So I want to merge overlapping ranges and it should like the following:
Input: ranges = [(1..2), (3..6), (5..8)]
Output: expected = [(1..2), (3..8)]
but when the code iterate over the intervals and goes to the else statement I just get a message "function_merge.rb:9:in block in merge': undefined methodend=' for 2..19:Range (NoMethodError)"
I tried to save merged.last.end and interval.end to variables, rewrote the if statement over couple of lines (if interval.end > merged.last.end merged.last.end = interval.end end) but all that didn't work :-(
def merge(intervals)
merged = []
intervals.sort_by! { |interval| interval.begin }
intervals.each do |interval|
if merged.empty? || merged.last.end < interval.begin
merged << interval
else
merged.last.end = interval.end if interval.end > merged.last.end
end
end
return merged
end
I don't understand why I get this error message since "end" is a range method? I just want to "update" the merged.last.end with the interval.end number.
If you have any tips how to solve it, would be very nice :-)
As Sebastian points out, Ranges are immutable. Instead of trying to change the Range you'll have to make a new one.
def merge(intervals)
merged = []
intervals.sort_by! { |interval| interval.begin }
intervals.each do |interval|
if merged.empty? || merged.last.end < interval.begin
merged << interval
else
merged[-1] = Range.new(merged.last.begin, interval.end, interval.exclude_end?)
end
end
return merged
end
It has been explained that ranges are immutable. The question implies the elements covered by the ranges are all comparable (e.g, not ['a'..'z', 1..10]). I assume that the array of ranges does not contain a mix of finite and infinite ranges.
Solution
Code
def distill(arr)
a = arr.reject { |r| r.exclude_end? ? (r.end <= r.begin) : r.end < r.begin }.
sort_by(&:begin)
return [] if a.empty?
combined = []
curr = a.shift
loop do
break (combined << curr) if a.empty?
nxt = a.shift
if nxt.begin > curr.end
combined << curr
curr = nxt
else
last = [curr, nxt].max_by { |r| [r.end, r.exclude_end? ? 0 : 1] }
curr = last.exclude_end? ? (curr.begin...last.end) :
curr.begin..last.end
end
end
end
Examples
distill [5..8, 7...9, 9..11, 1...4, 38..37]
#=> [1...4, 5..11]
distill [1.5...2.2, 2.2..3.0, 3.0...4.5, 4.7..5.3, 5.2..4.6]
#=> [1.5...4.5, 4.7..5.3]
distill ['a'..'d', 'c'..'f', 'b'..'g']
# 'a'..'g'
Explanation
See Range#exclude_end?.
The steps for the first example are as follows.
arr = [5..8, 7...9, 9..11, 1...4, 38..37]
a = arr.reject { |r| r.exclude_end? ? (r.end <= r.begin) : r.end < r.begin }.
sort_by(&:begin)
#=> [1...4, 5..8, 7...9, 9..11]
a.empty?
#=> false, so do not return
combined = []
curr = a.shift
#=> 1...4
a #=> [5..8, 7...9, 9..11]
The calculations within the loop can be best explained by salting the code with puts statements and displaying the results.
loop do
puts "a.empty? #=> true, so break #{combined + [curr]}" if a.empty?
break (combined << curr) if a.empty?
puts "a.empty? #=> false"
nxt = a.shift
puts "nxt=#{nxt}, a=#{a}"
puts "nxt.begin=#{nxt.begin} > #{curr.end} = curr.end = #{nxt.begin > curr.end}"
if nxt.begin > curr.end
combined << curr
puts "combined << #{curr} = #{combined}"
curr = nxt
puts "curr = nxt = #{curr}"
else
last = [curr, nxt].max_by { |r| [r.end, r.exclude_end? ? 0 : 1] }
puts "last=#{last}, last.exclude_end?=#{last.exclude_end?}"
curr = last.exclude_end? ? (curr.begin...last.end) :
curr.begin..last.end
puts "new value of curr=#{curr}"
end
puts
end
a.empty? #=> false
nxt=5..8, a=[7...9, 9..11]
nxt.begin=5 > 4 = curr.end = true
combined << 1...4 = [1...4]
curr = nxt = 5..8
a.empty? #=> false
nxt=7...9, a=[9..11]
nxt.begin=7 > 8 = curr.end = false
last=7...9, last.exclude_end?=true
new value of curr=5...9
a.empty? #=> false
nxt=9..11, a=[]
nxt.begin=9 > 9 = curr.end = false
last=9..11, last.exclude_end?=false
new value of curr=5..11
a.empty? #=> true, so break [1...4, 5..11]
It is sometimes convenient to be able to return an empty (but valid) range such as 38..37; one should not think of empty ranges as necessarily being an indication that something is amiss.
