How do I check an array for duplicates? [duplicate] - ruby

This question already has answers here:
How to find and return a duplicate value in array
(23 answers)
Closed 8 years ago.
I've got an array A. I'd like to check if it contains duplicate values. How would I do so?

Just call uniq on it (which returns a new array without duplicates) and see whether the uniqed array has less elements than the original:
if a.uniq.length == a.length
puts "a does not contain duplicates"
else
puts "a does contain duplicates"
end
Note that the objects in the array need to respond to hash and eql? in a meaningful for uniq to work properly.

In order to find the duplicated elements, I use this approach (with Ruby 1.9.3):
array = [1, 2, 1, 3, 5, 4, 5, 5]
=> [1, 2, 1, 3, 5, 4, 5, 5]
dup = array.select{|element| array.count(element) > 1 }
=> [1, 1, 5, 5, 5]
dup.uniq
=> [1, 5]

If you want to return the duplicates, you can do this:
dups = [1,1,1,2,2,3].group_by{|e| e}.keep_if{|_, e| e.length > 1}
# => {1=>[1, 1, 1], 2=>[2, 2]}
If you want just the values:
dups.keys
# => [1, 2]
If you want the number of duplicates:
dups.map{|k, v| {k => v.length}}
# => [{1=>3}, {2=>2}]

Might want to monkeypatch Array if using this more than once:
class Array
def uniq?
self.length == self.uniq.length
end
end
Then:
irb(main):018:0> [1,2].uniq?
=> true
irb(main):019:0> [2,2].uniq?
=> false

Related

Find all max of elements of an array [duplicate]

This question already has answers here:
Returning all maximum or minimum values that can be multiple
(3 answers)
Closed 8 years ago.
Suppose I have a array, namely arr: [1, 2, 3, 4, 8, 8], and I want to find all max elements in this array:
arr.allmax # => [8, 8]
Is there a built-in method combinations to solve this? I don't like to monkey patch as I am doing now:
class Array
def allmax
max = self.max
self.select { |e| e == max }
end
end
Monkey patch is not a good idea, I could just do:
some_array.select { |e| e == some_array.max }
and it will work as allmax. Thanks for all answers and comments for inspirations.
Here's a fun way to do it.
arr.sort!.slice arr.index(arr[-1]) || 0..-1
Sort the array, then find the leftmost index of the array which matches the rightmost index of the array, and take the subslice that matches that range (or the range 0..-1 if the array is empty).
This one is kind of fun in that it requires no intermediate arrays, though it does mutate the input to achieve the one-liner.
Here is one way :
2.1.0 :006 > arr = [1, 2, 3, 4, 8, 8]
=> [1, 2, 3, 4, 8, 8]
2.1.0 :007 > arr.group_by { |i| i }.max.last
=> [8, 8]
2.1.0 :008 >
Here is a method :-
def all_max(arr)
return [] if arr.empty?
arr.group_by { |i| i }.max.last
end
Another way:
def all_max(arr)
return [] if arr.empty?
mx = arr.max
[mx] * arr.count { |e| e == mx }
end
all_max([1, 2, 3, 4, 8, 8])
#=> [8, 8]
To construct the array in a single pass, you could do this:
arr.each_with_object([]) do |e,a|
if a.empty?
a << e
else
case e <=> a.first
when 0 then a << e
when 1 then a.replace([e])
end
end
end

Three ways to create a range, hash, array in ruby

I am doing a tutorial course on ruby and it asks for 3 ways to create range, hash, array.
I can only think of 2: (1..3) and Range.new(1,3) (and similarly for hash and array).
What is the third way?
The tutorial in question is The Odin Project
Ranges may be constructed using the s..e and s...e literals, or with ::new.
Ranges constructed using .. run from the beginning to the end inclusively.
Those created using ... exclude the end value. When used as an iterator, ranges return each value in the sequence.
(0..2) == (0..2) #=> true
(0..2) == Range.new(0,2) #=> true
(0..2) == (0...2) #=> false
Read More Here
For Arrays there's Array::[] (example taken directly from the docs):
Array.[]( 1, 'a', /^A/ ) # => [1, "a", /^A/]
Array[ 1, 'a', /^A/ ] # => [1, "a", /^A/]
[ 1, 'a', /^A/ ] # => [1, "a", /^A/]
Similarly there's Hash::[]. Not sure about Ranges; in fact, the docs (as far as I can tell) only mention literals and Range::new.
I can't see why you'd use these over a literal, but there you go.
You can also make a exclusive range, using (1...4), which if turned into an array would become [1, 2, 3]
(1..3) is an inclusive range, so it contains all numbers, from 1 to 3, but if you used (1...3), having 3 dots instead of 2 makes it exclusive, so it contains all numbers from 1, up to but not including 3.
As for arrays and hashes, #to_a, Array#[], #to_h, and Hash#[] will work.
(1..3).to_a
=> [1, 2, 3]
Array[1, 2, 3]
=> [1, 2, 3]
[[1, 2], [3, 4], [5, 6]].to_h
=> {1=>2, 3=>4, 5=>6}
Hash[ [[1, 2], [3, 4], [5, 6]] ]
=> {1=>2, 3=>4, 5=>6}
But they are probably looking for Array#[] and Hash#[] on the array and hash part.

