What is a good way to get from this:
['a','b','c',['d1','d2']]
to this:
[['a','b','c','d1']['a','b','c','d2']]
another example, from this:
[['a1','a2'],'b',['c1','c2']]
to this:
[['a1','b','c1'],['a1','b','c2'],['a2','b','c1'],['a2','b','c2']]
edit 1:
Sorry for the confusion and thanks for response so far, the individual contents of the array items doesn't matter but the order must be preserved. The method needs to work for both example because the nested array can be in any position of the outer array, and the nested array can have more the 2 elements.
It's sort of like a regex with multiple or conditions
ab(c|d)
expand to match abc and abd
It is a bit hard to know exactly what you want, but this produces something quite similar:
# Create a list:
a = [['a1','a2'],'b',['c1','c2']]
# Split it into sub-arrays and single values:
list, other = a.partition{|x|x.is_a? Array}
# Split the array in order to get the product:
list_first, list_rest = list.pop, list
# Get the product and add the others_values:
p list_first.product(*list_rest).map{|list| list+other}
#=> [["c1", "a1", "b"], ["c1", "a2", "b"], ["c2", "a1", "b"], ["c2", "a2", "b"]]
1st:
arr1 = ['a','b','c',['d1','d2']]
*a, b = arr1
# ["a", "b", "c", ["d1", "d2"]]
a
# ["a", "b", "c"]
b
# ["d1", "d2"]
b.map{|x| a+[x]}
# [["a", "b", "c", "d1"], ["a", "b", "c", "d2"]]
and 2nd:
a, b, c = [["a1", "a2"], "b", ["c1", "c2"] ]
a.product c
#=> [["a1", "c1"], ["a1", "c2"], ["a2", "c1"], ["a2", "c2"]]
a.product(c).map{|x| x<<b}
#=> [["a1", "c1", "b"], ["a1", "c2", "b"], ["a2", "c1", "b"], ["a2", "c2", "b"]]
#or little less readable:
a.product(c).map{|x| [ x[0], b, x[1] ]}
# [["a1", "b", "c1"], ["a1", "b", "c2"], ["a2", "b", "c1"], ["a2", "b", "c2"]]
hirolau got me really close, here is what I ended with so the order is preserved:
# given a sample list
sample = [['a','b'],'c','d',['e','f'],'g',['h','i']]
# partition out the arrays
arrs, non_arrays = sample.partition {|sample| sample.is_a? Array}
# work out all possible products
first_elem, *the_rest = arrs
products = first_elem.product(*the_rest)
# finally swap it back in to get all valid combinations with order preserved
combinations = []
products.each do |p|
combinations << sample.map {|elem| elem.is_a?(Array) ? p.shift : elem}
end
# combinations
=> [["a", "c", "d", "e", "g", "h"],
["a", "c", "d", "e", "g", "i"],
["a", "c", "d", "f", "g", "h"],
["a", "c", "d", "f", "g", "i"],
["b", "c", "d", "e", "g", "h"],
["b", "c", "d", "e", "g", "i"],
["b", "c", "d", "f", "g", "h"],
["b", "c", "d", "f", "g", "i"]]
Given an array of arrays
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"]]
What is the simplest way to merge the array items that contain members that are shared by any two or more arrays items. For example the above should be
[["A", "B", "C", "D","E", "F"], ["G"]] since "B" and "C" are shared by the first and second array items.
Here are some more test cases.
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]]
=> [["A", "B", "C", "D", "E", "F", "G"]]
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
=> [["A", "B", "C", "D", "E", "F"], ["G", "H,"]]
Here is my quick version which can be optimized I am sure :)
# array = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"]]
# array = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]]
array = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
array.collect! do |e|
t = e
e.each do |f|
array.each do |a|
if a.index(f)
t = t | a
end
end
end
e = t.sort
end
p array.uniq
Edit: Martin DeMello code was fixed.
When running Martin DeMello code (the accepted answer) I get:
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]] =>
[["B", "C", "E", "F", "A", "D", "G"], ["A", "B", "C", "D"], ["F", "G"]]
and
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]] =>
[["B", "C", "E", "F", "A", "D"], ["A", "B", "C", "D"], ["G", "H"], ["G", "H"]]
which does not seem to meet your spec.
