In ruby, given two date ranges, I want the range that represents the intersection of the two date ranges, or nil if no intersection. For example:
(Date.new(2011,1,1)..Date.new(2011,1,15)) & (Date.new(2011,1,10)..Date.new(2011,2,15))
=> Mon, 10 Jan 2011..Sat, 15 Jan 2011
Edit: Should have said that I want it to work for DateTime as well, so interval can be down to mins and secs:
(DateTime.new(2011,1,1,22,45)..Date.new(2011,2,15)) & (Date.new(2011,1,1)..Date.new(2011,2,15))
=> Sat, 01 Jan 2011 22:45:00 +0000..Tue, 15 Feb 2011
require 'date'
class Range
def intersection(other)
return nil if (self.max < other.begin or other.max < self.begin)
[self.begin, other.begin].max..[self.max, other.max].min
end
alias_method :&, :intersection
end
p (Date.new(2011,1,1)..Date.new(2011,1,15)) & (Date.new(2011,1,10)..Date.new(2011,2,15))
#<Date: 2011-01-10 ((2455572j,0s,0n),+0s,2299161j)>..#<Date: 2011-01-15 ((2455577j,0s,0n),+0s,2299161j)>
You can try this to get a range representing intersection
range1 = Date.new(2011,12,1)..Date.new(2011,12,10)
range2 = Date.new(2011,12,4)..Date.new(2011,12,12)
inters = range1.to_a & range2.to_a
intersected_range = inters.min..inters.max
Converting your example:
class Range
def intersection(other)
raise ArgumentError, 'value must be a Range' unless other.kind_of?(Range)
inters = self.to_a & other.to_a
inters.empty? ? nil : inters.min..inters.max
end
alias_method :&, :intersection
end
I found this: http://www.postal-code.com/binarycode/2009/06/06/better-range-intersection-in-ruby/ which is a pretty good start, but does not work for dates. I've tweaked a bit into this:
class Range
def intersection(other)
raise ArgumentError, 'value must be a Range' unless other.kind_of?(Range)
new_min = self.cover?(other.min) ? other.min : other.cover?(min) ? min : nil
new_max = self.cover?(other.max) ? other.max : other.cover?(max) ? max : nil
new_min && new_max ? new_min..new_max : nil
end
alias_method :&, :intersection
end
I've omitted the tests, but they are basically the tests from the post above changed for dates. This works for ruby 1.9.2.
Anyone got a better solution?
I baked this solution for ascending ranges, also taking care of the exclude end situations:
intersect_ranges = ->(r1, r2) do
new_end = [r1.end, r2.end].min
new_begin = [r1.begin, r2.begin].max
exclude_end = (r2.exclude_end? && new_end == r2.end) || (r1.exclude_end? && new_end == r1.end)
valid = (new_begin <= new_end && !exclude_end)
valid ||= (new_begin < new_end && exclude_end))
valid ? Range.new(new_begin, new_end, exclude_end) : nil
end
I'm also a bit worried by you guys adding it to the Range class itself, since the behavior of intersecting ranges is not uniformly defined.
(How about intersecting 1...4 and 4...1? Why nil when there is no intersection; we could also say this is an empty range: 1...1 )
You can use overlaps? with Range starting with Rails v3
# For dates, make sure you have the correct format
first_range = first_start.to_date..first_end.to_date
second_range = second_start.to_date..second_end.to_date
intersection = first_range.overlaps?(second_range) # => Boolean
# Example with numbers
(1..7).overlaps?(3..5) # => true
More details in the docs
Try something like this
require 'date'
sample = Date.parse('2011-01-01')
sample1 = Date.parse('2011-01-15')
sample2 = Date.parse('2010-12-19')
sample3 = Date.parse('2011-01-11')
puts (sample..sample1).to_a & (sample2..sample3).to_a
What this will give you is a array of intersection dates!!
