Array#Rotate Equivalent in Ruby 1.8.7

Array#rotate equivalent in ruby 1.8.7

If you require 'backports/1.9.2/array/rotate', you will get Array#rotate and rotate! in older versions of Ruby.

Either way, you avoid reinventing the wheel, and more importantly you gain the advantage of an implementation that passes RubySpec. It will work for all corner cases and ensure compatibility with Ruby 1.9.

For example, none of the two answers given work for []!

Array#rotate produces a NoMethod error

It's new to Ruby 1.9. You must be using Ruby 1.8.6 or 1.8.7.

Here is what you can do.

Printing an ASCII spinning cursor in the console

Yes, this works on Windows, OS X, and Linux. Improving on Niklas' suggestion, you can make this more general like so:

def show_wait_cursor(seconds,fps=10)
  chars = %w[| / - \\]
  delay = 1.0/fps
  (seconds*fps).round.times{ |i|
    print chars[i % chars.length]
    sleep delay
    print "\b"
  }
end

show_wait_cursor(3)

If you don't know how long the process will take, you can do this in another thread:

def show_wait_spinner(fps=10)
  chars = %w[| / - \\]
  delay = 1.0/fps
  iter = 0
  spinner = Thread.new do
    while iter do  # Keep spinning until told otherwise
      print chars[(iter+=1) % chars.length]
      sleep delay
      print "\b"
    end
  end
  yield.tap{       # After yielding to the block, save the return value
    iter = false   # Tell the thread to exit, cleaning up after itself…
    spinner.join   # …and wait for it to do so.
  }                # Use the block's return value as the method's
end

print "Doing something tricky..."
show_wait_spinner{
  sleep rand(4)+2 # Simulate a task taking an unknown amount of time
}
puts "Done!"

This one outputs:

Doing something tricky...|
Doing something tricky.../
Doing something tricky...-
Doing something tricky...\ 
(et cetera)
Doing something tricky...done!

Ruby method to cycle array to the right

You can use rotate method which does similar things

2.6.5 :012 > arr = [1,2,3,4]
 => [1, 2, 3, 4]
2.6.5 :013 > arr.rotate
 => [2, 3, 4, 1]
2.6.5 :014 > arr = [1,2,3,4,5]
 => [1, 2, 3, 4, 5]
2.6.5 :015 > arr.rotate(2)
 => [3, 4, 5, 1, 2]

For Ex :-

2.6.5 :021 > def array_rotation(array, shift = 1)
2.6.5 :022?>     array.rotate(- shift)
2.6.5 :023?>   end
 => :array_rotation
2.6.5 :024 > arr = [1,2,3,4,5]
 => [1, 2, 3, 4, 5]
2.6.5 :025 > array_rotation arr, 2
 => [4, 5, 1, 2, 3]
2.6.5 :026 > arr = [1,2,3,4]
 => [1, 2, 3, 4]
2.6.5 :027 > array_rotation arr, 1
 => [4, 1, 2, 3]

How do I iterate through an array and change characters efficiently?

I like the tr approach, but if you want something similar to next, then Array#rotate could be a good option; here's an example:

def letter_swap(full_name)
  vowels     = %w(a e i o u)
  consonants = %w(b c d f g h j k l m n p q r s t v w x y z)

  full_name.each_char.with_object("") do |char, new_name|
    if vowels.include?(char)
      new_name << vowels.rotate(vowels.index(char) + 1).first
    elsif consonants.include?(char)
      new_name << consonants.rotate(consonants.index(char) + 1).first
    else
      new_name << char
    end
  end
end

Of course you could DRY this code, but at the expense of readability (IMO); for example:

def letter_swap(full_name)
  letters = [%w(a e i o u), %w(b c d f g h j k l m n p q r s t v w x y z)]

  full_name.each_char.with_object("") do |char, new_name|
    group = letters.find { |group| group.include?(char) }
    new_name << (group.nil? ? char : group.rotate(group.index(char) + 1).first)
  end
end

How do I find the first two consecutive elements in my array of numbers?

You could use each_cons and find to get the first element from the array of pairs where the second element less the first one is equal to 1:

p [1, 7, 8, 12, 14, 15].each_cons(2).find { |a, b| b - a == 1 }
# => [7, 8]

How do you rotate a two dimensional array?

Here it is in C#

int[,] array = new int[4,4] {
    { 1,2,3,4 },
    { 5,6,7,8 },
    { 9,0,1,2 },
    { 3,4,5,6 }
};

int[,] rotated = RotateMatrix(array, 4);

static int[,] RotateMatrix(int[,] matrix, int n) {
    int[,] ret = new int[n, n];

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            ret[i, j] = matrix[n - j - 1, i];
        }
    }

    return ret;
}

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