开发者

How do you select every nth item in an array?

开发者 https://www.devze.com 2023-02-04 00:17 出处:网络
I\'m looking to find a way in Ruby to select every nth item in an array. For instance, selecting every second item 开发者_运维问答would transform:

I'm looking to find a way in Ruby to select every nth item in an array. For instance, selecting every second item 开发者_运维问答would transform:

["cat", "dog", "mouse", "tiger"]

into:

["dog", "tiger"]

Is there a Ruby method to do so, or is there any other way to do it?

I tried using something like:

[1,2,3,4].select {|x| x % 2 == 0}
# results in [2,4]

but that only works for an array with integers, not strings.


You can use Enumerable#each_slice:

["cat", "dog", "mouse", "tiger"].each_slice(2).map(&:last)
# => ["dog", "tiger"]

Update:

As mentioned in the comment, last is not always suitable, so it could be replaced by first, and skipping first element:

["cat", "dog", "mouse", "tiger"].drop(1).each_slice(2).map(&:first)

Unfortunately, making it less elegant.

IMO, the most elegant is to use .select.with_index, which Nakilon suggested in his comment:

["cat", "dog", "mouse", "tiger"].select.with_index{|_,i| (i+1) % 2 == 0}


You could also use step:

n = 2
a = ["cat", "dog", "mouse", "tiger"]
b = (n - 1).step(a.size - 1, n).map { |i| a[i] }


How about this -

arr = ["cat", "dog", "mouse", "tiger"]
n = 2
(0... arr.length).select{ |x| x%n == n-1 }.map { |y| arr[y] } 
    #=> ["dog", "tiger"]


You can simply use values_at method. You can find it easily in documentation.

Here are some examples:

array = ["Hello", 2, "apple", 3]
array.values_at(0,1) # pass any number of arguments you like
=> ["Hello", 2]

array.values_at(0..3) # an argument can be a range
=>["Hello", 2, "apple", 3]

I believe this would fix your problem with "dog" and "tiger"

array = ["cat", "dog", "mouse", "tiger"]
array.values_at(1,3)

and with your another array

[1,2,3,4].values_at(1,3)
=> [2, 4] 


If you need that in other places, you could add a method to Enumerable:

module Enumerable
   def select_with_index
      index = -1
      (block_given? && self.class == Range || self.class == Array)  ?  select { |x| index += 1; yield(x, index) }  :  self
   end
end

p ["cat", "dog", "mouse", "tiger"].select_with_index { |x, i| x if i % 2 != 0 }

Note: This is not my original code. I got it from here when I had had the same need as you.


Yet another ways:

xs.each_with_index.map { |x, idx| x if idx % 2 != 0 }.compact

xs.each_with_index.select { |x, idx| idx % 2 }.map(&:first)

xs.values_at(*(1...xs.length).step(2))


If what you are looking for, is to select only odd or even numbers, there's a very simple way:

animals.select.with_index{ |_, i| i.odd? }

e.g.

['a','b','c','d'].select.with_index{ |_,i| i.odd? }
# => ["b", "d"]


I like both Anshul's and Mu's answers and want to refine and simplify them a bit by submitting each as a monkeypatch to Enumerable:

Mu's

module Enumerable
  def every_nth(n)
    (n - 1).step(self.size - 1, n).map { |i| self[i] }
  end
end 

Anshul's

module Enumerable
  def every_nth(n)
    (0... self.length).select{ |x| x%n == n-1 }.map { |y| self[y] }
  end
end 

Then it is very easy to work with. For example, consider:

a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25]

a.every_nth(2)
 => [2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24] 

a.every_nth(3)
 => [3, 6, 9, 12, 15, 18, 21, 24]

a.every_nth(5)
 => [5, 10, 15, 20, 25]


my_array = ["cat", "dog", "mouse", "tiger"]

my_new_array = my_array.select {|x| index(x) % 2 == 0}


class Array
  def every(n)
    select {|x| index(x) % n == 0}
  end
end
0

精彩评论

暂无评论...
验证码 换一张
取 消