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I have a simple method that iterates through an array and returns a duplicate. (Or duplicates)
def find_dup(array)
duplicate = 0
array.each { |element| duplicate = element if array.count(element) > 1}
duplicate
end
It works, but I'd like to express this more elegantly.
The reason it is three lines is that the variable "duplicate", which the method must return, is not visible to the method if I introduce it inside the block, i.e,
def find_dup(array)
array.each { |element| duplicate = element if array.count(element) > 1}
duplicate
end
I've tried a few ways to define "duplicate" as the result of a block, but to no avail.
Any thoughts?
It's a little too much to do cleanly in a one-liner, but this is a more
efficient solution.
def find_dups(arr)
counts = Hash.new { |hash,key| hash[key] = 0 }
arr.each_with_object(counts) do |x, memo|
memo[x] += 1
end.select { |key,val| val > 1 }.keys
end
The Hash.new call instantiates a hash where the default value is 0.
each_with_object modifies this hash to track the count of each element in arr, then at the
end the filter is used to select only those having a count greater than one.
The benefit of this approach over a solution using Array#includes? or Array#count is that it only scans the array a single time. Thus it is a O(N) time instead of O(N^2).
Your method is only finding the last duplicate in the array. If you want all the duplicates, I would do something like this:
def find_dups(arr)
dups = Hash.new { |h, k| h[k] = 0 }
arr.each { |el| dups[el] += 1 }
dups.select { |k, v| v > 1 }.keys
end
If what you really want is a one-liner that isn't concerned with big-O complexity and only returns the last duplicate in the array, I would do this:
def find_last_dup(arr)
arr.reverse_each { |el| return el if arr.count(el) > 1 }
end
You can do this as one line and it flows a bit nicer. Though this would find the first instance of a duplicate whereas your code is returning the last instance of a duplicate, not sure if that's part of your requirement.
def find_dup(array)
array.group_by { |value| value }.find { |_, groups| groups.count > 1 }.first
end
Also, note that making things one line doesn't strictly mean is better. I'd find the code more readable split over more lines, but that's just my opinion.
def find_dup(array)
array.group_by { |value|
value
}.find { |_, groups|
groups.count > 1
}.first
end
Just want to add one more approach to the mix.
def find_last_dup(arr)
arr.reverse_each.detect { |x| arr.count(x) > 1 }
end
Alternatively, you can get linear time complexity in two lines.
def find_last_dup(arr)
freq = arr.each_with_object(Hash.new(0)) { |x, obj| obj[x] += 1 }
arr.reverse_each.detect { |x| freq[x] > 1 }
end
For the sake of argument, the latter approach can be reduced to one line as well, but this would be unidiomatic and confusing.
def find_last_dup(arr)
arr.each_with_object(Hash.new(0)) { |x, obj| obj[x] += 1 }
.tap do |freq| return arr.reverse_each.detect { |x| freq[x] > 1 } end
end
Given:
> a
=> [8, 5, 6, 6, 5, 8, 6, 1, 9, 7, 2, 10, 7, 7, 3, 4]
You can group the dups together:
> a.uniq.each_with_object(Hash.new(0)) {|e, h| c=a.count(e); h[e]=c if c>1}
=> {8=>2, 5=>2, 6=>3, 7=>3}
Or,
> a.group_by{ |e| e}.select{|k,v| v if v.length>1}
=> {8=>[8, 8], 5=>[5, 5], 6=>[6, 6, 6], 7=>[7, 7, 7]}
In each case, the order of the result is based on the order of the elements in a that have dups. If you just want the first:
> a.group_by{ |e| e}.select{|k,v| v if v.length>1}.first
=> [8, [8, 8]]
Or last:
> a.group_by{ |e| e}.select{|k,v| v if v.length>1}.to_a.last
=> [7, [7, 7, 7]]
If you want to 'fast forward' to the first value that has a dup, you can use drop_while:
> b=[1,2,3,4,5,4,5,6]
> b.drop_while {|e| b.count(e)==1 }[0]
=> 4
Or the last:
> b.reverse.drop_while {|e| b.count(e)==1 }[0]
=> 5
def find_duplicates(array)
array.dup.uniq.each { |element| array.delete_at(array.index(element)) }.uniq
end
The above method find_duplicates duplicated the input array and deletes the first occurrence of all the elements, leaving the array with only remaining occurrences of the duplicate elements.
Example:
array = [1, 2, 3, 4, 3, 4, 3]
=> [1, 2, 3, 4, 3, 4, 3]
find_duplicates(array)
=> [3, 4]
New to Ruby; When I am working with arrays, is there a difference between Enumerable#each_with_index and Array#index I have been working with a multidimensional array in which I am trying to find the location of a given value. My code is able to pass all the tests when I use either one. Coordinates is the array I am using to hold the location of the given value (in this case 1)
field=[[1,0],[0,0]]
coordinates=[]
field.each_with_index do |item|
if item.index(1)
coordinates.push(field.index(item)).push(item.index(1))
end
end
Thanks in advance for the help.
