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I was doing a quick read up on arrays and some basic methods. And one of the exercise questions at the end of the reading gave me an array and asked to get the following output
=> [10, 8, 4, 2]
Here's the array:
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
solution:1
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
numbers = numbers.select { |number| number.even? }.reverse
numbers.delete(6)
p numbers
But my question to you is why would the above code return the correct output but the following code won't?
solution: 2
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
numbers = numbers.select { |number| number.even? }
numbers.delete(6)
numbers.reverse
p numbers
I understand it's not the most fluent, but when I try to solve these exercises it's easier for me to separate everything and then clean up the code.
I expected it to pull the even numbers delete 6 from them and then print the reversed array.
Instead it pulls the even numbers, deletes 6, and prints the even numbers. Completely skips the .reverse
As max says, .reverse doesn't change the array. Try, instead:
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
numbers = numbers.select { |number| number.even? }
numbers.delete(6)
numbers.reverse!
p numbers
=> [10, 8, 4, 2]
As other commenters have mentioned, .reverse doesn't change the array.
You either have to declare numbers.reverse as a new variable (i.e. reversed_numbers = numbers.reverse) or use numbers.reverse! (as demonstrated by jvillian) to change the value of the numbers variable itself at invocation.
Between the two, the latter method is more suitable.
Hope this helped!
I've had a tough time finding an exact example of this.
If I have an array that contains 5 elements. E.g.
list = [5, 8, 10, 11, 15]
I would like to fetch what would be the 8th (for example) element of that array if it were to be looped. I don't want to just duplicate the array and fetch the 8th element because the nth element may change
Essentially the 8th element should be the number 10.
Any clean way to do this?
Math to the rescue
list[(8 % list.length) - 1]
A link about this modulo operator that we love
This should do:
def fetch_cycled_at_position(ary, num)
ary[(num % ary.length) - 1]
end
ary = _
=> [5, 8, 10, 11, 15]
fetch_cycled_at_position(ary, 1) # Fetch first element
=> 5
fetch_cycled_at_position(ary, 5) # Fetch 5th element
=> 15
fetch_cycled_at_position(ary, 8) # Fetch 8th element
=> 10
You could use rotate:
[5, 8, 10, 11, 15].rotate(7).first
#=> 10
It's 7 because arrays are zero based.
Just out of curiosity using Array#cycle:
[5, 8, 10, 11, 15].cycle.take(8).last
This is quite inefficient but fancy.
I ran these in my irb to get the output,
irb(main):006:0> list = [5, 8, 10, 11, 15]
=> [5, 8, 10, 11, 15]
irb(main):007:0> list[(8 % list.length) - 1]
=> 10
hope it will help you.
I have a hash whose keys are a range of integers (lets say [1..5]) and its corresponding 5 values are all nil. I have also an array of integers (lets say [1,2,3,4,5]. What I want to do is very specific: I want to take every single key and add it to every single of the array elements, giving me a hash that has the original keys, but has now for values the entire shifted array.
After spending a few hours I have concluded that this is impossible through a really laconic expression, because it is leading to .each shadowing statements.
I think that the only way to go through with this is to create 5 almost identical methods and call them separately.
def a1
array.each do |x|
x+1
end
end
def a2
array.each do |x|
x+2
end
end
and so on..
The end product I want to achieve is this:
{1=>[2,3,4,5,6],2=>[3,4,5,6,7],3=>[4,5,6,7,8],4=>[5,6,7,8,9],5=>[6,7,8,9,10]}
It feels like there should be a more DRY way to achieve this. Any ideas?
Assuming these initial conditions:
h = {1=>nil, 2=>nil, 3=>nil, 4=>nil, 5=>nil}
arr = [1,2,3,4,5]
...it's pretty straightforward:
h.keys.each do |key|
h[key] = arr.map {|i| i+key}
end
# h is now: {1=>[2, 3, 4, 5, 6], 2=>[3, 4, 5, 6, 7], 3=>[4, 5, 6, 7, 8], 4=>[5, 6, 7, 8, 9], 5=>[6, 7, 8, 9, 10]}
(However, it may be that your question is about achieving the initial conditions. If so, I didn't grasp that, and I didn't worry about it; I just started with what I took to be your initial conditions and ended up with your desired result.)
Why don't you do this
h = {}
rng.each{|i| h[i] = ary.map{|j| j + i}}
That should work where rng is the range and ary is the array.
