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
另一个选择:
a = [2,4,6,3,8]
a -= [3]
结果是
=> [2, 4, 6, 8]
其他回答
A .delete_at(3) 3在这里是位置。
另一个选择:
a = [2,4,6,3,8]
a -= [3]
结果是
=> [2, 4, 6, 8]
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.
你可以简单地运行:
[2,4,6,3,8].delete(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