如何使用Ruby在一步中初始化数组？(How to initialize an array in one step using Ruby?)

如何使用Ruby在一步中初始化数组？(How to initialize an array in one step using Ruby?)

我以这种方式初始化数组：

array = Array.new array << '1' << '2' << '3'

有可能一步做到吗？ 如果是这样，怎么样？

I initialize an array this way:

array = Array.new array << '1' << '2' << '3'

Is it possible to do that in one step? If so, how?

最满意答案

正如其他人所说，你可以使用数组文字：

array = [ '1', '2', '3' ]

您也可以使用范围：

array = ('1'..'3').to_a # parentheses are required # or array = *('1'..'3') # parentheses not required, but included for clarity

对于许多空格分隔的字符串的数组，最简单的是这个符号：

array = %w[ 1 2 3 ]

一般来说，您可以将块传递给Array.new，并使用它来确定每个条目的值为：

array = Array.new(3){ |i| (i+1).to_s }

最后，尽管它不会产生与上面其他答案相同的三个字符串数组，但也可以使用Ruby 1.8.7+中的枚举器来创建数组; 例如：

array = 1.step(17,3).to_a #=> [1, 4, 7, 10, 13, 16]

You can use an array literal:

array = [ '1', '2', '3' ]

You can also use a range:

array = ('1'..'3').to_a # parentheses are required # or array = *('1'..'3') # parentheses not required, but included for clarity

For arrays of whitespace-delimited strings, you can use Percent String syntax:

array = %w[ 1 2 3 ]

You can also pass a block to Array.new to determine what the value for each entry will be:

array = Array.new(3) { |i| (i+1).to_s }

Finally, although it doesn't produce the same array of three strings as the other answers above, note also that you can use enumerators in Ruby 1.8.7+ to create arrays; for example:

array = 1.step(17,3).to_a #=> [1, 4, 7, 10, 13, 16]