我在Ruby中有一个元素数组

[2,4,6,3,8]

例如,我需要删除值为3的元素

我怎么做呢?


当前回答

以下是一些基准:

require 'fruity'


class Array          
  def rodrigo_except(*values)
    self - values
  end    

  def niels_except value
    value = value.kind_of?(Array) ? value : [value]
    self - value
  end
end

ARY = [2,4,6,3,8]

compare do
  soziev  { a = ARY.dup; a.delete(3);               a }
  steve   { a = ARY.dup; a -= [3];                  a }
  barlop  { a = ARY.dup; a.delete_if{ |i| i == 3 }; a }
  rodrigo { a = ARY.dup; a.rodrigo_except(3);         }
  niels   { a = ARY.dup; a.niels_except(3);           }
end

# >> Running each test 4096 times. Test will take about 2 seconds.
# >> soziev is similar to barlop
# >> barlop is faster than steve by 2x ± 1.0
# >> steve is faster than rodrigo by 4x ± 1.0
# >> rodrigo is similar to niels

再次使用包含大量重复项的更大数组:

class Array          
  def rodrigo_except(*values)
    self - values
  end    

  def niels_except value
    value = value.kind_of?(Array) ? value : [value]
    self - value
  end
end

ARY = [2,4,6,3,8] * 1000

compare do
  soziev  { a = ARY.dup; a.delete(3);               a }
  steve   { a = ARY.dup; a -= [3];                  a }
  barlop  { a = ARY.dup; a.delete_if{ |i| i == 3 }; a }
  rodrigo { a = ARY.dup; a.rodrigo_except(3);         }
  niels   { a = ARY.dup; a.niels_except(3);           }
end

# >> Running each test 16 times. Test will take about 1 second.
# >> steve is faster than soziev by 30.000000000000004% ± 10.0%
# >> soziev is faster than barlop by 50.0% ± 10.0%
# >> barlop is faster than rodrigo by 3x ± 0.1
# >> rodrigo is similar to niels

甚至更大,有更多的副本:

class Array          
  def rodrigo_except(*values)
    self - values
  end    

  def niels_except value
    value = value.kind_of?(Array) ? value : [value]
    self - value
  end
end

ARY = [2,4,6,3,8] * 100_000

compare do
  soziev  { a = ARY.dup; a.delete(3);               a }
  steve   { a = ARY.dup; a -= [3];                  a }
  barlop  { a = ARY.dup; a.delete_if{ |i| i == 3 }; a }
  rodrigo { a = ARY.dup; a.rodrigo_except(3);         }
  niels   { a = ARY.dup; a.niels_except(3);           }
end

# >> Running each test once. Test will take about 6 seconds.
# >> steve is similar to soziev
# >> soziev is faster than barlop by 2x ± 0.1
# >> barlop is faster than niels by 3x ± 1.0
# >> niels is similar to rodrigo

其他回答

借用Travis的评论,这是一个更好的答案:

我个人喜欢[1,2,7,4,5]-[7],结果=>[1,2,4,5]从irb

我修改了他的答案,因为3是他示例数组中的第三个元素。对于那些没有意识到3在数组中的位置2的人来说,这可能会导致一些困惑。

如果你还想让这个删除操作具有可链性,这样你就可以删除一些项并继续在结果数组上进行链接操作,请使用tap:

[2, 4, 6, 3, 8].tap { |ary| ary.delete(3) }.count #=> 4

我不确定是否有人已经声明了这一点,但Array.delete()和-= value将删除数组中传递给它的每个值的实例。为了删除特定元素的第一个实例,您可以执行如下操作

arr = [1,3,2,44,5]
arr.delete_at(arr.index(44))

#=> [1,3,2,5]

有一种更简单的方法。我并不是说这是最佳实践,但这是应该被认可的。

我改进了尼尔斯的解决方案

class Array          
  def except(*values)
    self - values
  end    
end

现在你可以使用

[1, 2, 3, 4].except(3, 4) # return [1, 2]
[1, 2, 3, 4].except(4)    # return [1, 2, 3]

假设你想在数组的多个位置按值删除3, 我认为ruby执行这个任务的方法是使用delete_if方法:

[2,4,6,3,8,3].delete_if {|x| x == 3 } 

你也可以使用delete_if来删除'array of arrays'场景中的元素。

希望这能解决你的疑问