Let's take a closer look at your code:
field=[[1,0],[0,0]]
coordindates = []
field.each_with_index do |item|
if item.index(1)
coordinates.push(field.index(item)).push(item.index(1))
end
end
Let:
enum = field.each_with_index
#=> #<Enumerator: [[1, 0], [0, 0]]:each_with_index>
As you see this returns an enumerator.
Ruby sees your code like this:
enum.each do |item|
if item.index(1)
coordinates.push(field.index(item)).push(item.index(1))
end
end
The elements of the enumerator will be passed into the block by Enumerator#each, which will call Array#each since the receiver, field is an instance of the class Array.
We can see the elements of enum by converting it to an array:
enum.to_a
#=> [[[1, 0], 0], [[0, 0], 1]]
As you see, it has two elements, each being an array of two elements, the first being an array of two integers and the second being an integer.
We can simulate the operation of each by sending Enumerator#next to enum and assigning the block variables to the value returned by next. As there is but one block variable, item, we have:
item = enum.next
#=> [[1, 0], 0]
That is quite likely neither what you were expecting nor what you wanted.
Next, you invoke Array#index on item:
item.index(1)
#=> nil
index searches the array item for an element that equals 1. If it finds one it returns that element's index in the array. (For example, item.index(0) #=> 1). As neither [1,0] nor 0 equals 1, index returns nil.
Let's rewind (and recreate the enumerator). You need two block variables:
field.each_with_index do |item, index|...
which is the same as:
enum.each do |item, index|...
So now:
item, index = enum.next
#=> [[1, 0], 0]
item #=> [1, 0]
index #=> 0
and
item.index(1)
#=> 0
I will let you take it from here, but let me mention just one more thing. I'm not advocating it, but you could have written:
field.each_with_index do |(first, second), index|...
in which case:
(first, second), index = enum.next
#=> [[1, 0], 0]
first #=> 1
second #=> 0
index #=> 0
See Ruby doc: http://ruby-doc.org/core-2.2.0/Array.html#method-i-index and http://ruby-doc.org/core-2.2.1/Enumerable.html#method-i-each_with_index
index is a method of Array, it detects whether an item exists in a specific Array, return the index if the item exists, and nil if does not exist.
each_with_index is a method of Enumerable mixin, it usually takes 2 arguments, first one is item and the second one is index.
So your sample could be simplified as:
field = [[1, 0], [0, 0]]
coordinates = []
field.each_with_index do |item, index|
item_index = item.index(1)
coordinates << index << item_index if item_index
end
puts coordinates.inspect # => [0, 0]
Note your field.index(item) is just index.
Array#index and Enumerable#each_with_index are not related whatsoever (functionally speaking), one is used to get the index of an object within an Array and the other one is used to walk through a collection.
Actually each_with_index is a mutation of the each method that additionally yields the index of the actual object within the collection. This method is very useful if you need to keep track of the current position of the object you are in. It saves you the trouble of creating and incrementing an additional variable.
For example
['a', 'b', 'c', 'd'].each_with_index do|char, idx|
puts "#{idx}) #{char}"
end
output:
0) a
1) b
2) c
3) d
Without each_with_index
idx = 0
['a', 'b', 'c', 'd'].each do|char|
puts "#{idx}) #{char}"
idx += 1
end
output:
0) a
1) b
2) c
3) d
To find the index of an object within a multidimensional Array you will have to iterate at least through all the rows of the matrix, like this (using each_with_index and index)
def find_position(matrix, obj)
matrix.each_with_index do|row, i|
return [i, j] if j = row.index(obj)
end
return nil
end
for example:
find_position([[2,3],[4,5]], 5) # => [1, 1]
Given that:
fields = [[1, 0], [0, 0]]
Enumerable#each_with_index passes each index as well as each object to the block:
fields.each_with_index do |field, idx|
puts "#{field}, #{idx}"
end
#=> [1, 0], 0
#=> [0, 0], 1
Array#index returns the index of the matching array element:
puts fields.index([1, 0])
#=> 0
The object of my coding exercise is to get rid of duplicates in an array without using the uniq method. Here is my code:
numbers = [1, 4, 2, 4, 3, 1, 5]
def my_uniq(array)
sorted = array.sort
count = 1
while count <= sorted.length
while true
sorted.delete_if {|i| i = i + count}
count += 1
end
end
return sorted
end
When I run this, I get an infinite loop. What is wrong?
Can I use delete the way that I am doing with count?
How will it execute? Will count continue until the end of the array before the method iterates to the next index?
I did this with each or map, and got the same results. What is the best way to do this using each, delete_if, map, or a while loop (with a second loop that compares against the first one)?
