Summary: The basic question here was, I've discovered, whether you can pass a code block to a Ruby array which will actually reduce the contents of that array down to another array, not to a single value (the way inject does). The short answer is "no".
I'm accepting the answer that says this. Thanks to Squeegy for a great looping strategy to get streaks out of an array.
The Challenge: To reduce an array's elements without looping through it explicitly.
The Input: All integers from -10 to 10 (except 0) ordered randomly.
The Desired Output: An array representing streaks of positive or negative numbers. For instance, a -3 represents three consecutive negative numbers. A 2 represents two consecutive positive numbers.
Sample script:
original_array = (-10..10).to_a.sort{rand(3)-1}
original_array.reject!{|i| i == 0} # remove zero
streaks = (-1..1).to_a # this is a placeholder.
# The streaks array will contain the output.
# Your code goes here, hopefully without looping through the array
puts "Original Array:"
puts original_array.join(",")
puts "Streaks:"
puts streaks.join(",")
puts "Streaks Sum:"
puts streaks.inject{|sum,n| sum + n}
Sample outputs:
Original Array:
3,-4,-6,1,-10,-5,7,-8,9,-3,-7,8,10,4,2,5,-2,6,-1,-9
Streaks:
1,-2,1,-2,1,-1,1,-2,5,-1,1,-2
Streaks Sum:
0
Original Array:
-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,7,8,9,10
Streaks:
-10,10
Streaks Sum:
0
Note a few things:
The streaks array has alternating positive and negative values.
The sum of the elements streaks array is always 0 (as is the sum of the original).
The sum of the absolute values of the streak array is always 20.
Hope that's clear!
Edit: I do realize that such constructs as reject! are actually looping through the array in the background. I'm not excluding looping because I'm a mean person. Just looking to learn about the language. If explicit iteration is necessary, that's fine.
Well, here's a one-line version, if that pleases you more:
streaks = original_array.inject([]) {|a,x| (a.empty? || x * a[-1] < 0 ? a << 0 : a)[-1] += x <=> 0; a}
And if even inject is too loopy for you, here's a really silly way:
streaks = eval "[#{original_array.join(",").gsub(/((\-\d+,?)+|(\d+,?)+)/) {($1[0..0] == "-" ? "-" : "") + $1.split(/,/).size.to_s + ","}}]"
But I think it's pretty clear that you're better off with something much more straightforward:
streaks = []
original_array.each do |x|
xsign = (x <=> 0)
if streaks.empty? || x * streaks[-1] < 0
streaks << xsign
else
streaks[-1] += xsign
end
end
In addition to being much easier to understand and maintain, the "loop" version runs in about two-thirds the time of the inject version, and about a sixth of the time of the eval/regexp one.
PS: Here's one more potentially interesting version:
a = [[]]
original_array.each do |x|
a << [] if x * (a[-1][-1] || 0) < 0
a[-1] << x
end
streaks = a.map {|aa| (aa.first <=> 0) * aa.size}
This uses two passes, first building an array of streak arrays, then converting the array of arrays to an array of signed sizes. In Ruby 1.8.5, this is actually slightly faster than the inject version above (though in Ruby 1.9 it's a little slower), but the boring loop is still the fastest.
new_array = original_array.dup
<Squeegy's answer, using new_array>
Ta da! No looping through the original array. Although inside dup it's a MEMCPY, which I suppose might be considered a loop at the assembler level?
http://www.ruby-doc.org/doxygen/1.8.4/array_8c-source.html
EDIT: ;)
original_array.each do |num|
if streaks.size == 0
streaks << num
else
if !((streaks[-1] > 0) ^ (num > 0))
streaks[-1] += 1
else
streaks << (num > 0 ? 1 : -1)
end
end
end
The magic here is the ^ xor operator.
true ^ false #=> true
true ^ true #=> false
false ^ false #=> false
So if the last number in the array is on the same side of zero as the number being processed, then add it to the streak, otherwise add it to the streaks array to start a new streak. Note that sine true ^ true returns false we have to negate the whole expression.
