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

我想我明白了:

a = [3, 2, 4, 6, 3, 8]
a.delete(3)
#=> 3
a
#=> [2, 4, 6, 8]

其他回答

我喜欢在其他答案中提到的-=[4]方式来删除值为4的元素。

但有这样一种方法:

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

在“基本数组操作”中提到的map函数之后。

我想我明白了:

a = [3, 2, 4, 6, 3, 8]
a.delete(3)
#=> 3
a
#=> [2, 4, 6, 8]

A .delete_at(3) 3在这里是位置。

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。

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