Alternative solution
If the ranges are all finite, as in the example, and the combined sizes of the ranges is not excessive, one could write the following.
Code
def distill(arr)
arr.flat_map(&:to_a).
uniq.
sort.
chunk_while { |x,y| y == x.next }.
map { |a| a.first..a.last }
end
Examples
distill [5..8, 7...9, 9..11, 1...4, 38..37]
#=> [1..3, 5..11]
distill ['a'..'d', 'c'..'f', 'b'..'g']
# 'a'..'g'
Explanation
The steps for the first example are as follows.
arr = [5..8, 7...9, 9..11, 1...4, 38..37]
a = arr.flat_map(&:to_a)
#=> => [5, 6, 7, 8, 7, 8, 9, 10, 11, 1, 2, 3]
b = a.uniq
#=> [5, 6, 7, 8, 9, 10, 11, 1, 2, 3]
c = b.sort
#=> [1, 2, 3, 5, 6, 7, 8, 9, 10, 11]
d = c.chunk_while { |x,y| y == x.next }
#=> #<Enumerator: #<Enumerator::Generator:0x00005c2683af8dd0>:each>
e = d.map { |a| a.first..a.last }
#=> [1..3, 5..11]
One can convert the enumerator d to an array to see the elements it will generate and pass to chunk_while's block:
d.to_a
#=> [[1, 2, 3], [5, 6, 7, 8, 9, 10, 11]]
See Enumerable#chunk_while. One could alternatively use Enumerable#slice_when.
Related
I am trying to iterate over an array and count the number of positive, negative and zeros in an array. Right now I am doing it like this
arr = [1, -1, 0, 2, 3, -2, -5]
pos = arr.select { |i| i > 0 }.count
neg = arr.select { |i| i < 0 }.count
zero = arr.select { |i| i == 0 }.count
puts pos
puts neg
puts zero
But is there any way where I can do this in one line? Something like this?
pos, neg, zero = arr.select { |i| i > 0; i < 0; i == 0; }.count
Use inject and the <=> operator:
neg, zero, pos = arr.inject([0,0,0]) { |a,b| a[(b<=>0)+1] += 1; a }
Alternatively, as #HolgerJust mentioned:
neg, zero, pos = arr.each_with_object([0,0,0]) { |a,b| b[(a<=>0)+1] += 1 }
is slightly longer but doesn't have the extra ; a in the block.
Inspired by #steenslag's use of tally:
neg, zero, pos = arr.map { |x| x<=>0 }.tally.values_at(-1,0,1)
If you use a counting hash the code is short and the results are returned in a hash, which may be convenient.
arr = [1, -1, 0, 2, 3, -2, -5, 4]
You could write
arr.each_with_object(Hash.new(0)) { |n,h| h[n<=>0] += 1 }
#=> {1=>4, -1=>3, 0=>1}
or perhaps you would prefer
labels = { -1=>:neg, 0=>:zero, 1=>:pos }
arr.each_with_object(Hash.new(0)) { |n,h| h[labels[n<=>0]] += 1 }
#=> {:pos=>4, :neg=>3, :zero=>1}
the last line of which could alternatively be written
arr.each_with_object({}) { |n,h| h[labels[n<=>0]] = (h[labels[n<=>0]] ||= 0) + 1 }
See Hash::new, specifically the (second) form that takes an argument called the default value (here zero), and no block. If a hash is defined h = Hash.new(0), then if h has no key k, h[k] returns 0 (and h is not changed).
arr = [1, -1, 0, 2, 3, -2, -5]
neg, zero, pos = arr.map{|n| n <=> 0}.tally.values_at(-1, 0, 1)
Using the new tally method.