Since Ruby 1.9 there's a much simpler way to solve this problem:
original_array.chunk{|x| x <=> 0 }.map{|a,b| a * b.size }
Enumerable.chunk will group all consecutive elements of an array together by the output of a block:
>> original_array.chunk{|x| x <=> 0 }
=> [[1, [3]], [-1, [-4, -6]], [1, [1]], [-1, [-10, -5]], [1, [7]], [-1, [-8]], [1, [9]], [-1, [-3, -7]], [1, [8, 10, 4, 2, 5]], [-1, [-2]], [1, [6]], [-1, [-1, -9]]]
This is almost exactly what OP asks for, except the resulting groups need to be counted up to get the final streaks array.
More string abuse, a la Glenn McDonald, only different:
runs = original_array.map do |e|
if e < 0
'-'
else
'+'
end
end.join.scan(/-+|\++/).map do |t|
"#{t[0..0]}#{t.length}".to_i
end
p original_array
p runs
# => [2, 6, -4, 9, -8, -3, 1, 10, 5, -7, -1, 8, 7, -2, 4, 3, -5, -9, -10, -6]
# => [2, -1, 1, -2, 3, -2, 2, -1, 2, -4]
Related
I am asked to write some code in Ruby that iterates over every n-th element of an array and prints it until all elements of the array are printed.
The question reads:
Imagine an iterator that accesses an array in strides and runs some code at each stride. If the strides reach the end of the array then they simply begin anew from the array's beginning.
For example:
x = [0,1,2,3,4]
x.stride(1) do |elem|; puts elem; end # prints 0,1,2,3,4
x.stride(2) do |elem|; puts elem; end # prints 0,2,4,1,3
x.stride(8) do |elem|; puts elem; end # prints 0,3,1,4,2
[].stride(2) do |elem|; puts elem; end # does not print anything, but the code is correct
Assume that the stride is equal or greater than 1, and that both the stride and the array's size are not a integral/whole multiple of each other, meaning that the whole array can be printed using a given stride. Fill in the code that's missing:
class Array
def stride(step)
numelems = ... # size of the array
...
end
end
It is obvious that numelemns = self.length(). However am having trouble with the rest.
I am going to try writing some code in Python that accomplishes this task, but I am afraid that I will not be able to translate it to Ruby.
Any ideas? The answer should not be more than 4-5 lines long as the question is one that our proffessor gave us to solve in a couple of minutes.
A solution to this is provided below (thanks #user3574603):
class Array
def stride(step)
yield self[0]
(self * step).map.with_index do |element, index|
next element if index == 0
yield element if index % step == 0
end
end
end
The following assumes that arr.size and n are not both even numbers and arr.size is not a multiple of n.
def striding(arr, n)
sz = arr.size
result = '_' * sz
idx = 0
sz.times do
result[idx] = arr[idx].to_s
puts "S".rjust(idx+1)
puts result
idx = (idx + n) % sz
end
end
striding [1,2,3,4,5,6,7,8,9,1,2,3,4,5,6], 7
S
1______________
S
1______8_______
S
1______8______6
S
1_____78______6
S
1_____78_____56
S
1____678_____56
S
1____678____456
S
1___5678____456
S
1___5678___3456
S
1__45678___3456
S
1__45678__23456
S
1_345678__23456
S
1_345678_123456
S
12345678_123456
S
123456789123456
Here is an example where arr.size is a multiple of n.
striding [1,2,3,4,5,6], 3
S
1_____
S
1__4__
S
1__4__
S
1__4__
S
1__4__
S
1__4__
Here is an example where arr.size and n are both even numbers.
striding [1,2,3,4,5,6,7,8], 6
S
1_______
S
1_____7_
S
1___5_7_
S
1_3_5_7_
S
1_3_5_7_
S
1_3_5_7_
S
1_3_5_7_
S
1_3_5_7_
Imagine an iterator that accesses an array in strides and runs some code at each stride. If the strides reach the end of the array then they simply begin anew from the array's beginning.
Based on this specification, stride will always iterate forever, unless the array is empty. But that is not a problem, since we can easily take only the amount of elements we need.
In fact, that is a good design: producing an infinite stream of values lets the consumer decide how many they need.
A simple solution could look like this:
module CoreExtensions
module EnumerableExtensions
module EnumerableWithStride
def stride(step = 1)
return enum_for(__callee__, step) unless block_given?
enum = cycle
loop do
yield(enum.next)
(step - 1).times { enum.next }
end
self
end
end
end
end
Enumerable.include(CoreExtensions::EnumerableExtensions::EnumerableWithStride)
A couple of things to note here:
I chose to add the stride method to Enumerable instead of Array. Enumerable is Ruby's work horse for iteration and there is nothing in the stride method that requires self to be an Array. Enumerable is simply the better place for it.