Here is a clearly written example.
numbers = [1, 4, 2, 4, 3, 1, 5]
def remove_duplicates(array)
response = Array.new
array.each do |number|
response << number unless response.include?(number)
end
return response
end
remove_duplicates(numbers)
As others pointed out, your inner loop is infinite. Here's a concise solution with no loops:
numbers.group_by{|n| n}.keys
You can sort it if you want, but this solution doesn't require it.
the problem is that the inner loop is an infinite loop:
while true
sorted.delete_if {|i| i = i + count}
count += 1
end #while
you can probably do what you are doing but it's not eliminating duplicates.
one way to do this would be:
numbers = [1, 4, 2, 4, 3, 1, 5]
target = []
numbers.each {|x| target << x unless target.include?(x) }
puts target.inspect
to add it to the array class:
class ::Array
def my_uniq
target = []
self.each {|x| target << x unless target.include?(x) }
target
end
end
now you can do:
numbers = [1, 4, 2, 4, 3, 1, 5]
numbers.my_uniq
You count use Set that acts like an array with does not allow duplicates:
require 'set'
numbers = [1, 4, 2, 4, 3, 1, 5]
Set.new(numbers).to_a
#=> [1, 4, 2, 3, 5]
Try using Array#& passing the array itself as parameter:
x = [1,2,3,3,3]
x & x #=> [1,2,3]
This is one of the answer. However, I do not know how much of performance issue it takes to return unique
def my_uniq(ints)
i = 0
uniq = []
while i < ints.length
ints.each do |integers|
if integers == i
uniq.push(integers)
end
i += 1
end
end
return uniq
end
I build the array here:
def initialize
#names = []
end
#names << page.all('//*[#id="USERS_AVAIL"]/option').map {|result| result.text.split(", ")}
later on I'm trying to compile and visit url's by iterating through the names array like so:
#names.each do |name|
visit "https://example.com/k=#{name}&tpe=1"
end
Some puts statements show me that the each method is calling every element of the array all at once instead of iterating as intended. I.E.: "https://example.com/k=#{[[%22Adviento%22,%20%22Justin%22],%20[%22Asamoah%22,%20%22Nathan%22],%20[%22Baughman%22,%20%22Zachary%22],}&tpe=1". #names.length has a count of only 4 but a puts of the #names array shows the proper output? I'm not sure what could be wrong, thanks in advance for any assist.
Replace << with +=. The << is inserting the entire array as a single element of its own, whereas += will concatenate the array, which seems to be your intention.
For example:
a = [1,2,3]
# => [1, 2, 3]
a << [4,5,6]
# => [1, 2, 3, [4, 5, 6]] # WRONG
a = [1,2,3]
# => [1, 2, 3]
a += [4,5,6]
# => [1, 2, 3, 4, 5, 6] # CORRECT
Try:
#names += page.all('//*[#id="USERS_AVAIL"]/option')
.map { |r| r.text.split(',').map(&:strip) }.flatten
If the quotes are in the literal form %22 and you want to capture the strings in between them:
#names += page.all('//*[#id="USERS_AVAIL"]/option')
.map { |r| r.text.scan(/%22([^%]+)%22/) }.flatten
def peel array
output = []
while ! array.empty? do
output << array.shift
mutate! array
end
output.flatten
end
I have not included the mutate! method, because I am only interested in removing the output variable. The mutate! call is important because we cannot iterate over the array using each because array is changing.
EDIT: I am getting an array as output, which is what I want. The method works correctly, but I think there is a way to collect the array.shift values without using a temp variable.
EDIT #2: OK, here is the mutate! method and test case:
def mutate! array
array.reverse!
end
a = (1..5).to_a
peel( a ).should == [ 1, 5, 2, 4, 3 ]
It doesn't matter if peel modifies array. I guess it should be called peel!. Yes, mutate! must be called after each element is removed.
All this reversing makes me dizzy.
def peel(array)
indices = array.size.times.map do |i|
i = -i if i.odd?
i = i/2
end
array.values_at(*indices) # indices will be [0, -1, 1, -2, 2] in the example
end
a = (1..5).to_a
p peel(a) #=>[1, 5, 2, 4, 3]
Another approach:
def peel(array)
mid = array.size/2
array[0..mid]
.zip(array[mid..-1].reverse)
.flatten(1)
.take(array.size)
end
Usage:
peel [1,2,3,4,5,6]
#=> [1, 6, 2, 5, 3, 4]
peel [1,2,3,4,5]
#=> [1, 5, 2, 4, 3]
Here's a way using parallel assignment:
def peel array
n = array.size
n.times {|i| (n-2-2*i).times {|j| array[n-1-j], array[n-2-j] = array[n-2-j], array[n-1-j]}}
array
end
peel [1,2,3,4,5] # => [1,5,2,4,3]
peel [1,2,3,4,5,6] # => [1,6,2,5,3,4]
What I'm doing here is a series of pairwise exchanges. By way of example, for [1,2,3,4,5,6], the first 6-2=4 steps (6 being the size of the array) alter the array as follows:
[1,2,3,4,6,5]
[1,2,3,6,4,5]
[1,2,6,3,4,5]
[1,6,2,3,4,5]
The 1, 6 and the 2 are in now the right positions. We repeat these steps, but this time only 6-4=2 times, to move the 5 and 3 into the correct positions:
[1,6,2,3,5,4]
[1,6,2,5,3,4]
The 4 is pushed to the end, it's correct position, so we are finished.