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

因此，当3出现多次，而你只想删除第一次出现的3时，你可以简单地执行如下操作。

arr = [2, 4, 6, 3, 8, 10, 3, 12]

arr.delete_at arr.index 3

#This will modify arr as [2, 4, 6, 8, 10, 3, 12] where first occurrence of 3 is deleted. Returns the element deleted. In this case => 3.

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

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。

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。

你也可以猴子修补它。我一直不明白为什么Ruby对哈希有except方法，而对数组没有:

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

现在你可以做:

[1,3,7,"436",354,nil].except(354) #=> [1,3,7,"436",nil]

Or:

[1,3,7,"436",354,nil].except([354, 1]) #=> [3,7,"436",nil]