I have times as [[start, end], ...] and I want to remove the some time ranges from a each initial time range, here is what I did:
def exclude_intersecting_time_ranges(initial_times, other_times)
initial_times.map { |initial_time|
other_times.each do |other_time|
next unless initial_time
# Other started after initial ended
next if other_time.first >= initial_time.last
# Other ended before initial started
next if other_time.last <= initial_time.first
# if other time started before and ended after after, no hour is counted
if other_time.first <= initial_time.first && other_time.last >= initial_time.last
initial_time = nil
# if other time range is inside initial time range, split in two time ranges
elsif initial_time.first < other_time.first && initial_time.last > other_time.last
initial_times.push([other_time.last, initial_time.last])
initial_time = [initial_time.first, other_time.first]
# if start time of other time range is before initial time range
elsif other_time.first <= initial_time.first
initial_time = [other_time.last, initial_time.last]
# if end time of other time range if after initial time range
elsif other_time.last >= initial_time.last
initial_time = [initial_time.first, other_time.first]
end
end
initial_time
}.compact
end
Since this question is related to How to combine overlapping time ranges (time ranges union), I also wanted to post my finding of the gem range_operators here, because if helped me in the same situation.
I'd transfer them into an array, since arrays know the intersection-operation:
(Date.new(2011,1,1)..Date.new(2011,1,15)).to_a & (Date.new(2011,1,10)..Date.new(2011,2,15)).to_a
Of course this returns an Array. So if you want an Enumerator (Range doesn't seem to be possible since these are not consecutive values anymore) just throw to_enum at the end.
Related
I’m having trouble using Ruby to pass some tests that make the array too big and return an error.
Solution.rb: failed to allocate memory (NoMemoryError)
I have failed to pass it twice.
The problem is about scheduling meetings. The method receives two parameters in order: a matrix with all the first days that investors can meet in the company, and a matrix with all the last days.
For example:
firstDay = [1,5,10]
lastDay = [4,10,10]
This shows that the first investor will be able to find himself between the days 1..4, the second between the days 5..10 and the last one in 10..10.
I need to return the largest number of investors that the company will serve. In this case, all of them can be attended to, the first one on day 1, the second one on day 5, and the last one on day 10.
So far, the code works normally, but with some hidden tests with at least 1000 investors, the error I mentioned earlier appears.
Is there a best practice in Ruby to handle this?
My current code is:
def countMeetings(firstDay, lastDay)
GC::Profiler.enable
GC::Profiler.clear
first = firstDay.sort.first
last = lastDay.sort.last
available = []
#Construct the available days for meetings
firstDay.each_with_index do |d, i|
available.push((firstDay[i]..lastDay[i]).to_a)
end
available = available.flatten.uniq.sort
investors = {}
attended_day = []
attended_investor = []
#Construct a list of investor based in their first and last days
firstDay.each_index do |i|
investors[i+1] = (firstDay[i]..lastDay[i]).to_a
end
for day in available
investors.each do |key, value|
next if attended_investor.include?(key)
if value.include?(day)
next if attended_day.include?(day)
attended_day.push(day)
attended_investor.push(key)
end
end
end
attended_investor.size
end
Using Lazy as far as I could understand, I escaped the MemoryError, but I started receiving a runtime error:
Your code was not executed on time. Allowed time: 10s
And my code look like this:
def countMeetings(firstDay, lastDay)
loop_size = firstDay.size
first = firstDay.sort.first
last = lastDay.sort.last
daily_attendance = {}
(first..last).each do |day|
for ind in 0...loop_size
(firstDay[ind]..lastDay[ind]).lazy.each do |investor_day|
next if daily_attendance.has_value?(ind)
if investor_day == day
daily_attendance[day] = ind
end
end
end
end
daily_attendance.size
end
And it went through the cases with few investors. I thought about using multi-thread and the code became the following:
def countMeetings(firstDay, lastDay)
loop_size = firstDay.size
first = firstDay.sort.first
last = lastDay.sort.last
threads = []
daily_attendance = {}
(first..last).lazy.each_slice(25000) do |slice|
slice.each do |day|
threads << Thread.new do
for ind in 0...loop_size
(firstDay[ind]..lastDay[ind]).lazy.each do |investor_day|
next if daily_attendance.has_value?(ind)
if investor_day == day
daily_attendance[day] = ind
end
end
end
end
end
end
threads.each{|t| t.join}
daily_attendance.size
end
Unfortunately, it went back to the MemoryError.