For example
h = {}
(1..5).each{|i| h[i] = [1,2,3,4,5].map{|j| j+i}}
results in
h = {1=>[2, 3, 4, 5, 6], 2=>[3, 4, 5, 6, 7], 3=>[4, 5, 6, 7, 8], 4=>[5, 6, 7, 8, 9], 5=>[6, 7, 8, 9, 10]}
I have an array of numbers sorted either in ascending or descending order, and I want to find the index at which to insert a number while preserving the order of the array. If the array is [1, 5, 7, 11, 51] and the number to insert is 9, I would be expecting 3 so I could do [1, 5, 7, 11, 51].insert(3, 9). If the array is [49, 32, 22, 11, 10, 8, 3, 2] and the number to be inserted is 9, I would be expecting 5 so I could do [49, 32, 22, 11, 10, 8, 3, 2].insert(5, 9)
What would be the best/cleanest way to find the index at which to insert 9 in either of these two arrays while preserving the sorting of the array?
I wrote this code that works, but it's not very pretty:
array = [55, 33, 10, 7, 1]
num_to_insert = 9
index_to_insert = array[0..-2].each_with_index.map do |n, index|
range = [n, array[index.next]].sort
index.next if num_to_insert.between?(range[0], range[1])
end.compact.first
index_to_insert # => 3
Wand Maker's answer isn't bad, but it has two problems:
It sorts the entire array to determine whether it's ascending or descending. That's silly when all you have to do is find one element that's not equal to the one before it compare the first and last elements to determine this. That's O(n) O(1) in the worst case instead of O(n log n).
It uses Array#index when it should use bsearch. We can do a binary search instead of iterating over the whole array because it's sorted. That's O(log n) in the worst case instead of O(n).
I found it was clearer to split it into two methods, but you could of course turn it into one:
def search_proc(ary, n)
case ary.first <=> ary.last
when 1 then ->(idx) { n > ary[idx] }
when -1 then ->(idx) { n < ary[idx] }
else raise "Array neither ascending nor descending"
end
end
def find_insert_idx(ary, n)
(0...ary.size).bsearch(&search_proc(ary, n))
end
p find_insert_idx([1, 5, 7, 11, 51], 9)
#=> 3
p find_insert_idx([49, 32, 22, 11, 10, 8, 3, 2], 9)
#=> 5
(I use Range#bsearch here. Array#bsearch works the same, but it was more convenient to use a range to return an index, and more efficient since otherwise we'd have to do each_with_index.to_a or something.)
This is not a good way, but perhaps cleaner since you can use the method insert_sorted(number) on either an ascending or descending array without bothering about the index it will be placed on:
module SortedInsert
def insert_index(number)
self.each_with_index do |element, index|
if element > number && ascending?
return index
end
if element < number && descending?
return index
end
end
length
end
def insert_sorted(number)
insert(insert_index(number), number)
end
def ascending?
first <= last
end
def descending?
!ascending?
end
end
Use it on a array as follows:
array = [2, 61, 12, 7, 98, 64]
ascending = array.sort
descending = array.sort.reverse
ascending.extend SortedInsert
descending.extend SortedInsert
number_to_insert = 3
puts "Descending: "
p number_to_insert
p descending
p descending.insert_sorted(number_to_insert)
puts "Ascending: "
p number_to_insert
p ascending
p ascending.insert_sorted(number_to_insert)
This will give:
Descending:
3
[98, 64, 61, 12, 7, 2]
[98, 64, 61, 12, 7, 3, 2]
Ascending:
3
[2, 7, 12, 61, 64, 98]
[2, 3, 7, 12, 61, 64, 98]
Notes:
The module defines a few methods that will be added to the specific Array object alone.
The new methods provides a sorted array (either ascending/descending) a method insert_sorted(number) which enables to insert the number at sorted position.
In case the position of insertion is required, there is a method for that too: insert_index(number), which will provide the index to which the number needs to be inserted so that the resultant array remains sorted.
Caveat: The module assumes the array being extended is sorted either as ascending or descending.
Here is the simplest way I can think of doing.
def find_insert_idx(ary, n)
is_asc = (ary.sort == ary)
if (is_asc)
return ary.index { |i| i > n }
else
return ary.index { |i| i < n }
end
end
p find_insert_idx([1,5,7,11,51], 9)
#=> 3
p find_insert_idx([49,32,22,11,10,8,3,2], 9)
#=> 5
I am writing an appointment form that will let the user choose a date. It will then take the date and check against a Google Calendar what time slots are available for that date within a range of 30 minutes time intervals from 10:00am to 5:00pm.
