ruby 数组中的 1 到 100 个奇数
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1 to 100 odd numbers in array
提问by Johan S
Is there any cool way in Ruby to create an array with 1 to 100 with only odd entries (1, 3 etc). I now have a loop for this but that is obviously not a cool way to do it! Any suggestions?
在 Ruby 中有什么很酷的方法可以创建一个只有奇数条目（1、3 等）的 1 到 100 的数组。我现在有一个循环，但这显然不是一个很酷的方法！有什么建议？
My current code:
我目前的代码：
def create_1_to_100_odd_array
array = [1]
i = 3
while i < 100
array.push i
i += 2
end
array
end
Thanks in advance
提前致谢
回答by Vincent B.
回答by sawa
May not be efficient, but a short piece of code:
可能效率不高，但是一小段代码：
(1..100).select(&:odd?)
# => [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99]
回答by steenslag
Just toying...
玩玩而已...
(0...50).map(&:object_id)
#or
1.step(100,2).to_a
回答by FUT
Since you need a function, then:
既然你需要一个函数，那么：
def odd_to(n)
(1..n).step(2).to_a
end
回答by samuil
Not very effective solution, but quite elegant:
不是很有效的解决方案，但很优雅：
(1..100).select {|a| a%2 != 0}
回答by pjs
You can do it as a one-liner when you instantiate the array:
当您实例化数组时，您可以将其作为单行：
def create_array_of_odds_to(n)
Array.new((n + 1) / 2) {|i| 2 * i + 1}
end
create_array_of_odds_to 10 # => [1, 3, 5, 7, 9]