This can be done without consuming any more memory than the range of days. The key is to avoid Arrays and keep things as Enumerators as much as possible.
First, rather than the awkward pair of Arrays that need to be converted into Ranges, pass in an Enumerable of Ranges. This both simplifies the method, and it allows it to be Lazy if the list of ranges is very large. It could be read from a file, fetched from a database or an API, or generated by another lazy enumerator. This saves you from requiring big arrays.
Here's an example using an Array of Ranges.
p count_meetings([(1..4), (5..10), (10..10)])
Or to demonstrate transforming your firstDay and lastDay Arrays into a lazy Enumerable of Ranges...
firstDays = [1,5,10]
lastDays = [4,10,10]
p count_meetings(
firstDays.lazy.zip(lastDays).map { |first,last|
(first..last)
}
)
firstDays.lazy makes everything that comes after lazy. .zip(lastDays) iterates through both Arrays in pairs: [1,4], [5,10], and [10,10]. Then we turn them into Ranges. Because it's lazy it will only map them as needed. This avoids making another big Array.
Now that's fixed, all we need to do is iterate over each Range and increment their attendance for the day.
def count_meetings(attendee_ranges)
# Make a Hash whose default values are 0.
daily_attendance = Hash.new(0)
# For each attendee
attendee_ranges.each { |range|
# For each day they will attend, add one to the attendance for that day.
range.each { |day| daily_attendance[day] += 1 }
}
# Get the day/attendance pair with the maximum value, and only return the value.
daily_attendance.max[1]
end
Memory growth is limited to how big the day range is. If the earliest attendee is on day 1 and the last is on day 1000 daily_attendance is just 1000 entries which is a long time for a conference.
And since you've built the whole Hash anyway, why waste it? Write one function that returns the full attendance, and another that extracts the max.
def count_meeting_attendance(attendee_ranges)
daily_attendance = Hash.new(0)
attendee_ranges.each { |range|
range.each { |day| daily_attendance[day] += 1 }
}
return daily_attendance
end
def max_meeting_attendance(*args)
count_meeting_attendance(*args).max[1]
end
Since this is an exercise and you're stuck with the wonky arguments, we can do the same trick and lazily zip firstDays and lastDays together and turn them into Ranges.
def count_meeting_attendance(firstDays, lastDays)
attendee_ranges = firstDays.lazy.zip(lastDays).map { |first,last|
(first..last)
}
daily_attendance = Hash.new(0)
attendee_ranges.each { |range|
range.each { |day| daily_attendance[day] += 1 }
}
return daily_attendance
end
Is it possible to use destructuring in ruby to extract the end and beginning from a range?
module PriceHelper
def price_range_human( range )
"$%s to $%s" % [range.begin, range.end].map(:number_to_currency)
end
end
I know that I can use array coercion as a really bad hack:
first, *center, last = *rng
"$%s to $%s" % [first, last].map(:number_to_currency)
But is there a syntactical way to get begin and end without actually manually creating an array?
min, max = (1..10)
Would have been awesome.
You can use minmax to destructure ranges:
min, max = (1..10).minmax
min # => 1
max # => 10
If you are using Ruby before 2.7, avoid using this on large ranges.
The beginning and end? I'd use:
foo = 1..2
foo.min # => 1
foo.max # => 2
Trying to use destructuring for a range is a bad idea. Imagine the sizes of the array that could be generated then thrown away, wasting CPU time and memory. It's actually a great way to DOS your own code if your range ends with Float::INFINITY.
end is not the same as max: in 1...10, end is 10, but max is 9
That's because start_val ... end_val is equivalent to start_val .. (end_val - 1):
start_value = 1
end_value = 2
foo = start_value...end_value
foo.end # => 2
foo.max # => 1
foo = start_value..(end_value - 1)
foo.end # => 1
foo.max # => 1
max reflects the reality of the values actually used by Ruby when iterating over the range or testing for inclusion in the range.
