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If I have an array of arrays, A, and want to get rid of all arrays in A who also have a sub-array in A, how would I do that. In this context, array_1 is a sub-array of array_2 if array_1 - array_2 = []. In the case that multiple arrays are simply rearranged versions of the same elements, bonus points if you can get rid of all but one of them, but you can handle this however you want if it's easier.
In python, I could easily use comprehension, with A being a set of frozen sets :
A = {a for a in A if all(b-a for b in A-{a})}
Is there a simple way to write this in ruby? I don't care if the order of A or it's arrays are preserved at all. Also, in my program, none of the arrays have duplicate elements, if that makes things any easier/faster.
Example
A = [[1,6],[1,2],[2,4],[3,5],[1,3,6],[2,3,6]]
# [1,6] is a subarray of [1,3,6], so [1,3,6] should be removed
remove_super_arrays(A)
> A = [[1,6],[1,2],[2,4],[3,5],[2,3,6]]
A = [[1,2,4],[2,3,4],[1,4,5],[2,6]]
# although there is overlap, there are no subarrays, so nothing should be removed
remove_super_arrays(A)
> A = [[1,2,4],[2,3,4],[1,4,5],[2,6]]
A = [[1],[2,1,3],[2,4],[1,4]]
# [1] is a subarray of [2,1,3] and [1,4]
remove_super_arrays(A)
> A = [[1],[2,4]]
Code
def remove_super_arrays(arr)
order = arr.each_with_index.to_a.to_h
arr.sort_by(&:size).reject.with_index do |a,i|
arr[0,i].any? { |aa| (aa.size < a.size) && (aa-a).empty? }
end.sort_by { |a| order[a] }
end
Examples
remove_super_arrays([[1,6],[1,2],[2,4],[3,5],[1,3,6],[2,3,6]] )
#=> [[1,6],[1,2],[2,4],[3,5],[2,3,6]]
remove_super_arrays([[1,2,4],[2,3,4],[1,4,5],[2,6]])
#=> [[1,2,4],[2,3,4],[1,4,5],[2,6]]
remove_super_arrays([[1],[2,1,3],[2,4],[1,4]])
#=> [[1],[2,4]]
Explanation
Consider the first example.
arr = [[1,6],[1,2],[2,4],[3,5],[1,3,6],[2,3,6]]
We first save the positions of the elements of a
order = arr.each_with_index.to_a.to_h # save original order
#=> {[1, 6]=>0, [1, 2]=>1, [2, 4]=>2, [3, 5]=>3, [1, 3, 6]=>4, [2, 3, 6]=>5}
Then reject elements of arr:
b = arr.sort_by(&:size)
#=> [[1, 6], [1, 2], [2, 4], [3, 5], [1, 3, 6], [2, 3, 6]]
c = b.reject.with_index do |a,i|
arr[0,i].any? { |aa| (aa.size < a.size) && (aa-a).empty? }
end
#=> [[1, 6], [1, 2], [2, 4], [3, 5], [2, 3, 6]]
Lastly, reorder c to correspond to the original ordering of the elements of arr.
c.sort_by { |a| order[a] }
#=> [[1, 6], [1, 2], [2, 4], [3, 5], [2, 3, 6]]
which in this case happens to be the same order as the elements of c.
Let's look more carefully at the calculation of c:
enum1 = b.reject
#=> #<Enumerator: [[1, 6], [1, 2], [2, 4], [3, 5], [1, 3, 6],
# [2, 3, 6]]:reject>
enum2 = enum1.with_index
#=> #<Enumerator: #<Enumerator: [[1, 6], [1, 2], [2, 4], [3, 5],
# [1, 3, 6], [2, 3, 6]]:reject>:with_index>
The first element is generated by the enumerator enum2 and passed to the block and assigned as values of the block variables:
a, i = enum2.next
#=> [[1, 6], 0]
a #=> [1, 6]
i #=> 0
The block calculation is then performed:
d = arr[0,i]
#=> []
d.any? { |aa| (aa.size < a.size) && (aa-a).empty? }
#=> false
so a[0] is not rejected. The next pair passed to the block by enum2 is [[1, 2], 1]. That value is retained as well, but let's skip ahead to the last element passed to the block by enum2:
a, i = enum2.next
#=> [[1, 2], 1]
a, i = enum2.next
#=> [[2, 4], 2]
a, i = enum2.next
#=> [[3, 5], 3]
a, i = enum2.next
#=> [[1, 3, 6], 4]
a #=> [1, 3, 6]
i #=> 4
Perform the block calculation:
d = arr[0,i]
#=> [[1, 6], [1, 2], [2, 4], [3, 5]]
d.any? { |aa| (aa.size < a.size) && (aa-a).empty? }
#=> true
As true is returned, a is rejected. In the last calculation the first element of d is passed to the block and the following calculation is performed:
aa = [1, 6]
(aa.size < a.size)
#=> 2 < 3 => true
(aa-a).empty?
