(1..10).to_a
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
(10..1).to_a
=> []
所以
for i in 1..10 do puts i end
會印 1 到 10
for i in 10..1 do puts i end
卻不會印 10 到 1
雖然可以用 10.downto(1)
不過還是太 Orz 了….
- 留言者: Duncan
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- 日期: 2005-09-27 11:48:25
或許 Python 更適合你。
- 留言者: JiaYun
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- 日期: 2005-09-27 13:49:19
Python:
for i in reversed(xrange(1,11)):
print i
這學期有個 project 有點想用 Ruby on Rails 做
如果 Python 的話也許可以用 http://www.djangoproject.com/
不過有辦法學那麼多嗎…. Orz
- 留言者: T55555
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- 日期: 2005-09-28 02:36:30
Please check the ruby-talk: 8993 and 39342
http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/8993
http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/39342
And be aware of Range using on Array index:
For example:
ary = %w[a b c d e]
ary[1..-1] # => [‘b’, ‘c’, ‘d’, ‘e’]
If we *assume* Range support from “big” to “small”,
the example 1..-1 will like have 1, 0, -1 elements
and ary[1..-1] will returns [‘b’, ‘a’, ‘e’]
but the current design is not like this way …
The main raison Range do not support from “big” to “small” is
because the Range is define base on object’s “succ”( or alias “next”) and “”
BTW: You could “redefine” the Range to behaviour as you like.
- 留言者: T55555
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- 日期: 2005-09-28 02:41:17
“succ” and ” < = > ”
後面是 *宇宙飛碟* operator.
- 留言者: T55555
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- 日期: 2005-09-28 02:55:29
Not prefect at all ( need to take care of === also, string Range … etc),
but it seems to work for what you expect.
class Range
alias old_each each
def each(&block)
return old_each(&block) unless first > last
first.downto(last) { |e| yield e }
return self
end
end
p (1..10).to_a
p (10..1).to_a
(10..1).each { |e| p e }
- 留言者: JiaYun
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- 日期: 2005-09-28 03:18:35
也許它可以加個 pred 或 prev, 再把用在 array 的情形改一下, 功能上會更對稱一點….
我好像 柯南 – 引爆摩天樓 裡那個超執著對稱的森谷帝二教授…. XD
- 留言者: Duncan
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- 日期: 2005-09-28 16:19:31
可以用 range(1, 11)[::-1] 或是 xrange(10, 0, -1)。
- 留言者: JiaYun
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- 日期: 2005-09-29 01:18:24
我想搭紅寶石列車，Duncan 集團誘惑我乘巨蟒….