Within my Calendar class, I have an available_times method:
def available_times(appointment_date)
appointment_date_events = calendar.events.select { |event| Date.parse(event.start_time) == appointment_date }
conflicts = appointment_date_events.map { |event| [Time.parse(event.start_time), Time.parse(event.end_time)] }
results = resolve_time_conflicts(conflicts)
end
This method takes a date and grabs the start_time and end_time for each event on that date. It then calls resolve_time_conflicts(conflicts):
def resolve_time_conflicts(conflicts)
start_time = Time.parse('10:00am')
available_times = []
14.times do |interval_multiple|
appointment_time = (start_time + interval_multiple * (30 * 60))
available_times << appointment_time unless conflicts.each{ |conflict| (conflict[0]..conflict[1]).include?(appointment_time)}
end
available_times
end
A 'Can't iterate over Time' error is being thrown when I attempt to iterate over the conflicts array. I tried to call to_enum on the conflicts array but am still getting the same error.
All of the other questions I saw on SO were referencing the step method, which doesn't seem applicable to this case.
Update:
Thanks #caryswoveland and #fivedigit. I combined both of your answers, which were very helpful for different aspects of my solution:
def available_times(appointment_date)
appointment_date_events = calendar.events.select { |event| Date.parse(event.start_time) == appointment_date }
conflicts = appointment_date_events.map { |event| DateTime.parse(event.start_time)..DateTime.parse(event.end_time) }
results = resolve_time_conflicts(conflicts)
end
def resolve_time_conflicts(conflicts)
date = conflicts.first.first
start_time = DateTime.new(date.year, date.month, date.day, 10, 00).change(offset: date.zone)
available_times = []
14.times do |interval_multiple|
appointment_time = (start_time + ((interval_multiple * 30).minutes))
available_times << appointment_time unless conflicts.any? { |conflict| conflict.cover?(appointment_time)}
end
available_times
end
Exception
#fivedigit has explained why the exception was raised.
Other problems
You need any? where you have each:
appointment_times = []
#=> []
appointment = 4
#=> 4
conflicts = [(1..3), (5..7)]
#=> [1..3, 5..7]
appointment_times << 5 unless conflicts.each { |r| r.cover?(appointment) }
#=> nil
appointment_times
#=> []
appointment_times << 5 unless conflicts.any? { |r| r.include?(appointment) }
#=> [5]
appointment_times
#=> [5]
I suggest you covert appointment_time to a Time object, make conflicts and array of elements [start_time, end_time] and then compare appointment_time to the endpoints:
...unless conflicts.any?{ |start_time, end_time|
start_time <= appointment_time && appointment_time <= end_time }
Aside: Range#include? only looks at endpoints (as Range#cover? does) when the endpoints are "numeric". Range#include? need only look at endpoints when they are Time objects, but I don't know if Ruby regards Time objects as "numeric". I guess one could look at the source code. Anybody know offhand?
Alternative approach
I would like to suggest a different way to implement your method. I will do so with an example.
Suppose appointments were in blocks of 15 minutes, with the first block being 10:00am-10:15am and the last 4:45pm-5:00pm. (blocks could be shorter, of course, as small as 1 second in duration.)
Let 10:00am-10:15am be block 0, 10:15am-10:30am be block 1, and so on, until block 27, 4:45pm-5:00pm.
Next, express conflicts as an array of block ranges, given by [start, end]. Suppose there were appointments at:
10:45am-11:30am (blocks 3, 4 and 5)
1:00pm- 1:30pm (blocks 12 and 13)
2:15pm- 3:30pm (blocks 17, 18 and 19)
Then:
conflicts = [[3,5], [12,13], [17,19]]
You must write a method reserved_blocks(appointment_date) that returns conflicts.
