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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
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 have been working through Chris Pine's tutorial for Ruby and am currently working on a way to sort an array of names without using sort.
My code is below. It works perfectly but is a step further than I thought I had got!
puts "Please enter some names:"
name = gets.chomp
names = []
while name != ''
names.push name
name = gets.chomp
end
names.each_index do |first|
names.each_index do |second|
if names[first] < names[second]
names[first], names[second] = names[second], names[first]
end
end
end
puts "The names you have entered in alphabetical order are: " + names.join(', ')
It is the sorting that I am having trouble getting my head around.
My understanding of it is that each_index would look at the position of each item in the array. Then the if statement takes each item and if the number is larger than the next it swaps it in the array, continuing to do this until the biggest number is at the start. I would have thought that this would just have reversed my array, however it does sort it alphabetically.
Would someone be able to talk me through how this algorithm is working alphabetically and at what point it is looking at what the starting letters are?
Thanks in advance for your help. I'm sure it is something very straightforward but after much searching I can't quite figure it out!
I think the quick sort algorithm is one of the easier ones to understand:
def qsort arr
return [] if arr.length == 0
pivot = arr.shift
less, more = arr.partition {|e| e < pivot }
qsort(less) + [pivot] + qsort(more)
end
puts qsort(["George","Adam","Michael","Susan","Abigail"])
The idea is that you pick an element (often called the pivot), and then partition the array into elements less than the pivot and those that are greater or equal to the pivot. Then recursively sort each group and combine with the pivot.
I can see why you're puzzled -- I was too. Look at what the algorithm does at each swap. I'm using numbers instead of names to make the order clearer, but it works the same way for strings:
names = [1, 2, 3, 4]
names.each_index do |first|
names.each_index do |second|
if names[first] < names[second]
names[first], names[second] = names[second], names[first]
puts "[#{names.join(', ')}]"
end
end
end
=>
[2, 1, 3, 4]
[3, 1, 2, 4]
[4, 1, 2, 3]
[1, 4, 2, 3]
[1, 2, 4, 3]
[1, 2, 3, 4]
In this case, it started with a sorted list, then made a bunch of swaps, then put things back in order. If you only look at the first couple of swaps, you might be fooled into thinking that it was going to do a descending sort. And the comparison (swap if names[first] < names[second]) certainly seems to imply a descending sort.
The trick is that the relationship between first and second is not ordered; sometimes first is to the left, sometimes it's to the right. Which makes the whole algorithm hard to reason about.
This algorithm is, I guess, a strange implementation of a Bubble Sort, which I normally see implemented like this:
names.each_index do |first|
(first + 1...names.length).each do |second|
if names[first] > names[second]
names[first], names[second] = names[second], names[first]
puts "[#{names.join(', ')}]"
end
end
end
If you run this code on the same array of sorted numbers, it does nothing: the array is already sorted so it swaps nothing. In this version, it takes care to keep second always to the right of first and does a swap only if the value at first is greater than the value at second. So in the first pass (where first is 0), the smallest number winds up in position 0, in the next pass the next smallest number winds up in the next position, etc.
And if you run it on array that reverse sorted, you can see what it's doing:
[3, 4, 2, 1]
[2, 4, 3, 1]
[1, 4, 3, 2]
[1, 3, 4, 2]
[1, 2, 4, 3]
[1, 2, 3, 4]
Finally, here's a way to visualize what's happening in the two algorithms. First the modified version:
0 1 2 3
0 X X X
1 X X
2 X
3
The numbers along the vertical axis represent values for first. The numbers along the horizontal represent values for second. The X indicates a spot at which the algorithm compares and potentially swaps. Note that it's just the portion above the diagonal.
Here's the same visualization for the algorithm that you provided in your question:
0 1 2 3
0 X X X X
1 X X X X
2 X X X X
3 X X X X
This algorithm compares all the possible positions (pointlessly including the values along the diagonal, where first and second are equal). The important bit to notice, though, is that the swaps that happen below and to the left of the diagonal represent cases where second is to the left of first -- the backwards case. And also note that these cases happen after the forward cases.
