Is there any method to check if array A contains all the elements of array B?
8 Answers
You can try this
a.sort.uniq == b.sort.uniq
or
(a-b).empty?
And if [1,2,2] != [1,2] in your case you can:
a.group_by{|i| i} == b.group_by{|i| i}
6 Comments
Ryan Bigg
+1 for
(a-b).empty?
oseiskar
Just in case someone ends up here: these are all wrong. The correct answer is
(b-a).empty?, not the other way around. Others are just plain wrong. See also stackoverflow.com/questions/7387937/…
fl00r
@oseiskar no, they are not wrong. It depends on the task.
oseiskar
I don't think so. If the task is to answer the question does array A contain all elements of array B, which quite literally translates to Ruby as
B.map{|e|A.include?(e)}.all?, then none of these give the correct answer in the generic case (or, e.g., the first example I came up with: A = [1,2,3], B = [1,2]).fl00r
@oseiskar yes, this sounds reasonable. In generic case they all wrong.
|
This should work for what you need:
(a & b) == b
4 Comments
fl00r
this won't work as a rule.
a = [1,2,2,1,3]; b = [3,2,1]; (a & b) == b => false
the Tin Man
Yes, but
a = [1,2,2,1,3]; b = [3,2,1]; (a & b).sort == b.sort => truefl00r
@the Tin Man, Yes, but a = [1,2,2,1,3]; b = [3,2,2,1]; (a & b).sort == b.sort => false
Will
no need to sort just reverse the order of the binary AND
You could use Ruby's Set class:
>> require 'set' #=> true
>> a = [*1..5] #=> [1, 2, 3, 4, 5]
>> b = [*1..3] #=> [1, 2, 3]
>> a.to_set.superset? b.to_set #=> true
For small arrays I usually do the same as what fl00r suggested:
>> (b-a).empty? #=> true
Comments
I prefer to do this via: (b - a).blank? # tells that b is contained in a
Comments
The simplest way is this:
(b-a).empty?
Comments
There's also the Set class (part of the standard library) which would allow you to just check to see if B is a subset of A, e.g.
>> a = [1,2,3,4,5] => [1, 2, 3, 4, 5]
>> b = [3,4,5] => [3, 4, 5]
>> require 'set' => true
>> set_a = a.to_set => #<Set: {1, 2, 3, 4, 5}>
>> set_b = b.to_set => #<Set: {3, 4, 5}>
>> set_b.subset? set_a => true
Comments
Ruby 2.6+
Ruby's introduced difference in 2.6 for exactly this purpose.
Very fast, very readable, as follows:
a = [1, 2, 3, 4, 5, 6]
b = [1, 2, 3, 4, 5, 6]
a.difference(b).any?
# => false
a.difference(b.reverse).any?
# => false
a = [1, 2, 3, 4, 5, 6]
b = [1, 2, 3]
a.difference(b).any?
# => true
Hope that helps someone!
Comments
You may want to check out the Set class in the Ruby Standard library. The proper_subset? method will probably do what you want.
differencein 2.6 which provides a perfect fast and readable solution for this. More info here.