ruby - 如何从满足谓词的 Ruby 数组中选择前 n 个元素？

Question

我想从数组中获取满足谓词的所有项。一旦我看到一个不满足的元素，我就应该停止迭代。例如:

[1, 4, -9, 3, 6].select_only_first { |x| x > 0}

我期望得到:[1, 4]

Answer 1

这就是你想要的:

arup@linux-wzza:~> pry
[1] pry(main)> [1, 4, -9, 3, 6].take_while { |x| x > 0}
=> [1, 4]
[2] pry(main)>

这是文档:

arup@linux-wzza:~> ri Array#take_while

= Array#take_while

(from ruby site)
------------------------------------------------------------------------------
  ary.take_while { |arr| block }  -> new_ary
  ary.take_while                  -> Enumerator

------------------------------------------------------------------------------

Passes elements to the block until the block returns nil or false, then stops
iterating and returns an array of all prior elements.

If no block is given, an Enumerator is returned instead.

See also Array#drop_while

  a = [1, 2, 3, 4, 5, 0]
  a.take_while { |i| i < 3 }  #=> [1, 2]


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