#=> ([1, 6] - [1, 3, 6]).empty? => [].empty? => true
As true && true #=> true, a ([1, 3, 6]) is rejected.
Alternative calculation
The following is a closer match to the OP's Python equivalent, but less efficient:
def remove_super_arrays(arr)
arr.select do |a|
(arr-[a]).all? { |aa| aa.size > a.size || (aa-a).any? }
end
end
or
def remove_super_arrays(arr)
arr.reject do |a|
(arr-[a]).any? { |aa| (aa.size < a.size) && (aa-a).empty? }
end
end
This was a nice exercise for me. I have used the logic from here.
My code iterates over each subarray (except the first), then there is the magic substraction using the first index, when it is empty the other array contained both numbers.
def remove_super_arrays(arr)
arr.each_with_index.with_object([]) do |(sub_array, index), result|
next if index == 0
result << sub_array unless (arr.first - sub_array).empty?
end.unshift(arr.first)
end
arr = [[1,6],[1,2],[2,4],[3,5],[1,3,6],[2,3,6]]
p remove_super_arrays(arr)
#=> [[1, 6], [1, 2], [2, 4], [3, 5], [2, 3, 6]]
I have an n-depth array where n is a variable greater than or equal to 2:
[[1,1],[[1,1],[1,1]]]
I want to flatten this array to have exactly 2 depth, like this:
[[1,1],[1,1],[1,1]]
Can anyone think of a good way to achieve that?
This should do it.
def flatten_after_first(arr)
arr.flat_map { |a| a.first.is_a?(Array) ? a.map(&:flatten) : [a] }
end
flatten_after_first [[1,1],[[1,1],[1,1]]]
#=> [[1, 1], [1, 1], [1, 1]]
flatten_after_first [[1,1], [[2,2], [2,2]], [[[3,3], [3,3]], [[3,3], [3,3]]]]
#=> [[1, 1], [2, 2], [2, 2], [3, 3, 3, 3], [3, 3, 3, 3]]
May be this will help
def flat(array)
array.each do |item|
if item.is_a?(Array) && item.flatten.count != item.count
flat(item)
else
$arr << item
end
end
end
###
$arr = []
collection = [[1, 1], [[1, 1], [1, 1], [[1, 2], [1, 2, 3]]]]
flat(collection)
puts $arr.inspect
=> [[1, 1], [1, 1], [1, 1], [1, 2], [1, 2, 3]]
$arr = []
collection = [[1,1],[[[1,1],[1,1]],[1,1]]]
flat(collection)
$arr
=> [[1, 1], [1, 1], [1, 1], [1, 1]]
Try this:
def depth_first_flatten array
result = []
array.each do |element|
if element.first.is_a? Array
result += deph(element)
else
result << element
end
end
result
end
# array = [[1,2],[[3,4],[5,6]]]
# depth_first_flatten(array)
#
# OUTPUT: [[1, 2], [3, 4], [5, 6]]
I have four 2d arrays of varying lengths. Is there a way to compare them to make sure none of the arrays share a value? Is there a simple way to code it or a gem I could use?
To back my question up with an example:
array1 = [[2,3],[2,4]]
array2 = [[1,3],[2,3],[3,3]]
array3 = [[5,3],[6,3],[7,3],[8,3],[9,3]]
I want a checker that would let me know that array1 and array2 have an element the same.
Is this doable?
You might do it as follows.
Code
def shared_values?(*arr)
a = arr.map(&:uniq).reduce(:+)
a.size > a.uniq.size
end
Examples
array1 = [[2, 3], [2, 4]]
array2 = [[1, 3], [2, 3], [3, 3]]
array3 = [[5, 3], [6, 3], [7, 3],[8, 3],[9, 3]]
shared_values? array1, array2, array3 #=> true
array1 = [[2, 4]]
shared_values? array1, array2, array3 #=> false
array1 = [[2, 4], [2, 4]]
shared_values? array1, array2, array3 #=> false
array1 = [[2, 3], [2, 4], [2, 4]]
shared_values? array1, array2, array3 #=> false
Explanation
For the last example above:
arr = [array1, array2, array3]
#=> [ [[2, 3], [2, 4], [2, 4]],
# [[1, 3], [2, 3], [3, 3]],
# [[5, 3], [6, 3], [7, 3], [8, 3], [9, 3]] ]
a = arr.map(&:uniq)
#=> [ [[2, 3], [2 ,4]],
# [[1, 3], [2, 3], [3, 3]],
# [[5, 3], [6, 3], [7, 3], [8, 3], [9, 3]] ]
b = a.reduce(:+)
#=> [[2, 3], [2, 4], [1, 3], [2, 3], [3, 3],
# [5, 3], [6, 3], [7, 3], [8, 3], [9, 3]]
c = b.uniq
#=> [[2, 3], [2, 4], [1, 3], [3, 3],
# [5, 3], [6, 3], [7, 3], [8, 3], [9, 3]]
b.size > c.size
# 10 > 9 #=> true
arr.map(&:uniq) is the same as:
arr.map { |a| a.uniq }
a.reduce(:+) uses the form of Enumerable#reduce that takes an argument that is a symbol, naming a method to be applied to each element of arr, the result being the sum of the three arrays that comprise the elements of b.