The remaining code is as follows:
BLOCKS = 28
MINUTES = ["00", "15", "30", "45"]
BLOCK_TO_TIME = (BLOCKS-1).times.map { |i|
"#{i<12 ? 10+i/4 : (i-8)/4}:#{MINUTES[i%4]}#{i<8 ? 'am' : 'pm'}" }
#=> ["10:00am", "10:15am", "10:30am", "10:45am",
# "11:00am", "11:15am", "11:30am", "11:45am",
# "12:00pm", "12:15pm", "12:30pm", "12:45pm",
# "1:00pm", "1:15pm", "1:30pm", "1:45pm",
# "2:00pm", "2:15pm", "2:30pm", "2:45pm",
# "3:00pm", "3:15pm", "3:30pm", "3:45pm",
# "4:00pm", "4:15pm", "4:30pm", "4:45pm"]
def available_times(appointment_date)
available = [*(0..BLOCKS-1)]-reserved_blocks(appointment_date)
.flat_map { |s,e| (s..e).to_a }
last = -2 # any value will do, can even remove statement
test = false
available.chunk { |b| (test=!test) if b > last+1; last = b; test }
.map { |_,a| [BLOCK_TO_TIME[a.first],
(a.last < BLOCKS-1) ? BLOCK_TO_TIME[a.last+1] : "5:00pm"] }
end
def reserved_blocks(date) # stub for demonstration.
[[3,5], [12,13], [17,19]]
end
Let's see what we get:
available_times("anything")
#=> [["10:00am", "10:45am"],
# ["11:30am", "1:00pm"],
# [ "1:45pm", "2:15pm"],
# [ "3:00pm", "5:00pm"]]
Explanation
Here is what's happening:
appointment_date = "anything" # dummy for demonstration
all_blocks = [*(0..BLOCKS-1)]
#=> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
# 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27]
reserved_ranges = reserved_blocks(appointment_date)
#=> [[3, 5], [12, 13], [17, 19]]
reserved = reserved_ranges.flat_map { |s,e| (s..e).to_a }
#=> [3, 4, 5, 12, 13, 17, 18, 19]
available = ALL_BLOCKS - reserved
#=> [0, 1, 2, 6, 7, 8, 9, 10, 11, 14, 15, 16, 20, 21, 22, 23, 24, 25, 26, 27]
last = -2
test = false
enum1 = available.chunk { |b| (test=!test) if b > last+1; last = b; test }
#=> #<Enumerator: #<Enumerator::Generator:0x00000103063570>:each>
We can convert it to an array to see what values it would pass into the block if map did not follow:
enum1.to_a
#=> [[true, [0, 1, 2]],
# [false, [6, 7, 8, 9, 10, 11]],
# [true, [14, 15, 16]],
# [false, [20, 21, 22, 23, 24, 25, 26, 27]]]
Enumerable#chunk groups consecutive values of the enumerator. It does so by grouping on the value of test and flipping its value between true and false whenever a non-consecutive value is encountered.
enum2 = enum1.map
#=> #<Enumerator: #<Enumerator: (cont.)
#<Enumerator::Generator:0x00000103063570>:each>:map>
enum2.to_a
#=> [[true, [0, 1, 2]],
# [false, [6, 7, 8, 9, 10, 11]],
# [true, [14, 15, 16]],
# [false, [20, 21, 22, 23, 24, 25, 26, 27]]]
You might think of enum2 as a "compound" enumerator.
Lastly, we convert the second element of each value of enum2 that is passed into the block (the block variable a, which equals [0,1,2] for the first element passed) to a range expressed as a 12-hour time. The first element of each value of enum2 (true or false) is not used, so so I've replaced its block variable with an underscore. This provides the desired result:
enum2.each { |_,a|[BLOCK_TO_TIME[a.first], \
(a.last < BLOCKS-1) ? BLOCK_TO_TIME[a.last+1] : "5:00pm"] }
#=> [["10:00am", "10:45am"],
# ["11:30am", "1:00pm"],
# [ "1:45pm", "2:15pm"],
# [ "3:00pm", "5:00pm"]]
The issue comes from this bit:
(conflict[0]..conflict[1]).include?(appointment_time)
# TypeError: can't iterate from Time
You're creating a range of times and then checking if appointment_time falls within the range. This is what causes the error you're experiencing.
Instead of include?, you should use cover?:
(conflict[0]..conflict[1]).cover?(appointment_time)
This assumes that conflict[0] is the earliest time.
Convert your range from a range of times to a range of integers:
range = (conflict[0].to_i..conflict[1].to_i)
Then use the === operator as you used the include?:
conflict === appointment_time
EDIT: You can also obviously convert appointment_time to integer and still use include? since the range is now just an integer range.