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?
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