Here is my approach using a few of his ideas:
a = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]]
b = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
def reduce(array)
h = Hash.new {|h,k| h[k] = []}
array.each_with_index do |x, i|
x.each do |j|
h[j] << i
if h[j].size > 1
# merge the two sub arrays
array[h[j][0]].replace((array[h[j][0]] | array[h[j][1]]).sort)
array.delete_at(h[j][1])
return reduce(array)
# recurse until nothing needs to be merged
end
end
end
array
end
puts reduce(a).to_s #[["A", "B", "C", "D", "E", "F", "G"]]
puts reduce(b).to_s #[["A", "B", "C", "D", "E", "F"], ["G", "H"]]
Different algorithm, with a merge-as-you-go approach rather than taking two passes over the array (vaguely influenced by the union-find algorithm). Thanks for a fun problem :)
A = [["A", "G"],["B", "C", "E", "F"], ["A", "B", "C", "D"], ["B"], ["H", "I"]]
H = {}
B = (0...(A.length)).to_a
def merge(i,j)
A[j].each do |e|
if H[e] and H[e] != j
merge(i, H[e])
else
H[e] = i
end
end
A[i] |= A[j]
B[j] = i
end
A.each_with_index do |x, i|
min = A.length
x.each do |j|
if H[j]
merge(H[j], i)
else
H[j] = i
end
end
end
out = B.sort.uniq.map {|i| A[i]}
p out
Not the simplest ,may be the longest :)
l = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"]]
puts l.flatten.inject([[],[]]) {|r,e| if l.inject(0) {|c,a| if a.include?(e) then c+1 else c end} >= 2 then r[0] << e ; r[0].uniq! else r[1] << e end ; r}.inspect
#[["B", "C"], ["E", "F", "A", "D", "G"]]
l = [["B", "C", "E", "F"], ["A", "B","C", "D"], ["G"]]
p l.inject([]){|r,e|
r.select{|i|i&e!=[]}==[]&&(r+=[e])||(r=r.map{|i|(i&e)!=nil&&(i|e).sort||i})
}
im not sure about your cond.
The simplest way to do it would be to take the powerset of an array (a set containing every possible combination of elements of the array), throw out any of the resulting sets if they don't have a common element, flatten the remaining sets and discard subsets and duplicates.
Or at least it would be if Ruby had proper Set support. Actually doing this in Ruby is horribly inefficient and an awful kludge:
power_set = array.inject([[]]){|c,y|r=[];c.each{|i|r<<i;r<<i+[y]};r}.reject{|x| x.empty?}
collected_powerset = power_set.collect{|subset| subset.flatten.uniq.sort unless
subset.inject(subset.last){|acc,a| acc & a}.empty?}.uniq.compact
collected_powerset.reject{|x| collected_powerset.any?{|c| (c & x) == x && x.length < c.length}}
Power set operation comes from here.
Straightforward rather than clever. It's destructive of the original array. The basic idea is:
go down the list of arrays, noting which array each element appears in
for every entry in this index list that shows an element in more than one array, merge all those arrays into the lowest-indexed array
when merging two arrays, replace the lower-indexed array with the merged result, and the higher-indexed array with a pointer to the lower-indexed array.
It's "algorithmically cheaper" than intersecting every pair of arrays, though the actual running speed will depend on what ruby hands over to the C layer.
a = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
h = Hash.new {|h,k| h[k] = []}
a.each_with_index {|x, i| x.each {|j| h[j] << i}}
b = (0...(a.length)).to_a
h.each_value do |x|
x = x.sort_by {|i| b[i]}
if x.length > 1
x[1..-1].each do |i|
b[i] = [b[i], b[x[0]]].min
a[b[i]] |= a[i]
end
end
end
a = b.sort.uniq.map {|i| a[i]}
def merge_intersecting(input, result=[])
head = input.first
tail = input[1..-1]
return result if tail.empty?
intersection = tail.select { |arr| !(head & arr).empty? }
unless intersection.empty?
merged = head | intersection.flatten
result << merged.sort
end
merge_intersecting(tail, result)
end
require 'minitest/spec'
require 'minitest/autorun'
describe "" do
it "merges input array" do
input = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]]
output = [["A", "B", "C", "D", "E", "F", "G"]]
merge_intersecting(input).must_equal output
end
it "merges input array" do
input = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
output = [["A", "B", "C", "D", "E", "F"], ["G", "H"]]
merge_intersecting(input).must_equal output
end
end