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。
其他回答
你可以简单地运行:
[2,4,6,3,8].delete(3)
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]
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]
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.