Ruby inject with index and brackets - ruby

I try to clean my Code. The first Version uses each_with_index. In the second version I tried to compact the code with the Enumerable.inject_with_index-construct, that I found here.
It works now, but seems to me as obscure as the first code.
Add even worse I don't understand the brackets around element,index in
.. .inject(groups) do |group_container, (element,index)|
but they are necessary
What is the use of these brackets?
How can I make the code clear and readable?
FIRST VERSION -- WITH "each_with_index"
class Array
# splits as good as possible to groups of same size
# elements are sorted. I.e. low elements go to the first group,
# and high elements to the last group
#
# the default for number_of_groups is 4
# because the intended use case is
# splitting statistic data in 4 quartiles
#
# a = [1, 8, 7, 5, 4, 2, 3, 8]
# a.sorted_in_groups(3) # => [[1, 2, 3], [4, 5, 7], [8, 8]]
#
# b = [[7, 8, 9], [4, 5, 7], [2, 8]]
# b.sorted_in_groups(2) {|sub_ary| sub_ary.sum } # => [ [[2, 8], [4, 5, 7]], [[7, 8, 9]] ]
def sorted_in_groups(number_of_groups = 4)
groups = Array.new(number_of_groups) { Array.new }
return groups if size == 0
average_group_size = size.to_f / number_of_groups.to_f
sorted = block_given? ? self.sort_by {|element| yield(element)} : self.sort
sorted.each_with_index do |element, index|
group_number = (index.to_f / average_group_size).floor
groups[group_number] << element
end
groups
end
end
SECOND VERSION -- WITH "inject" AND index
class Array
def sorted_in_groups(number_of_groups = 4)
groups = Array.new(number_of_groups) { Array.new }
return groups if size == 0
average_group_size = size.to_f / number_of_groups.to_f
sorted = block_given? ? self.sort_by {|element| yield(element)} : self.sort
sorted.each_with_index.inject(groups) do |group_container, (element,index)|
group_number = (index.to_f / average_group_size).floor
group_container[group_number] << element
group_container
end
end
end

What is the use of these brackets?
It's a very nice feature of ruby. I call it "destructuring array assignment", but it probably has an official name too.
Here's how it works. Let's say you have an array
arr = [1, 2, 3]
Then you assign this array to a list of names, like this:
a, b, c = arr
a # => 1
b # => 2
c # => 3
You see, the array was "destructured" into its individual elements. Now, to the each_with_index. As you know, it's like a regular each, but also returns an index. inject doesn't care about all this, it takes input elements and passes them to its block as is. If input element is an array (elem/index pair from each_with_index), then we can either take it apart in the block body
sorted.each_with_index.inject(groups) do |group_container, pair|
element, index = pair
# or
# element = pair[0]
# index = pair[1]
# rest of your code
end
Or destructure that array right in the block signature. Parentheses there are necessary to give ruby a hint that this is a single parameter that needs to be split in several.
Hope this helps.

lines = %w(a b c)
indexes = lines.each_with_index.inject([]) do |acc, (el, ind)|
acc << ind - 1 if el == "b"
acc
end
indexes # => [0]

What is the use of these brackets?
To understand the brackets, first you need to understand how destruction works in ruby. The simplest example I can think of this this:
1.8.7 :001 > [[1,3],[2,4]].each do |a,b|
1.8.7 :002 > puts a, b
1.8.7 :003?> end
1
3
2
4
You should know how each function works, and that the block receives one parameter. So what happens when you pass two parameters? It takes the first element [1,3] and try to split (destruct) it in two, and the result is a=1 and b=3.
Now, inject takes two arguments in the block parameter, so it is usually looks like |a,b|. So passing a parameter like |group_container, (element,index)| we are in fact taking the first one as any other, and destructing the second in two others (so, if the second parameter is [1,3], element=1 and index=3). The parenthesis are needed because if we used |group_container, element, index| we would never know if we are destructing the first or the second parameter, so the parenthesis there works as disambiguation.
9In fact, things works a bit different in the bottom end, but lets hide this for this given question.)

Seems like there already some answers given with good explanation. I want to add some information regards the clear and readable.
Instead of the solution you chose, it is also a possibility to extend Enumerable and add this functionality.
module Enumerable
# The block parameter is not needed but creates more readable code.
def inject_with_index(memo = self.first, &block)
skip = memo.equal?(self.first)
index = 0
self.each_entry do |entry|
if skip
skip = false
else
memo = yield(memo, index, entry)
end
index += 1
end
memo
end
end
This way you can call inject_with_index like so:
# m = memo, i = index, e = entry
(1..3).inject_with_index(0) do |m, i, e|
puts "m: #{m}, i: #{i}, e: #{e}"
m + i + e
end
#=> 9
If you not pass an initial value the first element will be used, thus not executing the block for the first element.

