Looping through an array with step

问题:

I want to look at every n-th elements in an array. In C++, I'd do this:

for(int x = 0; x<cx; x+=n){
    value_i_care_about = array[x];
    //do something with the value I care about.  
}

I want to do the same in Ruby, but can't find a way to "step". A while loop could do the job, but I find it distasteful using it for a known size, and expect there to be a better (more Ruby) way of doing this.

回答1:

Ranges have a step method which you can use to skip through the indexes:

(0..array.length - 1).step(2).each do |index|
  value_you_care_about = array[index]
end

Or if you are comfortable using ... with ranges the following is a bit more concise:

(0...array.length).step(2).each do |index|
  value_you_care_about = array[index]
end

回答2:

array.each_slice(n) do |e, *_|
  value_i_care_about = e
end

回答3:

Just use step() method from Range class which returns an enumerator

(1..10).step(2) {|x| puts x}

回答4:

We can iterate while skipping over a range of numbers on every iteration e.g.:

1.step(10, 2) { |i| print "#{i} "}

http://www.skorks.com/2009/09/a-wealth-of-ruby-loops-and-iterators/

So something like:

array.step(n) do |element|
  # process element
end

回答5:

This is a great example for the use of the modulo operator %

When you grasp this concept, you can apply it in a great number of different programming languages, without having to know them in and out.

step = 2
["1st","2nd","3rd","4th","5th","6th"].each_with_index do |element, index|
  puts element if index % step == 1
end

#=> "2nd"
#=> "4th"
#=> "6th"

回答6:

class Array
def step(interval, &block)
    ((interval -1)...self.length).step(interval) do |value|
        block.call(self[value])
    end
end
end

You could add the method to the class Array

回答7:

What about:

> [1, 2, 3, 4, 5, 6, 7].select.each_with_index { |_,i| i % 2 == 0 }
=> [1, 3, 5, 7]

Chaining of iterators is very useful.