I find a lot of reference about removing duplicates in ruby but I cannot find how to create duplicate.
If I have an array like [1,2,3] how can I map it to an array with dubbed items? [1,1,2,2,3,3]
Is there a method?
Try this one
[1, 2, 3].flat_map { |i| [i, i] }
=> [1, 1, 2, 2, 3, 3]
Here's yet another way, creating the array directly with Array#new :
array = [1, 2, 3]
repetitions = 2
p Array.new(array.size * repetitions) { |i| array[i / repetitions] }
# [1, 1, 2, 2, 3, 3]
According to fruity, #ursus's answer, #ilya's first two answers and mine have comparable performance. transpose.flatten is slower than any of the others.
#Ursus answer is the most clean, there are possible solutions:
a = [1, 2, 3]
a.zip(a).flatten
#=> [1, 1, 2, 2, 3, 3]
Or
a.inject([]) {|a, e| a << e << e} # a.inject([]) {|a, e| n.times {a << e}; a}
=> [1, 1, 2, 2, 3, 3]
Or
[a, a].transpose.flatten # ([a] * n).transpose.flatten
=> [1, 1, 2, 2, 3, 3]
Try this:
[1, 2, 3] * 2
=> [1, 2, 3, 1, 2, 3]
You might want it sorted:
([1, 2, 3] * 2).sort
=> [1, 1, 2, 2, 3, 3]
Related
I want to delete a value from an Array and get remain of it like this:
a = [1, 2, 3, 4]
=> [1, 2, 3, 4]
a.delete_at(2)
=> 3
a
=> [1, 2, 4]
How can I get [1, 2, 4] without destruction of the variable?
There are a couple of ways you can do, admittedly, they don't seem very elegant:
a[0..1] + a[3..-1]
# => [1, 2, 4]
a.dup.tap { |x| x.delete_at(2) }
# => [1, 2, 4]
a.values_at(0..1, 3..-1)
# => [1, 2, 4]
Personally, I think the way which conveys your intention best is:
a.reject.with_index { |_, i| i == 2 }
# => [1, 2, 4]
In Coffeescript I can do this:
[one..., two] = [1, 2, 3, 4, 5]
# one = [1, 2, 3, 4]
# two = 5
Is there any way to do this (oneliner) in Ruby?
EDIT
I know that I can do this:
one = [1, 2, 3, 4, 5]
two = one.slice!(-1)
EDIT 2
Oneliner could look like this:
two = (one = [1, 2, 3, 4, 5]).slice!(-1)
but this seems too ugly for me.
This should do it:
*one, two = [1, 2, 3, 4, 5]
one
# => [1, 2, 3, 4]
two
# => 5
You can see some more explanations on splat and array destructuring here
irb(main):001:0> a = [1,2,3,4,5]
=> [1, 2, 3, 4, 5]
irb(main):002:0> *one, two = a
=> [1, 2, 3, 4, 5]
irb(main):003:0> one
=> [1, 2, 3, 4]
irb(main):004:0> two
=> 5
I'd like to get [[2, 1, 3], [1, 3, 2]] from [1, 2, 3] in Ruby.
For [1, 2, 3, 4], I'd like to get [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
Rule: Within two numbers, if left one is smaller then it swap the position.
I have the following codes so far but it returns [[2, 3, 1], [2, 3, 1]]
What am I doing wrong here? I appreciate any inputs.
In amidakuji.rb
class Amidakuji
def initialize(column, rung)
#column = column
#rung = rung
#myarr = []
#per_arr = []
#build_arr = []
end
def build_initial
#arr = (1..#column).to_a
end
def swap_element
i = 0
arr = build_initial
while i < #column - 1 do
#build_arr << swap(arr, i)
i += 1
end
#build_arr
end
def swap(arr, a)
if arr[a] < arr[a + 1]
arr[a], arr[a + 1] = arr[a + 1], arr[a]
end
arr
end
end
In amidakuji_spec.rb
it 'should create an array with swapped elements' do
expect(#kuji1.swap_element).to eq ([[2, 1, 3], [1, 3, 2]])
end
Results
Failures:
expected: [[2, 1, 3], [1, 3, 2]]
got: [[2, 3, 1], [2, 3, 1]]
You can do this quite compactly by using the methods Enumerable#each_cons and Enumerable#map.
