Ruby-on-rails 如何选择数组中的每第 n 个项目？
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How do you select every nth item in an array?
提问by sjsc
I'm looking to find a way in Ruby to select every nth item in an array. For instance, selecting every second item would transform:
我希望在 Ruby 中找到一种方法来选择数组中的每个第 n 项。例如，选择每隔一个项目将转换：
["cat", "dog", "mouse", "tiger"]
into:
进入：
["dog", "tiger"]
Is there a Ruby method to do so, or is there any other way to do it?
有没有 Ruby 方法可以做到这一点，或者有没有其他方法可以做到这一点？
I tried using something like:
我尝试使用类似的东西：
[1,2,3,4].select {|x| x % 2 == 0}
# results in [2,4]
but that only works for an array with integers, not strings.
但这仅适用于带有整数的数组，而不适用于字符串。
采纳答案by mu is too short
You could also use step:
您还可以使用步骤：
n = 2
a = ["cat", "dog", "mouse", "tiger"]
b = (n - 1).step(a.size - 1, n).map { |i| a[i] }
回答by Mladen Jablanovi?
You can use Enumerable#each_slice:
您可以使用Enumerable#each_slice：
["cat", "dog", "mouse", "tiger"].each_slice(2).map(&:last)
# => ["dog", "tiger"]
Update:
更新：
As mentioned in the comment, lastis not always suitable, so it could be replaced by first, and skipping first element:
正如评论中提到的，last并不总是合适的，所以它可以替换为first, 并跳过第一个元素：
["cat", "dog", "mouse", "tiger"].drop(1).each_slice(2).map(&:first)
Unfortunately, making it less elegant.
不幸的是，使它不那么优雅。
IMO, the most elegant is to use .select.with_index, which Nakilon suggested in his comment:
IMO，最优雅的是使用.select.with_index，Nakilon 在他的评论中建议：
["cat", "dog", "mouse", "tiger"].select.with_index{|_,i| (i+1) % 2 == 0}
回答by anshul
How about this -
这个怎么样 -
arr = ["cat", "dog", "mouse", "tiger"]
n = 2
(0... arr.length).select{ |x| x%n == n-1 }.map { |y| arr[y] }
#=> ["dog", "tiger"]
回答by Antonija ?imi?
You can simply use values_atmethod. You can find it easily in documentation.
您可以简单地使用values_at方法。您可以在文档中轻松找到它。
Here are some examples:
这里有些例子：
array = ["Hello", 2, "apple", 3]
array.values_at(0,1) # pass any number of arguments you like
=> ["Hello", 2]
array.values_at(0..3) # an argument can be a range
=>["Hello", 2, "apple", 3]
I believe this would fix your problem with "dog" and "tiger"
我相信这会解决你的“狗”和“老虎”问题
array = ["cat", "dog", "mouse", "tiger"]
array.values_at(1,3)
and with your another array
和你的另一个阵列
[1,2,3,4].values_at(1,3)
=> [2, 4]
回答by coorasse
If what you are looking for, is to select only odd or even numbers, there's a very simple way:
如果您要寻找的是仅选择奇数或偶数，则有一种非常简单的方法：
animals.select.with_index{ |_, i| i.odd? }
e.g.
例如
['a','b','c','d'].select.with_index{ |_,i| i.odd? }
# => ["b", "d"]
回答by Zabba
If you need that in other places, you could add a method to Enumerable:
如果您在其他地方需要它，您可以添加一个方法Enumerable：
module Enumerable
def select_with_index
index = -1
(block_given? && self.class == Range || self.class == Array) ? select { |x| index += 1; yield(x, index) } : self
end
end
p ["cat", "dog", "mouse", "tiger"].select_with_index { |x, i| x if i % 2 != 0 }
Note: This is not my original code. I got it from herewhen I had had the same need as you.
注意：这不是我的原始代码。当我和你有同样的需求时，我从这里得到了它。
回答by tokland
Yet another ways:
还有一种方式：
xs.each_with_index.map { |x, idx| x if idx % 2 != 0 }.compact
xs.each_with_index.select { |x, idx| idx % 2 }.map(&:first)
xs.values_at(*(1...xs.length).step(2))
回答by Darren Hicks
I like both Anshul's and Mu's answers and want to refine and simplify them a bit by submitting each as a monkeypatch to Enumerable:
我喜欢 Anshul 和 Mu 的答案，并希望通过将每个答案作为 Monkeypatch 提交给 Enumerable 来改进和简化它们：
Mu's
穆氏
module Enumerable
def every_nth(n)
(n - 1).step(self.size - 1, n).map { |i| self[i] }
end
end
Anshul's
安舒尔
module Enumerable
def every_nth(n)
(0... self.length).select{ |x| x%n == n-1 }.map { |y| self[y] }
end
end
Then it is very easy to work with. For example, consider:
然后它很容易使用。例如，考虑：
a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25]
a.every_nth(2)
=> [2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24]
a.every_nth(3)
=> [3, 6, 9, 12, 15, 18, 21, 24]
a.every_nth(5)
=> [5, 10, 15, 20, 25]
回答by dawn
my_array = ["cat", "dog", "mouse", "tiger"]
my_array = [“猫”、“狗”、“老鼠”、“老虎”]
my_new_array = my_array.select {|x| index(x) % 2 == 0}
my_new_array = my_array.select {|x| 索引（x）% 2 == 0}
回答by JCLL
class Array
def every(n)
select {|x| index(x) % n == 0}
end
end
