> a.count) == 0 ? (b.last <=> a.last): (b.count <=> a.count) }
That should give you:
[[4, 5, 6, 7], [1, 2, 9], [1, 2, 3]]
How this works: we pass a block to sort function, which first checks if the array length is same, if not it continues to check for last element.
You could use
a.sort_by {|i| [i.length, i.last] }.reverse
# => [[4, 5, 6, 7], [1, 2, 9], [1, 2, 3]]
How does one merge N sorted arrays (or other list-like data structures) lazily in Ruby? For example, in Python you would use heapq.merge. There must be something like this built into Ruby, right?
Here's a (slightly golfed) solution that should work on arrays of any 'list-like' collections that support #first, #shift, and #empty?. Note that it is destructive - each call to lazymerge removes one item from one collection.
def minheap a,i
r=(l=2*(m=i)+1)+1 #get l,r index
m = l if l< a.size and a[l].first < a[m].first
m = r if r< a.size and a[r].first < a[m].first
(a[i],a[m]=a[m],a[i];minheap(a,m)) if (m!=i)
end
def lazymerge a
(a.size/2).downto(1){|i|minheap(a,i)}
r = a[0].shift
a[0]=a.pop if a[0].empty?
return r
end
p arrs = [ [1,2,3], [2,4,5], [4,5,6],[3,4,5]]
v=true
puts "Extracted #{v=lazymerge (arrs)}. Arr= #{arrs.inspect}" while v
Output:
[[1, 2, 3], [2, 4, 5], [4, 5, 6], [3, 4, 5]]
Extracted 1. Arr= [[2, 3], [2, 4, 5], [4, 5, 6], [3, 4, 5]]
Extracted 2. Arr= [[3], [2, 4, 5], [4, 5, 6], [3, 4, 5]]
Extracted 2. Arr= [[4, 5], [3], [4, 5, 6], [3, 4, 5]]
Extracted 3. Arr= [[4, 5], [3, 4, 5], [4, 5, 6]]
Extracted 3. Arr= [[4, 5], [4, 5], [4, 5, 6]]
Extracted 4. Arr= [[5], [4, 5], [4, 5, 6]]
Extracted 4. Arr= [[5], [5], [4, 5, 6]]
Extracted 4. Arr= [[5, 6], [5], [5]]
Extracted 5. Arr= [[6], [5], [5]]
Extracted 5. Arr= [[5], [6]]
Extracted 5. Arr= [[6]]
Extracted 6. Arr= [[]]
Extracted . Arr= [[]]
Note also that this algorithm is also lazy about maintaining the heap property - it is not maintained between calls. This probably causes it to do more work than needed, since it does a complete heapify on each subsequent call. This could be fixed by doing a complete heapify once up front, then calling minheap(a,0) before the return r line.
I ended up writing it myself using the data structures from the 'algorithm' gem. It wasn't as bad as I expected.
require 'algorithms'
class LazyHeapMerger
def initialize(sorted_arrays)
#heap = Containers::Heap.new { |x, y| (x.first <=> y.first) == -1 }
sorted_arrays.each do |a|
q = Containers::Queue.new(a)
#heap.push([q.pop, q])
end
end
def each
while #heap.length > 0
value, q = #heap.pop
#heap.push([q.pop, q]) if q.size > 0
yield value
end
end
end
m = LazyHeapMerger.new([[1, 2], [3, 5], [4]])
m.each do |o|
puts o
end
Here's an implementation which should work on any Enumerable, even infinite ones. It returns Enumerator.
def lazy_merge *list
list.map!(&:enum_for) # get an enumerator for each collection
Enumerator.new do |yielder|
hash = list.each_with_object({}){ |enum, hash|
begin
hash[enum] = enum.next
rescue StopIteration
# skip empty enumerators
end
}
loop do
raise StopIteration if hash.empty?
enum, value = hash.min_by{|k,v| v}
yielder.yield value
begin
hash[enum] = enum.next
rescue StopIteration
hash.delete(enum) # remove enumerator that we already processed
end
end
end
end
Infinity = 1.0/0 # easy way to get infinite range
p lazy_merge([1, 3, 5, 8], (2..4), (6..Infinity), []).take(12)
#=> [1, 2, 3, 3, 4, 5, 6, 7, 8, 8, 9, 10]
No, there's nothing built in to do that. At least, nothing that springs instantly to mind. However, there was a GSoC project to implement the relevant data types a couple of years ago, which you could use.