我有一个包含重复元素的Ruby数组。
array = [1,2,2,1,4,4,5,6,7,8,5,6]
我如何从这个数组中删除所有重复的元素,同时保留所有唯一的元素,而不使用for循环和迭代?
我有一个包含重复元素的Ruby数组。
array = [1,2,2,1,4,4,5,6,7,8,5,6]
我如何从这个数组中删除所有重复的元素,同时保留所有唯一的元素,而不使用for循环和迭代?
当前回答
只是为了提供一些见解:
require 'fruity'
require 'set'
array = [1,2,2,1,4,4,5,6,7,8,5,6] * 1_000
def mithun_sasidharan(ary)
ary.uniq
end
def jaredsmith(ary)
ary & ary
end
def lri(ary)
counts = Hash.new(0)
ary.each { |v| counts[v] += 1 }
counts.select { |v, count| count == 1 }.keys
end
def finks(ary)
ary.to_set
end
def santosh_mohanty(ary)
result = ary.reject.with_index do |ele,index|
res = (ary[index+1] ^ ele)
res == 0
end
end
SHORT_ARRAY = [1,1,2,2,3,1]
mithun_sasidharan(SHORT_ARRAY) # => [1, 2, 3]
jaredsmith(SHORT_ARRAY) # => [1, 2, 3]
lri(SHORT_ARRAY) # => [3]
finks(SHORT_ARRAY) # => #<Set: {1, 2, 3}>
santosh_mohanty(SHORT_ARRAY) # => [1, 2, 3, 1]
puts 'Ruby v%s' % RUBY_VERSION
compare do
_mithun_sasidharan { mithun_sasidharan(array) }
_jaredsmith { jaredsmith(array) }
_lri { lri(array) }
_finks { finks(array) }
_santosh_mohanty { santosh_mohanty(array) }
end
当运行时,结果是:
# >> Ruby v2.7.1
# >> Running each test 16 times. Test will take about 2 seconds.
# >> _mithun_sasidharan is faster than _jaredsmith by 2x ± 0.1
# >> _jaredsmith is faster than _santosh_mohanty by 4x ± 0.1 (results differ: [1, 2, 4, 5, 6, 7, 8] vs [1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, ...
# >> _santosh_mohanty is similar to _lri (results differ: [1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, ...
# >> _lri is similar to _finks (results differ: [] vs #<Set: {1, 2, 4, 5, 6, 7, 8}>)
注意:这些返回的结果不好:
lri(SHORT_ARRAY) # => [3] > . finks(SHORT_ARRAY) # => #<Set: {1,2,3 santosh_mohanty(SHORT_ARRAY) # => [1,2,3,1]
其他回答
你可以返回交叉点。
a = [1,1,2,3]
a & a
这也将删除重复项。
你可以用uniq方法删除重复的元素:
array.uniq # => [1, 2, 4, 5, 6, 7, 8]
还有一个可能有用的是,uniq接受一个块,所以如果你有一个键数组:
["bucket1:file1", "bucket2:file1", "bucket3:file2", "bucket4:file2"]
你想知道什么是唯一的文件,你可以用:
a.uniq { |f| f[/\d+$/] }.map { |p| p.split(':').last }
尝试使用XOR操作符,不使用内置函数:
a = [3,2,3,2,3,5,6,7].sort!
result = a.reject.with_index do |ele,index|
res = (a[index+1] ^ ele)
res == 0
end
print result
内置功能:
a = [3,2,3,2,3,5,6,7]
a.uniq
只是为了提供一些见解:
require 'fruity'
require 'set'
array = [1,2,2,1,4,4,5,6,7,8,5,6] * 1_000
def mithun_sasidharan(ary)
ary.uniq
end
def jaredsmith(ary)
ary & ary
end
def lri(ary)
counts = Hash.new(0)
ary.each { |v| counts[v] += 1 }
counts.select { |v, count| count == 1 }.keys
end
def finks(ary)
ary.to_set
end
def santosh_mohanty(ary)
result = ary.reject.with_index do |ele,index|
res = (ary[index+1] ^ ele)
res == 0
end
end
SHORT_ARRAY = [1,1,2,2,3,1]
mithun_sasidharan(SHORT_ARRAY) # => [1, 2, 3]
jaredsmith(SHORT_ARRAY) # => [1, 2, 3]
lri(SHORT_ARRAY) # => [3]
finks(SHORT_ARRAY) # => #<Set: {1, 2, 3}>
santosh_mohanty(SHORT_ARRAY) # => [1, 2, 3, 1]
puts 'Ruby v%s' % RUBY_VERSION
compare do
_mithun_sasidharan { mithun_sasidharan(array) }
_jaredsmith { jaredsmith(array) }
_lri { lri(array) }
_finks { finks(array) }
_santosh_mohanty { santosh_mohanty(array) }
end
当运行时,结果是:
# >> Ruby v2.7.1
# >> Running each test 16 times. Test will take about 2 seconds.
# >> _mithun_sasidharan is faster than _jaredsmith by 2x ± 0.1
# >> _jaredsmith is faster than _santosh_mohanty by 4x ± 0.1 (results differ: [1, 2, 4, 5, 6, 7, 8] vs [1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, ...
# >> _santosh_mohanty is similar to _lri (results differ: [1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, 7, 8, 5, 6, 1, 2, 1, 4, 5, 6, ...
# >> _lri is similar to _finks (results differ: [] vs #<Set: {1, 2, 4, 5, 6, 7, 8}>)
注意:这些返回的结果不好:
lri(SHORT_ARRAY) # => [3] > . finks(SHORT_ARRAY) # => #<Set: {1,2,3 santosh_mohanty(SHORT_ARRAY) # => [1,2,3,1]
如果有人在乎的话,这只是另一个选择。
您还可以使用数组的to_set方法将array转换为Set,并且根据定义,Set元素是唯一的。
[1,2,3,4,5,5,5,6].to_set => [1,2,3,4,5,6]