Instead of directly monkey-patching Enumerable, I put the method in a separate module. That makes it easier to debug code for others. If they see a stride method they don't recognize, and inspect the inheritance chain of the object, they will immediately see a module named EnumerableWithStride in the inheritance chain and can make the reasonable assumption that the method is probably coming from here:
[].stride
# Huh, what is this `stride` method? I have never seen it before.
# And it is not documented on https://ruby-doc.org/
# Let's investigate:
[].class.ancestors
#=> [
# Array,
# Enumerable,
# CoreExtensions::EnumerableExtensions::EnumerableWithStride,
# Object,
# Kernel,
# BasicObject
# ]
# So, we're confused about a method named `stride` and we
# found a module whose name includes `Stride`.
# We can reasonably guess that somewhere in the system,
# there must be a file named
# `core_extensions/enumerable_extensions/enumerable_with_stride.rb`.
# Or, we could ask the method directly:
meth = [].method(:stride)
meth.owner
#=> CoreExtensions::EnumerableExtensions::EnumerableWithStride
meth.source_location
#=> [
# 'core_extensions/enumerable_extensions/enumerable_with_stride.rb',
# 6
# ]
For an empty array, nothing happens:
[].stride(2, &method(:p))
#=> []
stride just returns self (just like each does) and the block is never executed.
For a non-empty array, we get an infinite stream of values:
x.stride(&method(:p))
# 0
# 1
# 2
# 3
# 4
# 0
# 1
# …
x.stride(2, &method(:p))
# 0
# 2
# 4
# 1
# 3
# 0
# 2
# …
x.stride(8, &method(:p))
# 0
# 3
# 1
# 4
# 2
# 0
# 3
# …
The nice thing about this infinite stream of values is that we, as the consumer can freely choose how many elements we want. For example, if I want 10 elements, I simply take 10 elements:
x.stride(3).take(10)
#=> [0, 3, 1, 4, 2, 0, 3, 1, 4, 2]
This works because, like all well-behaved iterators, our stride method returns an Enumerator in case no block is supplied:
enum = x.stride(2)
#=> #<Enumerator: ...>
enum.next
#=> 0
enum.next
#=> 2
enum.next
#=> 4
enum.next
#=> 1
enum.next
#=> 3
enum.next
#=> 0
enum.next
#=> 2
So, if we want to implement the requirement "until all the elements of the array are printed":
I am asked to write some code in Ruby that iterates over every n-th element of an array and prints it until all elements of the array are printed.
We could implement that something like this:
x.stride.take(x.length).each(&method(:p))
x.stride(2).take(x.length).each(&method(:p))
x.stride(8).take(x.length).each(&method(:p))
This is a pretty simplistic implementation, though. Here, we simply print as many elements as there are elements in the original array.
We could implement a more sophisticated logic using Enumerable#take_while that keeps track of which elements have been printed and which haven't, and only stops if all elements are printed. But we can easily prove that after x.length iterations either all elements have been printed or there will never be all elements printed (if the stride size is an integral multiple of the array length or vice versa). So, this should be fine.
This almost does what I think you want but breaks if the step is array.length + 1 array.length (but you mention that we should assume the stride is not a multiply of the array length).
class Array
def exhaustive_stride(step)
(self * step).map.with_index do |element, index|
next element if index == 0
element if index % step == 0
end.compact
end
end
x.exhaustive_stride 1
#=> [0, 1, 2, 3, 4]
x.exhaustive_stride 2
#=> [0, 2, 4, 1, 3]
x.exhaustive_stride 8
#=> [0, 3, 1, 4, 2]
[].exhaustive_stride 2
#=> []
Using the example array, it breaks when the stride is 5.
[0,1,2,3,4].exhaustive_stride 5
#=> [0, 0, 0, 0, 0]
Note
This works but the intermediate array makes it highly inefficient. Consider other answers.