In my opinion, end should reflect the actual maximum value that will be considered inside the range, not the value used at the end of the definition of the range, but I doubt that'll change otherwise it'd affect existing code.
... is more confusing and leads to increased maintenance problems so its use is not recommended.
No, Until I am proven incorrect by Cary Swoveland, Weekly World News or another tabloid, I'll continue believing without any evidence that the answer is "no"; but it's easy enough to make.
module RangeWithBounds
refine Range do
def bounds
[self.begin, self.end]
end
end
end
module Test
using RangeWithBounds
r = (1..10)
b, e = *r.bounds
puts "#{b}..#{e}"
end
Then again, I'd just write "#{r.begin.number_to_currency}..#{r.end.number_to_currency}" in the first place.
Amadan's answer is fine. you just need to remove the splat (*) when using it since it is not needed
eg,
> "%s to %s" % (1..3).bounds.map{|x| number_to_currency(x)}
=> "$1.00 to $3.00"
def Summer
#summer = true
puts "Your fruit are ripe for the picking."
if #tree_age == 1..5 && #tree_age > 0
#oranges = 5
elsif #tree_age == 6..15
#oranges = 20
else
#oranges = 50
end
end
I'm trying to ensure a tree between a certain age range gives x oranges, however I'm stuck with the following error referring to my elsif statement:
Orange_tree.rb:14: warning: integer literal in conditional range
I have also tried using an if greater than && less than conditional statement, can somebody please explain what this error means, and how to reach my solution.
You have a few problems:
You'll want to put your ranges in parenthesis when other operators or methods are nearby. Your current error comes from Ruby parsing elsif #tree_age == 6..15 differently than you expect - it's treating it as (1 == 6)..15, and false..15 obviously doesn't make any sense.
To test a number is within a range, use (1..5) === num, not num == (1..5). Range#=== is defined to test that the Range includes the right hand side, while Fixnum#== and Fixnum#=== both just test that the right hand side is numerically equivalent.
You don't need to test #tree_age > 0. You're already testing that it's in 1..5.
You could also consider a case statement for this, which can be a bit easier to read. case does its comparisons using ===.
#oranges = case #tree_age
when 1..5 then 5
when 6..15 then 20
else 50
end
You should use include? instead of == to determine if the given number is within the range:
def Summer
#summer = true
puts "Your fruit are ripe for the picking."
if (1..5).include?(#tree_age) && #tree_age > 0
#oranges = 5
elsif (6..15).include? #tree_age
#oranges = 20
else
#oranges = 50
end
end
==:
Returns true only if obj is a Range, has equivalent begin and end
items (by comparing them with ==), and has the same exclude_end?
setting as the range.
Which is obviously not the case.
The problem is with the lines that say == with a range.
if ( 10 == 1..11) # throws integer literal in conditional range warning
puts "true"
end
If you did this instead
if ( 10.between?(1, 11))
puts "true"
end
I want to know if a time belongs to an schedule or another.
In my case is for calculate if the time is in night schedule or normal schedule.
I have arrived to this solution:
NIGHT = ["21:00", "06:00"]
def night?( date )
date_str = date.strftime( "%H:%M" )
date_str > NIGHT[0] || date_str < NIGHT[1]
end
But I think is not very elegant and also only works for this concrete case and not every time range.
(I've found several similar question is SO but all of them make reference to Date ranges no Time ranges)
Updated
Solution has to work for random time ranges not only for this concrete one. Let's say:
"05:00"-"10:00"
"23:00"-"01:00"
"01:00"-"01:10"
This is actually more or less how I would do it, except maybe a bit more concise:
def night?( date )
!("06:00"..."21:00").include?(date.strftime("%H:%M"))
end
or, if your schedule boundaries can remain on the hour:
def night?(date)
!((6...21).include? date.hour)
end
Note the ... - that means, basically, "day time is hour 6 to hour 21 but not including hour 21".
