code:
c = 0
items.each { |i|
puts i.to_s
# if c > 9 escape the each iteration early - and do not repeat
c++
}
I want to grab t开发者_StackOverflow社区he first 10 items then leave the "each" loop.
What do I replace the commented line with? is there a better approach? something more Ruby idiomatic?
While the break
solution works, I think a more functional approach really suits this problem. You want to take
the first 10 elements and print them so try
items.take(10).each { |i| puts i.to_s }
There is no ++
operator in Ruby. It's also convention to use do
and end
for multi-line blocks. Modifying your solution yields:
c = 0
items.each do |i|
puts i.to_s
break if c > 9
c += 1
end
Or also:
items.each_with_index do |i, c|
puts i.to_s
break if c > 9
end
See each_with_index
and also Programming Ruby Break, Redo, and Next.
Update: Chuck's answer with ranges is more Ruby-like, and nimrodm's answer using take
is even better.
break
works for escaping early from a loop, but it's more idiomatic just to do items[0..9].each {|i| puts i}
. (And if all you're doing is literally printing the items with no changes at all, you can just do puts items[0..9]
.)
Another option would be
items.first(10).each do |i|
puts i.to_s
end
That reads a little more easily to me than breaking on an iterator, and first will return only as many items as available if there aren't enough.
Another variant:
puts items.first(10)
Note that this works fine with arrays of less than 10 items:
>> nums = (1..5).to_a
=> [1, 2, 3, 4, 5]
>> puts nums.first(10)
1
2
3
4
5
(One other note, a lot of people are offering some form of puts i.to_s
, but in such a case, isn't .to_s
redundant? puts
will automatically call .to_s
on a non-string to print it out, I thought. You would only need .to_s
if you wanted to say puts 'A' + i.to_s
or the like.)
Does this look like what you want?
10.times { |i|
puts items[i].to_s
}
items.each_with_index { |i, c| puts i and break if c <= 9 }
It was asked:
I want to grab the first 10 items then leave the "each" loop.
Use throw
and catch
to accomplish this, with few changes to the example:
catch(:done) do
c = 0
collected = []
items.each do |item|
collected << item
throw(:done, collected) if c == 9 # started at 0
c += 1
end
collected # if the list is less than 10 long, return what was collected
end
Simply throw
the label :done
with collected
and the catch
which is waiting for :done
will return collected
.
And to "ruby" this up a bit:
catch(:done) do
items.inject([]) do |collected, item|
throw(:done, collected) if collected.size == 10
collected << item # collected gets returned here and populates the first argument of this block
end
end
I do not know why some people refuse to use inject
and use reduce
instead (they are equivalent) when clearly the empty array given to inject([])
is being injected with item
s! Anyhow, the inject
will return collected
if there are less than 10 items.
Most answers are trying to answer what might be the intent of the question instead of what was asked and items.take(10)
does make perfect sense in that case. But I can imagine wanting to grab the first items that fit within my $100 budget. Then you can simply:
catch(:done) do
items.inject({items: [], budget: 100}) do |ledger, item|
remainder = ledger[:budget] - item.price
if remainder < 0
throw(:done, ledger)
else
ledger.tap do |this|
this[:items] << item
this[:budget] = remainder
end # tap just returns what is being tapped into, in this case, ledger
end
end
end
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