Here's another solution that uses recursion. Not the most efficient but one way of doing it.
class Array
def exhaustive_stride(x, r = [])
return [] if self.empty?
r << self[0] if r.empty?
while x > self.length
x -= self.length
end
r << self[x]
x += x
return r if r.count == self.count
stride(x, r)
end
end
[0,1,2,3,4].exhaustive_stride 1
#=> [0, 1, 2, 4, 3]
[0,1,2,3,4].exhaustive_stride 2
#=> [0, 2, 4, 3, 1]
[0,1,2,3,4].exhaustive_stride 8
#=> [0, 3, 1, 2, 4]
[].exhaustive_stride 2
#=> []
[0,1,2,3,4].exhaustive_stride 100_000_001
#=> [0, 1, 2, 4, 3]
This would work:
def stride(ary, step)
raise ArgumentError unless step.gcd(ary.size) == 1
Array.new(ary.size) { |i| ary[(i * step) % ary.size] }
end
Example:
x = [0, 1, 2, 3, 4]
stride(x, 1) #=> [0, 1, 2, 3, 4]
stride(x, 2) #=> [0, 2, 4, 1, 3]
stride(x, 8) #=> [0, 3, 1, 4, 2]
stride(x, -1) #=> [0, 4, 3, 2, 1]
First of all, the guard clause checks whether step and ary.size are coprime to ensure that all elements can be visited via step.
Array.new(ary.size) creates a new array of the same size as the original array. The elements are then retrieved from the original array by multiplying the element's index by step and then performing a modulo operation using the array's size.
Having % arr.size is equivalent to fetching the elements from a cyclic array, e.g. for a step value of 2:
0 1 2 3 4
| | | | |
[0, 1, 2, 3, 4, 0, 1, 2, 3, 4, ...
To turn this into an instance method for Array you merely replace ary with self (which can be omitted most of the time):
class Array
def stride(step)
raise ArgumentError unless step.gcd(size) == 1
Array.new(size) { |i| self[(i * step) % size] }
end
end
Lets say I have an integer 98.
The binary representation of this string would be:
(98).to_s(2) # 1100010
Now I want to convert this binary string to an integer array of all the bits that are set. This would give me:
[64,32,2]
How would I go about this?
Update: The conversion of int to int array does not necessarily need to involve String, it is just what I knew. I assume non String operations would also be faster.
Ruby is amazing seeing all these different ways to handle this!
This would work:
i = 98
(0...i.bit_length).map { |n| i[n] << n }.reject(&:zero?)
#=> [2, 32, 64]
Fixnum#bit_length returns the position of the highest "1" bit
Fixnum#[n] returns the integer's nth bit, i.e. 0 or 1
Fixnum#<< shifts the bit to the left. 1 << n is equivalent to 2n
Step by step:
(0...i.bit_length).map { |n| i[n] }
#=> [0, 1, 0, 0, 0, 1, 1]
(0...i.bit_length).map { |n| i[n] << n }
#=> [0, 2, 0, 0, 0, 32, 64]
(0...i.bit_length).map { |n| i[n] << n }.reject(&:zero?)
#=> [2, 32, 64]
You might want to reverse the result.
In newer versions of Ruby (2.7+) you could also utilize filter_map and nonzero? to remove all 0 values:
(0...i.bit_length).filter_map { |n| (i[n] << n).nonzero? }
#=> [2, 32, 64]
Reverse the string, map it to binary code values of each digit, reject zeros. Optionally reverse it again.
s.reverse.chars.map.with_index{ |c, i| c.to_i * 2**i }.reject{ |b| b == 0 }.reverse
Or you could push the values to array with each_with_index
a = []
s.reverse.each_with_index do |c, i|
a.unshift c.to_i * 2**i
end
what is probably faster and more readable, but less idiomatic.
(98).to_s(2).reverse.chars.each_with_index.
map {|x,i| x=="1" ? 2**i : nil }.compact.reverse
Phew! Let's break that down:
First get the binary String as your example (98).to_s(2)
We will need to start 0-index from right hand side, hence .reverse
.chars.each_with_index gives us pairs such as [ '1', 4 ] for character at bit position
The .map converts "1" characters to their value 2 ** i (i.e. 2 to the power of current bit position) and "0" to nil so it can be removed
.compact to discard the nil values that you don't want
.reverse to have descending powers of 2 as your example
Here are a couple of ways:
#1
s = (98).to_s(2)
sz = s.size-1
s.each_char.with_index.with_object([]) { |(c,i),a| a << 2**(sz-i) if c == '1' }
# => [64, 32, 2]
#2
n = 2**(98.to_s(2).size-1)
arr = []
while n > 0
arr << n if 90[n]==1
n /= 2
end
arr
#=> [64, 32, 2]
I have no clue how to call this in correct math-terms. Consider a method which takes two digits:
def num_of_sum(total, group_count)
end
where total is an integer and group_count is an integer.
