# Isolating non-unique items in an array

I’m basically trying to the opposite of .uniq Let’s say I have an array:

[“a”, “a”, “a”, “b”, “c”, “d”, “d”]

I would like to boil this down by first tossing the values that only
appear once. Which leaves me with:

[“a”, “a”, “a”, “d”, “d”]

Then I need to somehow determine that “a” appears 3 times and “d”
appears 2 times. Any help would be great. Thanks.

Jason B. wrote:

I’m basically trying to the opposite of .uniq Let’s say I have an array:

[“a”, “a”, “a”, “b”, “c”, “d”, “d”]

I would like to boil this down by first tossing the values that only
appear once. Which leaves me with:

[“a”, “a”, “a”, “d”, “d”]

Then I need to somehow determine that “a” appears 3 times and “d”
appears 2 times. Any help would be great. Thanks

This should work:

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
a.uniq.map{|i| i if (a.map{|j| j if j==i}.length > 1)}

Drew O. wrote:

Jason B. wrote:

I’m basically trying to the opposite of .uniq Let’s say I have an array:

[“a”, “a”, “a”, “b”, “c”, “d”, “d”]

I would like to boil this down by first tossing the values that only
appear once. Which leaves me with:

[“a”, “a”, “a”, “d”, “d”]

Then I need to somehow determine that “a” appears 3 times and “d”
appears 2 times. Any help would be great. Thanks

This should work:

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
a.uniq.map{|i| i if (a.map{|j| j if j==i}.length > 1)}

Actually, so that you have them appear more than once you’d need to do:

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
a.map{|i| i if (a.map{|j| j if j==i}.length > 1)}

Tested, and it doesn’t work, so scratch that…

Jason B. wrote:

I’m basically trying to the opposite of .uniq Let’s say I have an array:

[“a”, “a”, “a”, “b”, “c”, “d”, “d”]

I would like to boil this down by first tossing the values that only
appear once. Which leaves me with:

[“a”, “a”, “a”, “d”, “d”]

Then I need to somehow determine that “a” appears 3 times and “d”
appears 2 times. Any help would be great. Thanks.

a = %w( a a a b c d d )
want_and_counts = a.inject( Hash.new(0) ) { |h,x| h[x]+=1; h }.reject {
|k,v| v == 1 }

On 10/12/06, Jason B. [email protected] wrote:

appears 2 times. Any help would be great. Thanks.

Posted via http://www.ruby-forum.com/.

class Array
def count( item )
return 0 unless self.include? item

``````c = 0
self.each {|i| c += 1 if i == item}
return c
``````

end
end

ary = %w(a a a b c d d)
new_ary = ary.select {|item| ary.count(item) > 1}

counts = Hash.new {|h,k| h[k] = new_ary.count(k)}

It’s not going to be fast, but it should work.

Blessings,
TwP

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
puts a.uniq

Try it, it works!

Jason B. wrote:

appears 2 times. Any help would be great. Thanks.

# {“a”=>3, “d”=>2}

def get_counts(keys)
counts = Hash.new(0)
keys.each {|k| counts[k] += 1 }
counts
end

def non_uniq(elements)
counts = get_counts(elements)
counts.delete_if {|k, v| v < 2 }
elements.select {|e| counts.key?(e) }
end

elements = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]

p non_uniq(elements)
p get_counts(elements).delete_if {|k, v| v < 2 }

Tim P. wrote:

Then I need to somehow determine that “a” appears 3 times and “d”

c = 0
self.each {|i| c += 1 if i == item}
return c
end
end

ary = %w(a a a b c d d)
new_ary = ary.select {|item| ary.count(item) > 1}

Better:

class Array
def counts # could be golfed, but clarity is king
result = Hash.new{0}
self.each {|a| result[a] += 1}
result
end
end

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
c = a.counts
a.reject{|b| c[b] == 1} # => [“a”, “a”, “a”, “d”, “d”]

Jeffrey S. wrote:

Then I need to somehow determine that “a” appears 3 times and “d”
end

def non_uniq(elements)
counts = get_counts(elements)
counts.delete_if {|k, v| v < 2 }
elements.select {|e| counts.key?(e) }
end

# Better:

def non_uniq(elements)
counts = get_counts(elements)
elements.select {|e| counts[e] > 1 }
end

Jason B. wrote:

appears 2 times. Any help would be great. Thanks.

Here’s what I whipped up:

letters = [‘a’,‘a’,‘a’,‘b’,‘c’,‘d’,‘d’]
counts = {}

letters.each do |letter|
count = letters.find_all { |x| x == letter }.size
counts[letter] = count if count > 1
end

non_unique_letters = letters.find_all { |x| counts.has_key?(x) }

p counts
p non_unique_letters

Jamey

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Hi,

Jason B. [email protected] writes:

appears 2 times. Any help would be great. Thanks.
% irb

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
=> [“a”, “a”, “a”, “b”, “c”, “d”, “d”]

a.select{|i| a.grep(i).size > 1}
=> [“a”, “a”, “a”, “d”, “d”]

WATANABE Hirofumi wrote:

=> [“a”, “a”, “a”, “d”, “d”]
Taking that one step further to meet his second need:

class Array
def duplicates_count
uniq.map{ |e|
if ( count = grep(e).size ) > 1
{ e => count }
end
}.compact
end
end

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
p a.duplicates_count
#=> [{“a”=>3}, {“d”=>2}]

On Fri, 2006-10-13 at 00:24 +0900, Jason B. wrote:

appears 2 times. Any help would be great. Thanks.

There are bound to be better ways, but you could try something like:

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]

# => [“a”, “a”, “a”, “b”, “c”, “d”, “d”]

hsh = a.inject(Hash.new{|h,k| h[k] = 0}) {|h,v| h[v] += 1 ; h}.reject!
{|k,v| v == 1}

hsh[‘a’]

hsh[‘d’]

# => 2

Depending on how you’re using the results, You could also run a second
inject to reverse the lookup:

rhsh = hsh.inject(Hash.new { |h,k| h[k] = [] }) { |h,(k,v)| h[v] << k ;
h }

# => {2=>[“d”], 3=>[“a”]}

(2…rhsh.keys.max).each { |i| puts “count #{i}: #{rhsh[i]}” }

# count 3: a

If you went with this way, I guess you could forget about getting rid of
the 1s with reject!, since you could just ignore the 1 key in the
rhsh…

On Fri, 13 Oct 2006, Phrogz wrote:

a.select{|i| a.grep(i).size > 1}
}.compact
end
end

a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
p a.duplicates_count
#=> [{“a”=>3}, {“d”=>2}]

that’s O(n^2) though… the others are O(n)

-a

On 12.10.2006 17:24, Jason B. wrote:

appears 2 times. Any help would be great. Thanks.
Did anyone suggest inject + partition yet? If not:

irb(main):001:0> a = [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
=> [“a”, “a”, “a”, “b”, “c”, “d”, “d”]
irb(main):002:0> stat = a.inject(Hash.new(0)) do |h, e|
irb(main):003:1* h[e] += 1
irb(main):004:1> h
irb(main):005:1> end
=> {“a”=>3, “b”=>1, “c”=>1, “d”=>2}
irb(main):006:0> one, multi = stat.partition {|e,c| c == 1}
=> [[[“b”, 1], [“c”, 1]], [[“a”, 3], [“d”, 2]]]
irb(main):007:0> one
=> [[“b”, 1], [“c”, 1]]
irb(main):008:0> multi
=> [[“a”, 3], [“d”, 2]]
irb(main):009:0> one.map! {|e,c| e}
=> [“b”, “c”]

Kind regards

``robert``