Code
def doit(arr)
(0...arr.size).each_cons(2).map do |i,j|
a = arr.dup
a[i], a[j] = a[j], a[i]
a
end
end
Examples
doit([1,2,3]) #=> [[2, 1, 3], [1, 3, 2]]
doit([1,2,3,4]) #=> [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
doit([1,2,3,4,5]) #=> [[2, 1, 3, 4, 5], [1, 3, 2, 4, 5],
#=> [1, 2, 4, 3, 5], [1, 2, 3, 5, 4]]
Explanation
arr = [1,2,3,4]
b = (0...arr.size).each_cons(2)
#=> #<Enumerator: 0...4:each_cons(2)>
To view the contents of this enumerator:
b.to_a
#=> [[0, 1], [1, 2], [2, 3]]
Lastly
b.map do |i,j|
a = arr.dup
a[i], a[j] = a[j], a[i]
a
end
#=> [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
In the last step, consider the first element of b that is passed to map, which assigns the following values to the block variables:
i => 0
j => 1
We then make a copy of arr, swap the elements offsets 0 and 1, making
a => [2, 1, 3, 4]
and then enter a at the end of the block, causing map to replace [0, 1] with that array.
Given what you're trying to accomplish and the output you're getting, it looks like you're reusing the same array when you want distinct arrays instead. Specifically this line:
#build_arr << swap(arr, i)
is always passing the same 'arr' to swap.
So first time, it swaps the 1 and the 2 to give you [2, 1, 3]
Second time, it swaps the 1 and the 3 give you [2, 3, 1]
You push the same array onto #build_arr twice, which is why it repeats.
I couldn't find a way to build an array such as
[ [1,2,3] , [1,2,3] , [1,2,3] , [1,2,3] , [1,2,3] ]
given [1,2,3] and the number 5. I guess there are some kind of operators on arrays, such as product of mult, but none in the doc does it. Please tell me. I missed something very simple.
Array.new(5, [1, 2, 3]) or Array.new(5) { [1, 2, 3] }
Array.new(size, default_object) creates an array with an initial size, filled with the default object you specify. Keep in mind that if you mutate any of the nested arrays, you'll mutate all of them, since each element is a reference to the same object.
array = Array.new(5, [1, 2, 3])
array.first << 4
array # => [[1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]]
Array.new(size) { default_object } lets you create an array with separate objects.
array = Array.new(5) { [1, 2, 3] }
array.first << 4
array #=> [[1, 2, 3, 4], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
Look up at the very top of the page you linked to, under the section entitled "Creating Arrays" for some more ways to create arrays.
Why not just use:
[[1, 2, 3]] * 5
# => [[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
The documentation for Array.* says:
...returns a new array built by concatenating the int copies of self.
Typically we'd want to create a repeating single-level array:
[1, 2, 3] * 2
# => [1, 2, 3, 1, 2, 3]
but there's nothing to say we can't use a sub-array like I did above.
It looks like mutating one of the subarrays mutates all of them, but that may be what someone wants.
It's like Array.new(5, [1,2,3]):
foo = [[1, 2, 3]] * 2
foo[0][0] = 4
foo # => [[4, 2, 3], [4, 2, 3]]
foo = Array.new(2, [1,2,3])
foo[0][0] = 4
foo # => [[4, 2, 3], [4, 2, 3]]
A work-around, if that's not the behavior wanted is:
foo = ([[1, 2, 3]] * 2).map { |a| [*a] }
foo[0][0] = 4
foo # => [[4, 2, 3], [1, 2, 3]]
But, at that point it's not as convenient, so I'd use the default Array.new(n) {…} behavior.
Is there a way to do:
a = b.map{ |e| #return multiple elements to be added to a }
Where rather than returning a single object for each iteration to be added to a, multiple objects can be returned.
I'm currently achieving this with:
a = []
b.map{ |e| a.concat([x,y,z]) }
Is there a way to this in a single line without having to declare a = [] up front?
Use Enumerable#flat_map
b = [0, 3, 6]
a = b.flat_map { |x| [x, x+1, x+2] }
a # => [0, 1, 2, 3, 4, 5, 6, 7, 8]
Use Enumerable#flat_map
Which is probably not much different than:
p [1, 2, 3].map{|num| [1, 2, 3]}.flatten
--output:-
[1, 2, 3, 1, 2, 3, 1, 2, 3]