How would I get a 'nicely' grouped Array of integers of group_count-length which sum up till total.
My spec would look like:
describe "number to sum of" do
it "grabs all numbers" do
expect(num_of_sum(10, 2)).to eq([5,5])
expect(num_of_sum(10, 3)).to eq([3,3,4])
expect(num_of_sum(20, 3)).to eq([6,7,7])
expect(num_of_sum(100, 3)).to eq([33,33,34])
expect(num_of_sum(100, 2)).to eq([50,50])
end
end
I tried this, which works:
def num_of_sum(total, in_groups_of)
result = []
section_count ||= (total.to_f / in_groups_of.to_f).round
while(total > 0)
total -= section_count
if (total - section_count) < 0 && (total + section_count).even?
section_count += total
total -= total
end
result << section_count
end
result
end
But, for instance, this spec doesn't work:
expect(num_of_sum(67,5)).to eq([13,13,13,14,14])
I need the array to contain numbers that are as close to each other as possible. But the array is limited to the length of the group_count.
Does someone know what the mathemetical name for this is, so I can search a bit more accurately?
The mathematical term for this is an integer partition
A more direct approach to this is to observe that if you do integer division (round down) of the total by the number of groups, then your sum would be short by total mod number_of_groups, so you just need to distribute that amount across the array:
def even_partition(total, number_of_groups)
quotient, remainder = total.divmod(number_of_groups)
(number_of_groups-remainder).times.collect {quotient} +
remainder.times.collect { quotient + 1}
end
def n_parts(num, groupcount)
div, mod = num.divmod(groupcount)
Array.new(groupcount-mod, div) + Array.new(mod, div+1)
end
n_parts(100,3) => [33, 33, 34]
Docs to Array.new and Fixnum.divmod
A naive implementation is like this:
Let's take example of (20, 3). You want three numbers as a result.
20 / 3 # => 6
This is your "base" value. Create an array of three sixes, [6, 6, 6]. That'll get you 18. Now you have to distribute remaining 2 as equally as possible. For example, enumerate array elements and increment each one by 1, until you have no value to distribute. Result is [7, 7, 6]. Good enough, I think.
Possible (working) implementation:
def breakdown(total, group_count)
avg_value, extra = total.divmod(group_count)
result = Array.new(group_count, avg_value)
extra.times do |i|
result[i] += 1
end
result
end
breakdown(10, 2) == [5, 5] # => true
breakdown(10, 3) == [4, 3, 3] # => true
breakdown(20, 3) # => [7, 7, 6]
I have no clue how it’s called, but here is a solution:
def num_of_sum sum, count
result = [i = sum / count] * count # prepare an array e.g. [3,3,3] for 10,3
result[sum - i * count..-1] + # these should be left intact
result[0...sum - i * count].map { |i| i + 1 } # these are ++’ed
end
Hope it helps.
Another way:
def floors_then_ceils(n, groups)
floor, ceils = n.divmod(groups)
groups.times.map { |i| (i < groups-ceils) ? floor : floor + 1 }
end
floors_then_ceils(10, 3)
#=> [3, 3, 4]
floors_then_ceils(9, 3)
#=> [3, 3, 3]
Alternatively, groups.times.map... could be replaced with:
Array.new(groups-ceils, floor).concat(Array.new(ceils, floor+1))
I am trying to write a function in ruby that returns "Arithmetic" when the input is an arithmetic series array(the delta between a number and the consecutive number in the array is the same across the array). Here's what I have for so far:
array = [1,3,5,7,9] #arithmetic series input example
array.each_cons(2).map {|x,y| y - x == array[2] - array[1]}
This returns the following
[true,true,true,true,true]
My goal is to write a function that identifies what kind of series the array is, if there is one at all. Here's what I am trying to build
def isarithemetic(array)
if array.each_cons(2).map {|x,y| y - x == array[2] - array[1]} == true
return "arithmetic"
I am open to new suggestions as well.
