我有哈希数组:
[
{:name=>"King Summer", :number=>"0034242342"},
{:name=>"Max Snow", :number=>"899899080"},
{:name=>"Duck Doe", :number=>"90897688"},
{:name=>"Shark Jon", :number=>"0034242342"}
]
在上面,键:number
{:name=>"King Summer", :number=>"0034242342"}
的值与{:name=>"Shark Jon", :number=>"0034242342"}
重复
如何找到重复的:number 并在控制台中显示如下输出:
=== List with duplicate number ===
King Summer – 0034242342 : ok
Max Snow – 899899080 : ok
Duck Doe – 90897688 : ok
Shark Jon – 0034242342 : duplicate number
== List without duplicate number ===
1 - King Summer – 0034242342
2 - Max Snow – 899899080
3 - Duck Doe – 90897688
答案 0 :(得分:2)
您可以简单地使用inject
添加状态,其中inject的累加器将保存先前迭代中传递的数字,
arr.inject([]) { |m,a| m.include?(a[:number]) ? (a[:status] = 'duplicated number') : ((m << a[:number]) && (a[:status] = 'ok')); m}
arr
# [
# {:name=>"King Summer", :number=>"0034242342", :status=>"ok"},
# {:name=>"Max Snow", :number=>"899899080", :status=>"ok"},
# {:name=>"Duck Doe", :number=>"90897688", :status=>"ok"},
# {:name=>"Shark Jon", :number=>"0034242342", :status=>"duplicated number"}
# ]
稍后使用each
迭代此数据。
答案 1 :(得分:0)
您可以这样做。也许不是最干净的方法。
duplicated = []
non_duplicated = []
array = [
{:name=>"King Summer", :number=>"0034242342"},
{:name=>"Max Snow", :number=>"899899080"},
{:name=>"Duck Doe", :number=>"90897688"},
{:name=>"Shark Jon", :number=>"0034242342"}
]
array.each do |a|
is_duplicated = non_duplicated.find do |i|
i == a[:number]
end
if is_duplicated
duplicated.push(a[:number])
non_duplicated.delete(a[:number])
next
end
non_duplicated.push(a[:number])
end
替代
array.partition do |value|
number = value[:number]
array.map do |i|
i[:number]
end.count(number) > 1
end
这将返回一个包含2个部分的数组(重复的和其余的):
[
[
{:name=>"King Summer", :number=>"0034242342"},
{:name=>"Shark Jon", :number=>"0034242342"}
],
[
{:name=>"Max Snow", :number=>"899899080"},
{:name=>"Duck Doe", :number=>"90897688"}
]
]
答案 2 :(得分:0)
其他选择是建立一个散列,该散列将重复的(true)而不是(false)分组。
给出数组:
ary = [
{name: "King Summer", number: "0034242342"},
{name: "Max Snow", number: "899899080"},
{name: "Duck Doe", number: "90897688"},
{name: "Shark Jon", number: "0034242342"},
{name: "Jim Kirk", number: "90897688"},
{name: "Mr. Spock", number: "10897688"}
]
构建哈希:
res = ary.group_by { |h| h[:number] }.group_by { |_,v| v.size > 1 }.transform_values(&:to_h)
#=> {true=>{"0034242342"=>[{:name=>"King Summer", :number=>"0034242342"}, {:name=>"Shark Jon", :number=>"0034242342"}], "90897688"=>[{:name=>"Duck Doe", :number=>"90897688"}, {:name=>"Jim Kirk", :number=>"90897688"}]}, false=>{"899899080"=>[{:name=>"Max Snow", :number=>"899899080"}], "10897688"=>[{:name=>"Mr. Spock", :number=>"10897688"}]}}
然后列出重复项:
res[true].each do |number, people|
puts "- ph. number: #{number}"
people.each { |person| puts "\t"*2 + person[:name] }
end
# - ph. number: 0034242342
# King Summer
# Shark Jon
# - ph. number: 90897688
# Duck Doe
# Jim Kirk
res.each do |_, list|
puts "-"*10
list.each do |number, people|
puts "- ph. number: #{number}"
people.each { |person| puts "\t"*2 + person[:name] }
end
end
# ----------
# - ph. number: 0034242342
# King Summer
# Shark Jon
# - ph. number: 90897688
# Duck Doe
# Jim Kirk
# ----------
# - ph. number: 899899080
# Max Snow
# - ph. number: 10897688
# Mr. Spock