As others have already said, you should just use inject and count using the <=> operator. If you plan to use similar logic frequently, you could monkey patch a #tally_by method into Enumerable like so:
class Enumerable
def my_tally(*keys, &proc)
proc ||= -> e {e} # Default identity proc
h = keys.empty? ? Hash.new(0) : Hash[keys.map{|k|[k, 0]}]
inject(h){|a, e| a[proc.call(e)] += 1; a}
end
end
This allows you to write:
neg, zero, pos = arr.my_tally(-1, 0, 1){|e| e <=> 0}
While this is certainly more upfront code than the others, it may be nice to have if you find yourself using similar logic frequently. You could also just make this a regular method somewhere if you don't like monkey-patching.
I have a hash:
a = {"Q1"=>1, "Q2"=>2, "Q5"=>3, "Q8"=>3}
I want to retrieve a set of keys from it such that the sum of their values equals a certain number, for example 5. In such case, the output should be:
Q2 Q5
Please help me on how to get this.
def find_combo(h, tot)
arr = h.to_a
(1..arr.size).find do |n|
enum = arr.combination(n).find { |e| e.map(&:last).sum == tot }
return enum.map(&:first) unless enum.nil?
end
end
h = {"Q1"=>1, "Q2"=>2, "Q5"=>3, "Q8"=>3}
find_combo(h, 5) #=> ["Q2", "Q5"]
find_combo(h, 2) #=> ["Q2"]
find_combo(h, 6) #=> ["Q5", "Q8"]
find_combo(h, 4) #=> ["Q1", "Q5"]
find_combo(h, 8) #=> ["Q2", "Q5", "Q8"]
find_combo(h, 9) #=> ["Q1", "Q2", "Q5", "Q8"]
find_combo(h, 10) #=> nil
Just out of curiosity:
hash = {"Q1"=>1, "Q2"=>2, "Q5"=>3, "Q8"=>3}
arr = hash.to_a
1.upto(hash.size).
lazy.
find do |i|
res = arr.combination(i).find do |h|
h.map(&:last).sum == 5
end
break res if res
end.tap { |result| break result.to_h if result }
#⇒ {"Q2" => 2, "Q5" => 3}
I have a unique sorted array: [2,4,6,8,10].
I have a variable called i. If i is 5, I want to return the elements in the array that 5 falls between. In this case [4,6]. If i is 8, then [8,10].
How should I go about this?
I've tried with partition, to some extent. If i happens to be a number directly equal to one of the values in the array. This seems to work:
a=[2,4,6,8,10]
i = 6
a.partition { |v| v < i }.max[0..1] # returns [6,8]
However, if i is a number not directly equal to any of the values in the array. For example 5, it gets a little trickier.
I got it working for the last case:
a=[2,4,6,8,10]
i = 5
partition = a.partition { |v| v < i }
[].tap { |a| a << partition[0].max; a << partition[1].min } # returns [6,8]
While this works, I am looking to see if there is a better way to write this logic.
You could use Enumerable#each_cons.
def mind_the_gap(arr, n)
arr.each_cons(2).find { |l,u| l <= n && n < u }
end
arr = [2,4,6,8,10]
mind_the_gap(arr, 5) #=> [4,6]
mind_the_gap(arr, 8) #=> [8,10]
mind_the_gap(arr, 1) #=> nil
mind_the_gap(arr, 10) #=> nil
If you don't want the last two examples to return nil, you could change the method as follows.
def mind_the_gap(arr, n)
rv = arr.each_cons(2).find { |l,u| l <= n && n < u }
return rv unless rv.nil?
n < arr.first ? :low : :high
end
mind_the_gap(arr, 5) #=> [4,6]
mind_the_gap(arr, 8) #=> [8,10]
mind_the_gap(arr, 1) #=> :low
mind_the_gap(arr, 10) #=> :high
Another way is to use Enumerable#slice_when.
def mind_the_gap(arr, n)
a = arr.slice_when { |l,u| l <= n && n < u }.to_a
return [a.first.last, a.last.first] unless a.size == 1
n < arr.first ? :low : :high
end
mind_the_gap(arr, 5) #=> [4,6]
mind_the_gap(arr, 8) #=> [8,10]
mind_the_gap(arr, 1) #=> :low
mind_the_gap(arr, 10) #=> :high
If you're looking for elements inside a sorted array, the "better way" probably involves bsearch or bsearch_index.
