Lets say array look like below
city = ['london', 'new york', 'london', 'london', 'washington']
desired_location = ['london']
city & desired_location gives ['london']
but I want ['london', 'london', 'london']
3 Answers
You can use Enumerable#select
city.select {|c| desired_location.include?(c)}
# => ["london", "london", "london"]
4 Comments
Hari Krishnan
i think this will work, but both array will contain 1000 + element. is it feasible ?
fanta
compare with desired location ?, what Santhosh put works. city.select {|c| desired_location.include?(c)} will return ['london', 'london', 'london']. If desired_location is desired_location = ['london', 'new york'], you would get ["london", "new york", "london", "london"]
jrochkind
What didn't work? What happened when you tried it? Did you have a typo or syntax error? This should work.
Hari Krishnan
when i tried this. it didn't get repeated element from first array. lemme try again.
cities = ['london', 'new york', 'london', 'london', 'washington']
If desired_location contains a single element:
desired_location = ['london']
I recommend @santosh's solution, but this also works:
desired_location.flat_map { |c| [c]*cities.count(c) }
#=> ["london", "london", "london"]
Suppose desired_location contains multiple elements (which I assume is a possibility, for otherwise there would be no need for it to be an array):
desired_location = ['london', 'new york']
@Santosh' method returns:
["london", "new York", "london", "london"]
which is quite possibly what you want. If you'd prefer that they be grouped:
desired_location.flat_map { |c| [c]*cities.count(c) }
#=> ["london", "london", "london", "new york"]
or:
desired_location.map { |c| [c]*cities.count(c) }
#=> [["london", "london", "london"], ["new york"]]
Depending on your requirements, you might find it more useful to produce a hash:
Hash[desired_location.map { |c| [c, cities.count(c)] }]
#=> {"london"=>3, "new york"=>1}
2 Comments
Hari Krishnan
thanks @cary, let me try and reply you back. seems it will work. from final hash, is it finally sorted. I mean, first element will always be top ?
Cary Swoveland
As long as you are using Ruby v1.9+, the keys of the hash at the end are guaranteed to be ordered as they are in
desired_location.Another way:
cities = ['london', 'new york', 'london', 'london', 'washington']
puts cities.select{|city| cities.count(city) > 1}
Comments
city, which is also present indesired_locations. That is exactly what my solution does. You have not mentioned anything about top repeated elements, in your question. We could not have guessed that.