Ruby

marcandre (Marc-Andre Lafortune) wrote:

Currently, Enumerable#drop works only for non-negative arguments.

It could be extended so that negative arguments means dropping from the end:

[:hello, :world].drop(-1) # => [:hello]

This could especially be interesting for Lazy#drop, which would keep a circular buffer of elements before yielding them.

(1..6).lazy.drop(-3).each{|x| puts x} # -> prints 1, 2 and 3

Thoughts?

How Enumerable#drop can know the total number of elements?
The source of elements might be IO. Besides that, the total number of elements might be infinite.

Hi,

shugo (Shugo Maeda) wrote:

How Enumerable#drop can know the total number of elements?
The source of elements might be IO. Besides that, the total number of elements might be infinite.

drop would have to consume the whole iteration, indeed, which is why I was talking about a buffer. The buffer holds elements until we know they are not in the last (-n) elements. Here's a Ruby implementation:

class Lazy
def drop(n)
return to_enum :drop, n unless block_given?
if n < 0
buffer = []
each do |e|
buffer << e
yield buffer.shift if buffer.size > -n
end
else
# ...
end
end
end

For infinite sequences, drop with negative argument would not be very useful, but it would still yield all elements.

marcandre (Marc-Andre Lafortune) wrote:

How Enumerable#drop can know the total number of elements?
The source of elements might be IO. Besides that, the total number of elements might be infinite.

drop would have to consume the whole iteration, indeed, which is why I was talking about a buffer. The buffer holds elements until we know they are not in the last (-n) elements. Here's a Ruby implementation:

Ah, I see. It's interesting.

Do you have any use case in mind?

Using an existing feature, [1,2,3,4,5].lazy.drop(-n) could be written as [1,2,3,4,5].each_cons(n+1).lazy.map(&:first).
Enumerable#each_cons does not use a circular buffer, though. (It currently uses push & shift)