If the arrays are in a consistent format, like all numbers and not a mix of floating point and strings, you can do this:
array1 & array2
# => [[2, 3]]
That means to test if they overlap:
(array1 & array2).any?
If you can be assured each of the elements in all of the arrays are unique, then you can test quickly if there are any duplicates:
sum = array1 + array2 + array3 + array4
sum.length == sum.uniq.length
If each array may contain duplicates then you'd need to pair them off and compare A vs. B for all possible pairs.
i want to convert in ruby
[[1, 1], [2, 3], [3, 5], [4, 1], [1, 2], [2, 3], [3, 5], [4, 1]]
into
[{1=>1}, {2=>3}, {3=>5}, {4=>1}, {1=>2}, {2=>3}, {3=>5}, {4=>1}]
and after this to obtain sum of all different keys:
{1=>3,2=>6,3=>10,4=>2}
For the second question
sum = Hash.new(0)
original_array.each{|x, y| sum[x] += y}
sum # => {1 => 3, 2 => 6, 3 => 10, 4 => 2}
Functional approach:
xs = [[1, 1], [2, 3], [3, 5], [4, 1], [1, 2], [2, 3], [3, 5], [4, 1]]
Hash[xs.group_by(&:first).map do |k, pairs|
[k, pairs.map { |x, y| y }.inject(:+)]
end]
#=> {1=>3, 2=>6, 3=>10, 4=>2}
Using Facets is much simpler thanks to the abstractions map_by (a variation of group_by) and mash (map + Hash):
require 'facets'
xs.map_by { |k, v| [k, v] }.mash { |k, vs| [k, vs.inject(:+)] }
#=> {1=>3, 2=>6, 3=>10, 4=>2}
You don't need the intermediate form.
arrays = [[1, 1], [2, 3], [3, 5], [4, 1], [1, 2], [2, 3], [3, 5], [4, 1]]
aggregate = arrays.each_with_object Hash.new do |(key, value), hash|
hash[key] = hash.fetch(key, 0) + value
end
aggregate # => {1=>3, 2=>6, 3=>10, 4=>2}
arr= [[1, 1], [2, 3], [3, 5], [4, 1], [1, 2], [2, 3], [3, 5], [4, 1]]
final = Hash.new(0)
second_step = arr.inject([]) do |arr,inner|
arr << Hash[*inner]
final[inner.first] += inner.last
arr
end
second_step
#=> [{1=>1}, {2=>3}, {3=>5}, {4=>1}, {1=>2}, {2=>3}, {3=>5}, {4=>1}]
final
#=> {1=>3, 2=>6, 3=>10, 4=>2}
if you directly only need the last step
arr.inject(Hash.new(0)){|hash,inner| hash[inner.first] += inner.last;hash}
=> {1=>3, 2=>6, 3=>10, 4=>2}
Suppose I have
an_array = [[2, 3], [1, 4], [1, 3], [2, 1], [1, 2]]
I want to sort this array by the first value of each inner array, and then by the second (so the sorted array should look like this: [[1, 2], [1, 3], [1, 4], [2, 1], [2, 3]])
What's the most readable way to do this?
This is the default behavior for sorting arrays (see the Array#<=> method definition for proof). You should just be able to do:
an_array.sort
If you want some non-default behaviour, investigate sort_by (ruby 1.8.7+)
e.g. sort by the second element then by the first
a.sort_by {|e| [e[1], e[0]]} # => [[2, 1], [1, 2], [1, 3], [2, 3], [1, 4]]
or sort by the first element ascending and then the second element descending
a.sort_by {|e| [e[0], -e[1]]} # => [[1, 4], [1, 3], [1, 2], [2, 3], [2, 1]]
an_array.sort