The second element in the pair is the first element in the array that is greater than your variable, so bsearch_index can return it directly. You need to check it isn't nil or 0 before returning the found element and the previous one :
a = [2, 4, 6, 8, 10]
def find_surrounding_pair(array, element)
second_index = array.bsearch_index { |x| x > element }
array[second_index - 1, 2] if second_index && second_index > 0
end
puts find_surrounding_pair(a, 1).nil?
puts find_surrounding_pair(a, 2) == [2, 4]
puts find_surrounding_pair(a, 7) == [6, 8]
puts find_surrounding_pair(a, 8) == [8, 10]
puts find_surrounding_pair(a, 12).nil?
#=> true * 5
The complexity of this method should be O(log n).
what about this
val = 5
a = [2,4,6,8,10] # assuming it's sorted
a.slice(a.rindex {|e| e <= val}, 2)
It doesn't account for the case when the lookup value is equal or bigger the last element of the array. I'd probably append a nil element for this, if that would be appropriate for the problem.
This looks like a good use to check for the inclusion in a range:
a = [2,4,6,8,10]
b = 5
a.each_cons(2).select { |i, j| (i .. j) === b }
# => [[4, 6]]
It's not clear exactly what you mean by "falls between". In the code above 8 would fall between two sets of numbers:
b = 8
a.each_cons(2).select { |i, j| (i .. j) === b }
# => [[6, 8], [8, 10]]
if the test is i <= b <= j. If it's i <= b < j then use ... instead of ..:
a.each_cons(2).select { |i, j| (i ... j) === b }
# => [[8, 10]]
I'm not a big fan of using ... but it simplifies the code.
From the Range documentation:
Ranges constructed using .. run from the beginning to the end inclusively. Those created using ... exclude the end value.
You could change that to:
a.each_cons(2).select { |i, j| i <= b && b <= j }
or:
a.each_cons(2).select { |i, j| i <= b && b < j }
if those work better for your mind. Using a Range is a little slower, but not radically so.
The below code is my newbie take on a bubble sort method.
#For each element in the list, look at that element and the element
#directly to it's right. Swap these two elements so they are in
#ascending order.
def bubble_sort (array)
a = 0
b = 1
until (array.each_cons(2).all? { |a, b| (a <=> b) <= 0}) == true do
sort = lambda {array[a] <=> array[b]}
sort_call = sort.call
loop do
case sort_call
when -1 #don't swap
a += 1
b += 1
break
when 0 #don't swap
a += 1
b += 1
break
when 1 #swap
array.insert(a,array.delete_at(b))
a += 1
b += 1
break
else #end of array, return to start
a = 0
b = 1
break
end
end
end
puts array.inspect
end
array = [4, 2, 5, 6, 3, 23, 5546, 234, 234, 6]
bubble_sort(array)
I want to be able to alter this method so that it takes a block of code as an argument and uses this to determine how it sorts.
For example:
array = ["hello", "my", "name", "is", "daniel"]
bubble_sort(array) {array[#a].length <=> array[#b].length}
(When I've tried this I've turned a and b into instance variables throughout the code.)
I have tried using yield but I get undefined method 'length' for nil:NilClass once the end of the array is reached. I've tried adding in things such as
if array[#b+1] == nil
#a = 0
#b = 1
end
This helps but I still end up with weird problems like infinite loops or not being able to sort more than certain amount of elements.
Long story short, I have been at this for hours. Is there a simple way to do what I want to do? Thanks.