So essentially, what this algorithm does is reverse sort the array (as you had suspected) and then afterwards forward sort it. Probably not really what was intended, but the code sure is simple.
Your understanding is just a bit off.
You said:
Then the if statement takes each item and if the number is larger than the next it swaps it in the array
But this is not what the if statement is doing.
First, the two blocks enclosing it are simply setting up iterators first and second, which each count from the first to the last element of the array each time through the block. (This is inefficient but we'll leave discussion of efficient sorting for later. Or just see Brian Adkins' answer.)
When you reach the if statement, it is not comparing the indices themselves, but the names which are at those indices.
You can see what's going on by inserting this line just before the if. Though this will make your program quite verbose:
puts "Comparing names[#{first}] which is #{names[first]} to names[#{second}] which is #{names[second]}..."
Alternatively, you can create a new array and use a while loop to append the names in alphabetical order. Delete the elements that have been appended in the loop until there are no elements left in the old array.
sorted_names = []
while names.length!=0
sorted_names << names.min
names.delete(names.min)
end
puts sorted_names
This is the recursive solution for this case
def my_sort(list, new_array = nil)
return new_array if list.size <= 0
if new_array == nil
new_array = []
end
min = list.min
new_array << min
list.delete(min)
my_sort(list, new_array)
end
puts my_sort(["George","Adam","Michael","Susan","Abigail"])
Here is my code to sort items in an array without using the sort or min method, taking into account various forms of each item (e.g. strings, integers, nil):
def sort(objects)
index = 0
sorted_objects = []
while index < objects.length
sorted_item = objects.reduce do |min, item|
min.to_s > item.to_s ? item : min
end
sorted_objects << sorted_item
objects.delete_at(objects.find_index(sorted_item))
end
index += 1
sorted_objects
end
words_2 = %w{all i can say is that my life is pretty plain}
p sort(words_2)
=> ["all", "can", "i", "is", "is", "life", "my", "plain", "pretty", "say", "that"]
mixed_array_1 = ["2", 1, "5", 4, "3"]
p sort(mixed_array_1)
=> [1, "2", "3", 4, "5"]
mixed_array_2 = ["George","Adam","Michael","Susan","Abigail", "", nil, 4, "5", 100]
p sort(mixed_array_2)
=> ["", nil, 100, 4, "5", "Abigail", "Adam", "George", "Michael", "Susan"]
How do you create a for loop like
for (int x=0; x<data.length; x+=2)
in ruby? I want to iterate through an array but have my counter increment by two instead of one.
If what you really want is to consume 2 items from an array at a time, check out each_slice.
[1,2,3,4,5,6,7,8,9].each_slice(2) do |a, b|
puts "#{a}, #{b}"
end
# result
1, 2
3, 4
5, 6
7, 8
9,
Ruby's step is your friend:
0.step(data.length, 2).to_a
=> [0, 2, 4, 6]
I'm using to_a to show what values this would return. In real life step is an enumerator, so we'd use it like:
data = [0, 1, 2, 3, 4, 5]
0.step(data.length, 2).each do |i|
puts data[i]
end
Which outputs:
0
2
4
<== a nil
Notice that data contains six elements, so data.length returns 6, but an array is a zero-offset, so the last element would be element #5. We only get three values, plus a nil which would display as an empty line when printed, which would be element #6:
data[6]
=> nil
That's why we don't usually walk arrays and container using outside iterators in Ruby; It's too easy to fall off the end. Instead, use each and similar constructs, which always do the right thing.
To continue to use step and deal with the zero-offset for arrays and containers, you could use:
0.step(data.length - 1, 2)
but I'd still try working with each and other array iterators as a first choice, which #SergioTulentsev was giving as an example.
(0..data.length).step(2) do |x|
puts x
end
This seems like the closest substitute.
Using Range#step:
a = (1..50).to_a
(1..a.size).step(2).map{|i| a[i-1]} # [1, 3, 5, 7, 9 ...
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.