edit: here is a generic (and sadly much less pithy) solution:
class TimeRange
private
def coerce(time)
time.is_a? String and return time
return time.strftime("%H:%M")
end
public
def initialize(start,finish)
#start = coerce(start)
#finish = coerce(finish)
end
def include?(time)
time = coerce(time)
#start < #finish and return (#start..#finish).include?(time)
return !(#finish..#start).include?(time)
end
end
You can use it almost like a normal Range:
irb(main):013:0> TimeRange.new("02:00","01:00").include?(Time.mktime(2010,04,01,02,30))
=> true
irb(main):014:0> TimeRange.new("02:00","01:00").include?(Time.mktime(2010,04,01,01,30))
=> false
irb(main):015:0> TimeRange.new("01:00","02:00").include?(Time.mktime(2010,04,01,01,30))
=> true
irb(main):016:0> TimeRange.new("01:00","02:00").include?(Time.mktime(2010,04,01,02,30))
=> false
Note, the above class is ignorant about time zones.
In Rails 3.2 it has added Time.all_day and similars as a way of generating date ranges. I think you must see how it works. It may be useful.
First of all, for those of you, who don't know (or forgot) about Lychrel numbers, here is an entry from Wikipedia: http://en.wikipedia.org/wiki/Lychrel_number.
I want to implement the Lychrel number detector in the range from 0 to 10_000. Here is my solution:
class Integer
# Return a reversed integer number, e.g.:
#
# 1632.reverse #=> 2361
#
def reverse
self.to_s.reverse.to_i
end
# Check, whether given number
# is the Lychrel number or not.
#
def lychrel?(depth=30)
if depth == 0
return true
elsif self == self.reverse and depth != 30 # [1]
return false
end
# In case both statements are false, try
# recursive "reverse and add" again.
(self + self.reverse).lychrel?(depth-1)
end
end
puts (0..10000).find_all(&:lychrel?)
The issue with this code is the depth value [1]. So, basically, depth is a value, that defines how many times we need to proceed through the iteration process, to be sure, that current number is really a Lychrel number. The default value is 30 iterations, but I want to add more latitude, so programmer can specify his own depth through method's parameter. The 30 iterations is perfect for such small range as I need, but if I want to cover all natural numbers, I have to be more agile.
Because of the recursion, that takes a place in Integer#lychrel?, I can't be agile. If I had provided an argument to the lychrel?, there wouldn't have been any changes because of the [1] statement.
So, my question sounds like this: "How do I refactor my method, so it will accept parameters correctly?".
What you currently have is known as tail recursion. This can usually be re-written as a loop to get rid of the recursive call and eliminate the risk of running out of stack space. Try something more like this:
def lychrel?(depth=30)
val = self
first_iteration = true
while depth > 0 do
# Return false if the number has become a palindrome,
# but allow a palindrome as input
if first_iteration
first_iteration = false
else
if val == val.reverse
return false
end
# Perform next iteration
val = (val + val.reverse)
depth = depth - 1
end
return true
end
I don't have Ruby installed on this machine so I can't verify whether that 's 100% correct, but you get the idea. Also, I'm assuming that the purpose of the and depth != 30 bit is to allow a palindrome to be provided as input without immediately returning false.
By looping, you can use a state variable like first_iteration to keep track of whether or not you need to do the val == val.reverse check. With the recursive solution, scoping limitations prevent you from tracking this easily (you'd have to add another function parameter and pass the state variable to each recursive call in turn).
A more clean and ruby-like solution:
class Integer
def reverse
self.to_s.reverse.to_i
end
def lychrel?(depth=50)
n = self
depth.times do |i|
r = n.reverse
return false if i > 0 and n == r
n += r
end
true
end
end
puts (0...10000).find_all(&:lychrel?) #=> 249 numbers
bta's solution with some corrections:
class Integer
def reverse
self.to_s.reverse.to_i
end
def lychrel?(depth=30)
this = self
first_iteration = true
begin
if first_iteration
first_iteration = false
elsif this == this.reverse
return false
end
this += this.reverse
depth -= 1
end while depth > 0
return true
end
end
puts (1..10000).find_all { |num| num.lychrel?(255) }
Not so fast, but it works:
code/practice/ruby% time ruby lychrel.rb > /dev/null
ruby lychrel.rb > /dev/null 1.14s user 0.00s system 99% cpu 1.150 total