The way you're calling your lambda is a bit odd. It's actually completely unnecessary. I refactored your code and cleaned up a bit of the redundancy. The following works for me:
def sorted?(arr)
arr.each_cons(2).all? { |a, b| (a <=> b) <= 0 }
end
def bubble_sort (arr)
a = 0
b = 1
until sorted?(arr) do
# The yield call here passes `arr[a]` and `arr[b]` to the block.
comparison = if block_given?
yield(arr[a], arr[b])
else
arr[a] <=> arr[b]
end
if [-1, 0, 1].include? comparison
arr.insert(a, arr.delete_at(b)) if comparison == 1
a += 1
b += 1
else
a = 0
b = 1
end
end
arr
end
sample_array = [4, 2, 5, 6, 3, 23, 5546, 234, 234, 6]
# Sanity check:
100.times do
# `a` is the value of `arr[a]` in our function above. Likewise for `b` and `arr[b]`.
print bubble_sort(sample_array.shuffle) { |a, b| a <=> b }, "\n"
end
EDIT
A cleaner version:
# In place swap will be more efficient as it doesn't need to modify the size of the arra
def swap(arr, idx)
raise IndexError.new("Index #{idx} is out of bounds") if idx >= arr.length || idx < 0
temp = arr[idx]
arr[idx] = arr[idx + 1]
arr[idx + 1] = temp
end
def bubble_sort(arr)
loop do
sorted_elements = 0
arr.each_cons(2).each_with_index do |pair, idx|
comparison = if block_given?
yield pair.first, pair.last
else
pair.first <=> pair.last
end
if comparison > 0
swap(arr, idx)
else
sorted_elements += 1
end
end
return arr if sorted_elements >= arr.length - 1
end
end
# A simple test
sample_array = [4, 2, 2, 2, 2, 2, 5, 5, 6, 3, 23, 5546, 234, 234, 6]
sample_str_array = ["a", "ccc", "ccccc"]
100.times do
print bubble_sort(sample_array.shuffle) { |a, b| a <=> b }, "\n"
print bubble_sort(sample_str_array.shuffle) { |a, b| a.length <=> b.length }, "\n"
end
You're not too far off. Just a few things:
Make your function take a block argument
def bubble_sort (array, &block)
Check to see if the user has provided a block
if block_given?
# Call user's comparator block
else
# Use the default behavior
end
Call the user's comparator block
block.call(a, b)
In the user-provided block, accept block params for the elements to compare
bubble_sort(array) {|a,b| a.length <=> b.length}
That should put you in the right ballpark.
I would like to write a program in ruby 1.9.3 ver. which collects unique value ranges and then calculates amount of numbers in these ranges.
For example lets use 3 ranges (1..3), (6..8) and (2..4). It will return array with two ranges (1..4), (6..8) and amount of numbers - 7.
I wrote the following code:
z= []
def value_ranges(start, finish, z)
range = (start..finish)
arr = z
point = nil
if arr.empty?
point = nil
else
arr.each { |uniq|
if overlap?(uniq,range) == true
point = arr.index(uniq)
break
else
point = nil
end
}
end
if point != nil
if arr[point].first >= start && arr[point].end <= finish
range = (start..finish)
elsif arr[point].first >= start
range = (start..arr[point].end)
elsif arr[point].end <= finish
range = (arr[point].first..finish)
else
range = (arr[point].first..arr[point].end)
end
arr[point] = range
else
arr << range
end
print arr
end
def overlap?(x,y)
(x.first - y.end) * (y.first - x.end) >= 0
end
Problem comes when program meets a range which overlaps two already collected ranges.
For example (1..5) (7..9) (11..19) and the next given range is (8..11).
It should link both ranges and return the following result - (1..5),(7..19).
I don't have an idea how to recheck whole array without creating infinite loop. Also what is the best way to calculate amount of numbers in ranges?