If an array arr contains integer values, those (ordered) values comprise at most one arithmetic expression. If the average interval is integer-valued (i.e.,
((arr.last-arr.first) % (arr.size-1).zero?
is true), a unique progression is given by the values of arr.first, arr.last and arr.size, namely:
avg_interval = (arr.last-arr.first)/(arr.size-1)
(arr.first..arr.last).step(avg_interval).to_a
If the average interval is not integer-valued, it cannot be an arithmetic progression, but we don't need to check that condition. That's because by using integer arithmetic to compute avg_interval, the resulting arithmetic progression cannot equal arr if the average interval is not integer-valued.
Hence, rather than check if all the intervals are equal, we could just check to see if the above arithmetic progression equals arr.
Code
def arithmetic_progression?(arr)
return true if arr.size < 2
(arr.first..arr.last).step((arr.last-arr.first)/(arr.size-1)).to_a == arr
end
Examples
arithmetic_progression?([1,3,5,7,9]) #=> true
arithmetic_progression?([1,3,5,7,8]) #=> false
In the first example:
arr = [1,3,5,7,9]
avg_interval = (arr.last-arr.first)/(arr.size-1) #=> 2
(arr.first..arr.last).step(avg_interval).to_a #=> [1, 3, 5, 7, 9]
In the second:
arr = [1,3,5,7,8]
avg_interval = (arr.last-arr.first)/(arr.size-1) #=> 1
(arr.first..arr.last).step(avg_interval).to_a #=> [1, 2, 3, 4, 5, 6, 7, 8]
You could then write, for example:
arithmetic_progression?([1,3,5,7,9]) ? "Arithmetic progression" :
"Not arithmetic progression" #=> "Arithmetic progression"
Variant
The method could probably be made slightly more efficient by inserting the following line after the first line of the method:
return false unless ((arr.last-arr.first) % (arr.size-1)).zero?
You should be able to do this using all?:
array.each_cons(2).map {|x,y| y - x == array[2] - array[1]}.all?
or perhaps
array.each_cons(2).all? {|x,y| y - x == array[2] - array[1]}
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Turning long fixed number to array Ruby
Well, I have to iterate over the digits of a integer in Ruby. Right now I was just splitting it up into an array, and then iterating over that. However I was wondering if there was a faster way to do this?
The shortest solution probably is:
1234.to_s.chars.map(&:to_i)
#=> [1, 2, 3, 4]
A more orthodox mathematical approach:
class Integer
def digits(base: 10)
quotient, remainder = divmod(base)
quotient == 0 ? [remainder] : [*quotient.digits(base: base), remainder]
end
end
0.digits #=> [0]
1234.digits #=> [1, 2, 3, 4]
0x3f.digits(base: 16) #=> [3, 15]
You can use the old trick of modulus/divide by 10, but this won't be measurably faster unless you have huge numbers, and it will give the digits to you backwards:
i = 12345
while i > 0
digit = i % 10
i /= 10
puts digit
end
Output:
5
4
3
2
1
split=->(x, y=[]) {x < 10 ? y.unshift(x) : split.(x/10, y.unshift(x%10))}
split.(1000) #=> [1,0,0,0]
split.(1234) #=> [1,2,3,4]
Ruby has divmod, which will calculate both x%10and x/10 in one go:
class Integer
def split_digits
return [0] if zero?
res = []
quotient = self.abs #take care of negative integers
until quotient.zero? do
quotient, modulus = quotient.divmod(10) #one go!
res.unshift(modulus) #put the new value on the first place, shifting all other values
end
res # done
end
end
p 135.split_digits #=>[1, 3, 5]
For things like Project Euler, where speed is of some importance, this is nice to have. Defining it on Integer causes it to be available on Bignum too.
I like to use enumerators for this purpose:
class Integer
def digits
to_s.each_char.lazy.map(&:to_i)
end
end
This gives you access to all the good Enumerator stuff:
num = 1234567890
# use each to iterate over the digits
num.digits.each do |digit|
p digit
end
# make them into an array
p num.digits.to_a # => [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
# or take only some digits
p num.digits.take(5) # => [1, 2, 3, 4, 5]
# ...
Try mod by 10 (will give you the last digit), then divide by 10 (will give you the rest of digits), repeat this until you're down to the final digit. Of course, you'll have to reverse the order if you want to go through the digits from left to right.