从Ruby中的数组中删除重复元素
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Remove duplicate elements from array in Ruby
提问by Mithun Sasidharan
I have a Ruby array which contains duplicate elements.
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]
How can I remove all the duplicate elements from this array while retaining all unique elements without using for-loops and iteration?
如何在不使用 for 循环和迭代的情况下从该数组中删除所有重复元素，同时保留所有唯一元素？
回答by Mithun Sasidharan
回答by jaredsmith
You can return the intersection.
您可以返回交叉点。
a = [1,1,2,3]
a & a
This will also delete duplicates.
这也将删除重复项。
回答by Marek P?íhoda
You can remove the duplicate elements with the uniq method:
您可以使用 uniq 方法删除重复元素：
array.uniq # => [1, 2, 4, 5, 6, 7, 8]
What might also be useful to know is that uniqtakes a block, so if you have a have an array of keys:
知道它uniq需要一个块也是有用的，所以如果你有一个键数组：
["bucket1:file1", "bucket2:file1", "bucket3:file2", "bucket4:file2"]
and you want to know what the unique files are, you can find it out with:
并且您想知道独特的文件是什么，您可以通过以下方式找到它：
a.uniq { |f| f[/\d+$/] }.map { |p| p.split(':').last }
回答by Lri
If someone was looking for a way to remove all instances of repeated values, see "How can I efficiently extract repeated elements in a Ruby array?".
如果有人正在寻找删除所有重复值实例的方法，请参阅“如何有效地提取 Ruby 数组中的重复元素？”。
a = [1, 2, 2, 3]
counts = Hash.new(0)
a.each { |v| counts[v] += 1 }
p counts.select { |v, count| count == 1 }.keys # [1, 3]
回答by Finks
Just another alternative if anyone cares.
如果有人关心，这只是另一种选择。
You can also use the to_setmethod of an array which converts the Array into a Set and by definition, set elements are unique.
您还可以使用to_set数组的方法将数组转换为集合，并且根据定义，集合元素是唯一的。
[1,2,3,4,5,5,5,6].to_set => [1,2,3,4,5,6]
回答by the Tin Man
Just to provide some insight:
只是提供一些见解：
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
Which, when run, results in:
运行时，结果如下：
# >> 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}>)
Note: these returned bad results:
注意：这些返回了不好的结果：
lri(SHORT_ARRAY) # => [3]finks(SHORT_ARRAY) # => #<Set: {1, 2, 3}>santosh_mohanty(SHORT_ARRAY) # => [1, 2, 3, 1]
lri(SHORT_ARRAY) # => [3]finks(SHORT_ARRAY) # => #<Set: {1, 2, 3}>santosh_mohanty(SHORT_ARRAY) # => [1, 2, 3, 1]
回答by Santosh Mohanty
Try using the XOR operator, without using built-in functions:
尝试使用 XOR 运算符，而不使用内置函数：
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
With built-in functions:
具有内置功能：
a = [3,2,3,2,3,5,6,7]
a.uniq