In case, someone is here from 2013+ year, you have each_with_object and with_index for your needs:
records.each_with_object({}).with_index do |(record, memo), index|
memo[record.uid] = "#{index} in collection}"
end

Related

Ruby code to iterate over every n-th element of an array and print it until all elements 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.
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

How does to_enum(:method) receive its block here?

This code, from an example I found, counts the number of elements in the array which are equal to their index. But how ?
[4, 1, 2, 0].to_enum(:count).each_with_index{|elem, index| elem == index}
I could not have done it only with chaining, and the order of evaluation within the chain is confusing.
What I understand is we're using the overload of Enumerable#count which, if a block is given, counts the number of elements yielding a true value. I see that each_with_index has the logic for whether the item is equal to it's index.
What I don't understand is how each_with_index becomes the block argument of count, or why the each_with_index works as though it was called directly on [4,1,2,0]. If map_with_index existed, I could have done:
[4,1,2,0].map_with_index{ |e,i| e==i ? e : nil}.compact
but help me understand this enumerable-based style please - it's elegant!
Let's start with a simpler example:
[4, 1, 2, 0].count{|elem| elem == 4}
=> 1
So here the count method returns 1 since the block returns true for one element of the array (the first one).
Now let's look at your code. First, Ruby creates an enumerator object when we call to_enum:
[4, 1, 2, 0].to_enum(:count)
=> #<Enumerator: [4, 1, 2, 0]:count>
Here the enumerator is waiting to execute the iteration, using the [4, 1, 2, 0] array and the count method. Enumerators are like a pending iteration, waiting to happen later.
Next, you call the each_with_index method on the enumerator, and provide a block:
...each_with_index{|elem, index| elem == index}
This calls the Enumerator#each_with_index method on the enumerator object you created above. What Enumerator#each_with_index does is start the pending iteration, using the given block. But it also passes an index value to the block, along with the values from the iteration. Since the pending iteration was setup to use the count method, the enumerator will call Array#count. This passes each element from the array back to the enumerator, which passes them into the block along with the index. Finally, Array#count counts up the true values, just like with the simpler example above.
For me the key to understanding this is that you're using the Enumerator#each_with_index method.
The answer is but a click away: the documentation for Enumerator:
Most [Enumerator] methods [but presumably also Kernel#to_enum and Kernel#enum_for] have two forms: a block form where the contents are evaluated for each item in the enumeration, and a non-block form which returns a new Enumerator wrapping the iteration.
It is the second that applies here:
enum = [4, 1, 2, 0].to_enum(:count) # => #<Enumerator: [4, 1, 2, 0]:count>
enum.class # => Enumerator
enum_ewi = enum.each_with_index
# => #<Enumerator: #<Enumerator: [4, 1, 2, 0]:count>:each_with_index>
enum_ewi.class # => Enumerator
enum_ewi.each {|elem, index| elem == index} # => 2
Note in particular irb's return from the third line. It goes on say, "This allows you to chain Enumerators together." and gives map.with_index as an example.
Why stop here?
enum_ewi == enum_ewi.each.each.each # => true
yet_another = enum_ewi.each_with_index
# => #<Enumerator: #<Enumerator: #<Enumerator: [4, 1, 2, 0]:count>:each_with_index>:each_with_index>
yet_another.each_with_index {|e,i| puts "e = #{e}, i = #{i}"}
e = [4, 0], i = 0
e = [1, 1], i = 1
e = [2, 2], i = 2
e = [0, 3], i = 3
yet_another.each_with_index {|e,i| e.first.first == i} # => 2
(Edit 1: replaced example from docs with one pertinent to the question. Edit 2: added "Why stop here?)
Nice answer #Cary.. I'm not exactly sure how the block makes its way through the chain of objects, but despite appearances, the block is being executed by the count method, as in this stack trace, even though its variables are bound to those yielded by each_with_index
enum = [4, 1, 2, 0].to_enum(:count)
enum.each_with_index{|e,i| raise "--" if i==3; puts e; e == i}
4
1
2
RuntimeError: --
from (irb):243:in `block in irb_binding'
from (irb):243:in `count'
from (irb):243:in `each_with_index'
from (irb):243

Roll over on array in Ruby

I have an array in Ruby 2.0:
arr=[1,2,3,4,5]
I would like to be able to do something like:
arr[6] #=> 2
That is - roll over the end of the array and restart. Is that possible?
This would work:
arr = [1,2,3,4,5]
arr[6 % arr.size] #=> 2
Yes, it is possible to patch Array class in such way:
module RollOver
def [](index)
super index % size
end
end
Array.class_eval do
prepend RollOver
end
array = [1, 2, 3, 4, 5]
puts array[6] # => 2
But it is NOT recommended. Imagine how much code could be broken because of this patch. You better define another method for such operation.
UPDATE
If this behaviour is needed for one particular array only, then the best solution is:
array = [1, 2, 3, 4, 5]
def array.[](index)
super index % size
end
puts array[6] # => 2
Yes, Ruby allows that :-)

