Given:
a1 = [5, 1, 6, 14, 2, 8]
I would like to determine if it contains all elements of:
a2 = [2, 6, 15]
In this case the result is false
.
Are there any built-in Ruby/Rails methods to identify such array inclusion?
One way to implement this is:
a2.index{ |x| !a1.include?(x) }.nil?
Is there a better, more readable, way?
This question is related to
ruby-on-rails
arrays
ruby
Depending on how big your arrays are you might consider an efficient algorithm O(n log n)
def equal_a(a1, a2)
a1sorted = a1.sort
a2sorted = a2.sort
return false if a1.length != a2.length
0.upto(a1.length - 1) do
|i| return false if a1sorted[i] != a2sorted[i]
end
end
Sorting costs O(n log n) and checking each pair costs O(n) thus this algorithm is O(n log n). The other algorithms cannot be faster (asymptotically) using unsorted arrays.
If there are are no duplicate elements or you don't care about them, then you can use the Set class:
a1 = Set.new [5, 1, 6, 14, 2, 8]
a2 = Set.new [2, 6, 15]
a1.subset?(a2)
=> false
Behind the scenes this uses
all? { |o| set.include?(o) }
Most answers based on (a1 - a2) or (a1 & a2) would not work if there are duplicate elements in either array. I arrived here looking for a way to see if all letters of a word (split to an array) were part of a set of letters (for scrabble for example). None of these answers worked, but this one does:
def contains_all?(a1, a2)
try = a1.chars.all? do |letter|
a1.count(letter) <= a2.count(letter)
end
return try
end
This can be achieved by doing
(a2 & a1) == a2
This creates the intersection of both arrays, returning all elements from a2
which are also in a1
. If the result is the same as a2
, you can be sure you have all elements included in a1
.
This approach only works if all elements in a2
are different from each other in the first place. If there are doubles, this approach fails. The one from Tempos still works then, so I wholeheartedly recommend his approach (also it's probably faster).
Perhaps this is easier to read:
a2.all? { |e| a1.include?(e) }
You can also use array intersection:
(a1 & a2).size == a1.size
Note that size
is used here just for speed, you can also do (slower):
(a1 & a2) == a1
But I guess the first is more readable. These 3 are plain ruby (not rails).
You can monkey-patch the Array class:
class Array
def contains_all?(ary)
ary.uniq.all? { |x| count(x) >= ary.count(x) }
end
end
test
irb(main):131:0> %w[a b c c].contains_all? %w[a b c]
=> true
irb(main):132:0> %w[a b c c].contains_all? %w[a b c c]
=> true
irb(main):133:0> %w[a b c c].contains_all? %w[a b c c c]
=> false
irb(main):134:0> %w[a b c c].contains_all? %w[a]
=> true
irb(main):135:0> %w[a b c c].contains_all? %w[x]
=> false
irb(main):136:0> %w[a b c c].contains_all? %w[]
=> true
irb(main):137:0> %w[a b c d].contains_all? %w[d c h]
=> false
irb(main):138:0> %w[a b c d].contains_all? %w[d b c]
=> true
Of course the method can be written as a standard-alone method, eg
def contains_all?(a,b)
b.uniq.all? { |x| a.count(x) >= b.count(x) }
end
and you can invoke it like
contains_all?(%w[a b c c], %w[c c c])
Indeed, after profiling, the following version is much faster, and the code is shorter.
def contains_all?(a,b)
b.all? { |x| a.count(x) >= b.count(x) }
end
Source: Stackoverflow.com