Pad an array to be a certain size

There's probably a more efficient and more Rub开发者_Go百科y-ish way to do this:

# Pad array to size n by adding x's. Don't do anything if n <= a.length.
def padleft(a, n, x)
  return a if n <= a.length
  return padleft([x] + a, n, x)
end

What would you suggest?

Edited due to my misunderstanding of the question. Pervious version of my answer padded from the right side, but the question was asking to do it from the left side. I corrected it accordingly. This is due to naming convention. ljust, rjust are builtin methods for String, and I extended that convention to Array, but that corresponds to padright and padleft, respectively, in the terminology of the question.

Destructive methods

def padleft!(a, n, x)
  a.insert(0, *Array.new([0, n-a.length].max, x))
end
def padright!(a, n, x)
  a.fill(x, a.length...n)
end

It would be more natural to have it defined on Array class:

class Array
  def rjust!(n, x); insert(0, *Array.new([0, n-length].max, x)) end
  def ljust!(n, x); fill(x, length...n) end
end

Non-destructive methods

def padleft(a, n, x)
  Array.new([0, n-a.length].max, x)+a
end
def padright(a, n, x)
  a.dup.fill(x, a.length...n)
end

or

class Array
  def rjust(n, x); Array.new([0, n-length].max, x)+self end
  def ljust(n, x); dup.fill(x, length...n) end
end

FWIW:

def rpad(item, padding, num)
  Array(item).fill padding, Array(item).size, num
end
# rpad "initialize value(s)", 0, 3
# => ["initialize value(s)", 0, 0, 0]

Using 10 for the length to pad to, and 'x' to be what you're padding to, this pads right:

>> asdf = %w[0 1 2 3 ] #=> ["0", "1", "2", "3"]
>> asdf += (asdf.size < 10) ? ['x'] * (10 - asdf.size) : [] #=> ["0", "1", "2", "3", "x", "x", "x", "x", "x", "x"]

or

>> asdf = (asdf.size < 10) ? ['x'] * (10 - asdf.size) + asdf : asdf #=> ["x", "x", "x", "x", "x", "x", "0", "1", "2", "3"]

to padleft

If it makes sense to you to monkey-patch Array:

class Array
  def pad_right(s, char=nil)
    self + [char] * (s - size) if (size < s)
  end

  def pad_left(s, char=nil)
    (size < s) ? [char] * (s - size) + self : self
  end
end

%w[1 2 3].pad_right(5, 'x') # => ["1", "2", "3", "x", "x"]
%w[1 2 3].pad_left(5, 'x') # => ["x", "x", "1", "2", "3"]

Using the * operator to repeat a list.

# Pad array to size n by adding x's. Don't do anything if n <= a.length.
def padleft(a, n, x)
  return a if n <= a.length
  return [x] * (n - a.length) + a
end

Perhaps more Rubyish ;)

# Pad array to size n by adding x's. Don't do anything if n <= a.length.
def padleft(a, n, x)
  (n - a.size).times.inject(a) do |array, i|
    array << x
  end
end

Here's another fun one-liner to do it:

(non-destructive)

def padleft(a, n, x)
  a.dup.reverse.fill(x, a.length..n-1).reverse
end

(destructive)

def padleft(a, n, x)
  a.reverse.fill(x, a.length..n-1).reverse
end

If you're using Rails and want the padding on the right:

[1,2,3,4,5,6].in_groups_of(4)
=> [[1, 2, 3, 4], [5, 6, nil, nil]]

This doesn't come anywhere near answering the question but it's what I ended up needing after visiting this page. Hope it helps someone.