How to refactor this code to remove output variable?

def peel array
output = []
while ! array.empty? do
output << array.shift
mutate! array
end
output.flatten
end
I have not included the mutate! method, because I am only interested in removing the output variable. The mutate! call is important because we cannot iterate over the array using each because array is changing.
EDIT: I am getting an array as output, which is what I want. The method works correctly, but I think there is a way to collect the array.shift values without using a temp variable.
EDIT #2: OK, here is the mutate! method and test case:
def mutate! array
array.reverse!
end
a = (1..5).to_a
peel( a ).should == [ 1, 5, 2, 4, 3 ]
It doesn't matter if peel modifies array. I guess it should be called peel!. Yes, mutate! must be called after each element is removed.
All this reversing makes me dizzy.
def peel(array)
indices = array.size.times.map do |i|
i = -i if i.odd?
i = i/2
end
array.values_at(*indices) # indices will be [0, -1, 1, -2, 2] in the example
end
a = (1..5).to_a
p peel(a) #=>[1, 5, 2, 4, 3]
Another approach:
def peel(array)
mid = array.size/2
array[0..mid]
.zip(array[mid..-1].reverse)
.flatten(1)
.take(array.size)
end
Usage:
peel [1,2,3,4,5,6]
#=> [1, 6, 2, 5, 3, 4]
peel [1,2,3,4,5]
#=> [1, 5, 2, 4, 3]
Here's a way using parallel assignment:
def peel array
n = array.size
n.times {|i| (n-2-2*i).times {|j| array[n-1-j], array[n-2-j] = array[n-2-j], array[n-1-j]}}
array
end
peel [1,2,3,4,5] # => [1,5,2,4,3]
peel [1,2,3,4,5,6] # => [1,6,2,5,3,4]
What I'm doing here is a series of pairwise exchanges. By way of example, for [1,2,3,4,5,6], the first 6-2=4 steps (6 being the size of the array) alter the array as follows:
[1,2,3,4,6,5]
[1,2,3,6,4,5]
[1,2,6,3,4,5]
[1,6,2,3,4,5]
The 1, 6 and the 2 are in now the right positions. We repeat these steps, but this time only 6-4=2 times, to move the 5 and 3 into the correct positions:
[1,6,2,3,5,4]
[1,6,2,5,3,4]
The 4 is pushed to the end, it's correct position, so we are finished.

Skip over iteration in Enumerable#collect

(1..4).collect do |x|
next if x == 3
x + 1
end # => [2, 3, nil, 5]
# desired => [2, 3, 5]
If the condition for next is met, collect puts nil in the array, whereas what I'm trying to do is put no element in the returned array if the condition is met. Is this possible without calling delete_if { |x| x == nil } on the returned array?
My code excerpt is heavily abstracted, so looking for a general solution to the problem.
There is method Enumerable#reject which serves just the purpose:
(1..4).reject{|x| x == 3}.collect{|x| x + 1}
The practice of directly using an output of one method as an input of another is called method chaining and is very common in Ruby.
BTW, map (or collect) is used for direct mapping of input enumerable to the output one. If you need to output different number of elements, chances are that you need another method of Enumerable.
Edit: If you are bothered by the fact that some of the elements are iterated twice, you can use less elegant solution based on inject (or its similar method named each_with_object):
(1..4).each_with_object([]){|x,a| a << x + 1 unless x == 3}
I would simply call .compact on the resultant array, which removes any instances of nil in an array. If you'd like it to modify the existing array (no reason not to), use .compact!:
(1..4).collect do |x|
next if x == 3
x
end.compact!
In Ruby 2.7+, it’s possible to use filter_map for this exact purpose. From the docs:
Returns an array containing truthy elements returned by the block.
(0..9).filter_map {|i| i * 2 if i.even? } #=> [0, 4, 8, 12, 16]
{foo: 0, bar: 1, baz: 2}.filter_map {|key, value| key if value.even? } #=> [:foo, :baz]
For the example in the question: (1..4).filter_map { |x| x + 1 unless x == 3 }.
See this post for comparison with alternative methods, including benchmarks.
just a suggestion, why don't you do it this way:
result = []
(1..4).each do |x|
next if x == 3
result << x
end
result # => [1, 2, 4]
in that way you saved another iteration to remove nil elements from the array. hope it helps =)
i would suggest to use:
(1..4).to_a.delete_if {|x| x == 3}
instead of the collect + next statement.
You could pull the decision-making into a helper method, and use it via Enumerable#reduce:
def potentially_keep(list, i)
if i === 3
list
else
list.push i
end
end
# => :potentially_keep
(1..4).reduce([]) { |memo, i| potentially_keep(memo, i) }
# => [1, 2, 4]

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