What is the difference between `each` and `map` enumerators? [duplicate]

This question already has answers here:
How are array.each and array.map different? [duplicate]
(4 answers)
Closed 9 years ago.
Examining the following code
[1,2,3,4].map # => #<Enumerator: [1, 2, 3, 4]:map>
[1,2,3,4].each # => #<Enumerator: [1, 2, 3, 4]:each>
we can see that both are an enumerator. One is map and other is each. The method to_enum also returns an enumerator each. Is there any difference between these two enumerators?
Yes, when you iterate over the map enumerator, it will take the result and populate it into a new array:
[1,2,3,4].map.each { |n| n * 2 } # => [2, 4, 6, 8]
When you iterate over the each enumerator, it will return the original array
[1,2,3,4].each.each { |n| n * 2 } # => [1, 2, 3, 4]
Map exists to take an array of one type and convert it to an array of a different type, each exists to simply iterate over the array (doing something with a side effect for each element, such as printing it out).

How to drop the end of an array in Ruby

Array#drop removes the first n elements of an array. What is a good way to remove the last m elements of an array? Alternately, what is a good way to keep the middle elements of an array (greater than n, less than m)?
This is exactly what Array#pop is for:
x = [1,2,3]
x.pop(2) # => [2,3]
x # => [1]
You can also use Array#slice method, e.g.:
[1,2,3,4,5,6].slice(1..4) # => [2, 3, 4, 5]
or
a = [1,2,3,4,5,6]
a.take 3 # => [1, 2, 3]
a.first 3 # => [1, 2, 3]
a.first a.size - 1 # to get rid of the last one
The most direct opposite of drop (drop the first n elements) would be take, which keeps the first n elements (there's also take_while which is analogous to drop_while).
Slice allows you to return a subset of the array either by specifying a range or an offset and a length. Array#[] behaves the same when passed a range as an argument or when passed 2 numbers
this will get rid of last n elements:
a = [1,2,3,4,5,6]
n = 4
p a[0, (a.size-n)]
#=> [1, 2]
n = 2
p a[0, (a.size-n)]
#=> [1, 2, 3, 4]
regard "middle" elements:
min, max = 2, 5
p a.select {|v| (min..max).include? v }
#=> [2, 3, 4, 5]
I wanted the return value to be the array without the dropped elements. I found a couple solutions here to be okay:
count = 2
[1, 2, 3, 4, 5].slice 0..-(count + 1) # => [1, 2, 3]
[1, 2, 3, 4, 5].tap { |a| a.pop count } # => [1, 2, 3]
But I found another solution to be more readable if the order of the array isn't important (in my case I was deleting files):
count = 2
[1, 2, 3, 4, 5].reverse.drop count # => [3, 2, 1]
You could tack another .reverse on there if you need to preserve order but I think I prefer the tap solution at that point.
You can achieve the same as Array#pop in a non destructive way, and without needing to know the lenght of the array:
a = [1, 2, 3, 4, 5, 6]
b = a[0..-2]
# => [1, 2, 3, 4, 5]
n = 3 # if we want drop the last n elements
c = a[0..-(n+1)]
# => [1, 2, 3]
Array#delete_at() is the simplest way to delete the last element of an array, as so
arr = [1,2,3,4,5,6]
arr.delete_at(-1)
p arr # => [1,2,3,4,5]
For deleting a segment, or segments, of an array use methods in the other answers.
You can also add some methods
class Array
# Using slice
def cut(n)
slice(0..-n-1)
end
# Using pop
def cut2(n)
dup.tap{|x| x.pop(n)}
end
# Using take
def cut3(n)
length - n >=0 ? take(length - n) : []
end
end
[1,2,3,4,5].cut(2)
=> [1, 2, 3]

Is there a particular function to retrieve then delete random array element?

I know I can do this in a couple of steps, but was wondering if there is a function which can achieve this.
I want to array#sample, then remove the element which was retrieved.
How about this:
array.delete_at(rand(array.length))
Another inefficient one, but super obvious what's going on:
array.shuffle.pop
What would be nice would be a destructive version of the sample method on Array itself, something like:
class Array
def sample!
delete_at rand length
end
end
Linuxios's has it perfect. Here is another example:
array = %w[A B C]
item_deleted = array.delete_at(1)
Here it is in irb:
1.9.2p0 :043 > array = %w[A B C]
=> ["A", "B", "C"]
1.9.2p0 :044 > item_deleted = array.delete_at(1)
=> "B"
1.9.2p0 :045 > array
=> ["A", "C"]
1.9.2p0 :047 > item_deleted
=> "B"
An alternative to the rand(array.length) approach already mentioned, could be this one
element = array.delete array.sample
Eksample:
>> array = (1..10).to_a
>> element = array.delete array.sample
>> array # => [1, 2, 4, 5, 6, 7, 8, 9, 10]
>> element # => 3
This is also a set of two operations, but at least you won't have to move away from the array itself.
If you need to sample a number of items and the remove those from the original array:
array = (1..10).to_a
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
grab = array.sample(4)
=> [2, 6, 10, 5]
grab.each{ |a| array.delete a }
=> [2, 6, 10, 5]
array
=> [1, 3, 4, 7, 8, 9]

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