Here are two Ruby-like ways of doing it.
arr = [1..3, 6..8, 2..4]
#1 Efficient approach
First calculate the amalgamated ranges:
a = arr[1..-1].sort_by(&:first).each_with_object([arr.first]) do |r,ar|
if r.first <= ar.last.last
ar[-1] = ar.last.first..[ar.last.last,r.last].max
else
ar << r
end
end
#=> [1..4, 6..8]
Then compute the total number of elements in those ranges:
a.reduce(0) { |tot,r| tot + r.size }
#=> [1..4, 6..8].reduce(0) { |tot,r| tot + r.size }
#=> 7
Explanation
b = arr[1..-1]
#=> [6..8, 2..4]
c = b.sort_by(&:first)
#=> [2..4, 6..8]
enum = c.each_with_object([1..3])
#=> #<Enumerator: [2..4, 6..8]:each_with_object([1..3])>
The contents of the enumerator enum will be passed into the block and assigned to the block variables by Enumerator#each, which will call Array#each. We can see the contents of the enumerator by converting it to an array:
enum.to_a
#=> [[2..4, [1..3]], [6..8, [1..3]]]
and we can use Enumerator#next to step through the enumerator. The first element of the enumerator passed to the block by each is [2..4, [1..3]]. This is assigned to the block variables as follows:
r, ar = enum.next
#=> [2..4, [1..3]]
r #=> 2..4
ar #=> [1..3]
We now perform the block calculation
if r.first <= ar.last.last
#=> 2 <= (1..3).last
#=> 2 <= 3
#=> true
ar[-1] = ar.last.first..[ar.last.last,r.last].max
#=> ar[-1] = 1..[3,4].max
#=> ar[-1] = 1..4
#=> 1..4
else # not executed this time
ar << r
end
This is not so mysterious. So I don't have to keep saying "the last range of ar", let me define:
ar_last = ar.last
#=> 1..3
First of all, because we began by sorting the ranges by the beginning of each range, we know that when each element of enum is passed into the block:
ar_last.first <= r.first
For each element of enum passed into the block for which:
r.first <= ar_last.last
we compare r.last with ar_last.last. There are two possibilities to consider:
r.last <= ar_last.last, in which case the two ranges overlap and therefore ar_last would not change; and
r.last > ar_last.last, in which case the upper end of ar_last must be increased to r.last.
Here,
2 = r.first <= ar_last.last = 3
4 = r.last > ar_last.last = 3
so ar_last is changed from 1..3 to 1..4.
each now passes the last element of enum into the block:
r, ar = enum.next
#=> [6..8, [1..4]]
r #=> 6..8
ar #=> [1..4]
if r.first <= ar.last.last
#=> (6 <= 4) => false this time
...
else # executed this time
ar << r
#=> ar << (6..8)
#=> [1..4, 6..8]
end
and
a = ar #=> [1..4, 6..8]
This time, r.first > ar_last.last, meaning the range r does not overlap ar_last, so we append r to ar, and ar_last now equals r.
Lastly:
a.reduce(0) { |tot,r| tot + r.size }
#=> [1..4, 6..8].reduce(0) { |tot,r| tot + r.size }
#=> 7
which we could alternatively write:
a.map(&:size).reduce(:+)
#2 Easy but inefficient
Here is an easy, but not especially efficient, method that uses Enumerable#slice_when, newly-minted in v2.2.
arr = [(1..3), (6..8), (2..4)]
To calculate the amagamated ranges:
a = arr.flat_map(&:to_a)
.uniq
.sort
.slice_when { |i,j| i+1 != j }
.map { |ar| (ar.first..ar.last) }
#=> [1..4, 6..8]
The total number of elements in those ranges is calculated as in #1
Explanation
Here are the steps:
b = arr.flat_map(&:to_a)
#=> [1, 2, 3, 6, 7, 8, 2, 3, 4]
c = b.uniq
#=> [1, 2, 3, 6, 7, 8, 4]
d = c.sort
#=> [1, 2, 3, 4, 6, 7, 8]
e = d.slice_when { |i,j| i+1 != j }
#=> #<Enumerator: #<Enumerator::Generator:0x007f81629584f0>:each>
a = e.map { |ar| (ar.first..ar.last) }
#=> [1..4, 6..8]
We can see the contents of the enumerator e by converting it to an array:
e.to_a
#=> [[1, 2, 3, 4], [6, 7, 8]]