I am currently learning Ruby and am lost a bit.
This Python snippet:
n = iter([1,2,3,4,5])
for x in n:
print(x)
y = next(n)
print(y)
gives:
1
2
3
4
5
And I'm trying to do the same in Ruby, which is not working:
n = [1,2,3,4,5].each
for x in n do
puts x
y = n.next()
puts y
end
How do I need to write the Python example in Ruby?
You have the right idea except Ruby keeps track of the position of the enumerator internally. Try this:
enum = [1,2,3,4,5].each
#=> #<Enumerator: [1, 2, 3, 4, 5]:each>
loop do
puts enum.next
puts enum.next
end
1
2
3
4
5
Enumerator#next raises a StopInteration exception when invoked after all elements of the enumerator enum have been generated. Kernel#loop handles the exception by breaking out of the loop.
Note that Array#each, without the optional block, has the same effect as Object#to_enum. Also, though authors of Ruby books feel an obligation to cover for loops (seemingly, on page one), they are never used in practice.
In Python the iterator n obtained applying the function iter() can be consumed once.
array = [1,2,3,4,5]
n = iter(array)
for x in n:
print(x)
# 1
# 2
# 3
# 4
# 5
If you then call next(n) you get no output but error: StopIteration. If you try to iterate n again you get no output (no error, the for loop hanldes the exception).
But you can iterate over the array:
for x in array:
print(x)
If you check the output a code like yours, you get:
n = iter([1,2,3,4,5])
for x in n:
print('odd_', x)
y = next(n)
print('even', y)
# odd_ 1
# even 2
# odd_ 3
# even 4
# odd_ 5
# StopIteration
As pointed by #toniedzwiedz Object#to_enum is the equivalent for the Python iter(), so you can call Enumerator#next on it.
In Ruby you can consume more than once the enumerator:
array = [1, 2, 3, 4, 5]
n = array.to_enum
# n = array.each # each returns an Enumerable
for x in n do
puts x
end
for x in n do
puts x
end
Since you can consume more than once the enumerator you get "double" output:
for x in n do
p x
p n.next
end
Or using Array#each:
n.each do |x| # same as array.each.each
p x
p n.next
end
Also in Ruby you reach end the iteration signal:
6.times do
p n.next
end
# `next': iteration reached an end (StopIteration)
You're just trying to print each element of the Array?
[1,2,3,4,5].each {|n| puts n}
If you're looking for a direct equivalent of Python's iterator, you should check out Enumerator#next for a very similar API
arr = [1, 2, 3].to_enum
#<Enumerator: [1, 2, 3]:each>
arr.next
# 1
for x in arr do
puts x
puts arr.next
end
#1
#2
#2
#3
#3
#StopIteration: iteration reached an end
each_cons is what you need. But this method has a downside that is the current will not print last element of the array. Please keep in mind this.
[1,2,3,4,5].each_cons(2) { |current,next_val| puts next_val }
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
trio = Proc.new do |x|
tf = true
puts x
if tf
puts "ai yo"
end
end
trio.call([1, 2, 3, 4, 5])
output:
1 2 3 4 5 ai yo #its only doing the ai yo part only once when I believe it
should do it after every number
but What I am expecting of output is:
1 ai yo 2 ai yo 3 ai yo 4 ai yo 5 ai yo
I still cant wrap my head around why this is happening.
Im trying to get this program to work that i thought would be a cool way to use procs although in this specific problem i dont need to basically:
#The prime factors of 13195 are 5, 7, 13 and 29.
#What is the largest prime factor of the number 600851475143 ?
number = 13195
def factorsAndOptions(num, proc = Proc.new {|x|return x})
factorsArray = []
for i in 1..num
factorsArray.push(i) if num % i == 0
end
proc.call(factorsArray)
end
largestPrime = Proc.new do |x|
prime = true
for j in 2...x
if (x % x == 0)
prime = false
end
end
larger = x if prime && larger > x
puts larger
larger
end
factorsAndOptions(number, largestPrime)
call won't iterate over arguments. What you've written is, effectively:
puts [1, 2, 3, 4, 5]
puts "ai yo"
If you want to iterate, use each:
[1, 2, 3, 4, 5].each(&trio)
Or:
[1, 2, 3, 4, 5].each { |i| trio.call(i) }
As has been mentioned, you do not have any looping block. Your proc - trio is acting on the whole array as one single element.
In your example: x becomes [1, 2, 3, 4, 5] and not individual elements of the array as you are expecting.
To circumvent this you can either loop inside your Proc or create a separate Proc that will loop over the elements of the array and call the first Proc.
Example 1
trio = Proc.new do |arr|
arr.each do |elem|
puts elem
if tf
puts "ai yo"
end
end
end
This assumes that arr is an array
Example 2
trio = Proc.new do |x|
tf = true
puts x
if tf
puts "ai yo"
end
end
trio_helper = Proc.new do |x|
arr = x.to_a
arr.each do |elem|
trio.call(elem)
end
end
trio_helper.call([1, 2, 3, 4, 5])
This utilizes the original Proc you have written and uses another Proc to iterate over the array and call the first one on each element.
Why does this work:
a = [1, 2, 3]
while n = a.shift
puts n
end
while this doesn't:
a = [1, 2, 3]
puts n while n = a.shift
It works only if I initialize n in advance:
a = [1, 2, 3]
n = nil
puts n while n = a.shift
That is, in general, an interpreter problem, that could not appear in languages with local variable bubbling, like javascript.
The interpreter (reading from left to right) meets right-hand-operand n before any mention of it.
The more I think about it, the more I am convinced it is a bug in ruby interpreter. As #Cary pointed out, the control flow is in fact the same:
a = [2, 3]
n = 1
puts n while n = a.shift
#⇒ 2
#⇒ 3
No trail of 1 in the output above.
n is undefined at the time you attempt the first puts. The condition, and corresponding shift, is only checked after the puts has been evaluated. An alternative which will work as you expected would be
a = [1, 2, 3]
puts a.shift while a.length > 0
Regarding: puts n while n = a.shift,
it will pares puts n first, but n is undefined at that point. Ruby is a dynamically typed language; you don't declare variable type explicitly, but you should assign a value to variables.
For example:
irb(main):027:0> xyz
NameError: undefined local variable or method `xyz' for main:Object
irb(main):028:0> xyz = 1
=> 1
I'm writing a program that pushes Fibonacci numbers into an array, using Ruby. The code works, but I can't wrap my head around why it works.
This part I understand, it's the Fibonacci equation:
fib_array = []
def fib (n)
return n if n <= 1
fib(n - 1) + fib(n - 2)
end
This is what I don't understand:
10.times do |x|
fib_array << fib(x)
end
print fib_array
I wrote this grasping at straws, and it works. I don't understand why. I didn't feed it a number to start at, does Ruby take that to mean 0? Also, how did it know to compound the numbers instead of printing [0, 0, 0...]? I apologize if this is a dunderheaded question, but I'm at loss.
It looks like the bottom piece of code simply calls the fib function on x=0, x=1 ... x=9 and stores it's return value at the end of the array. When times is invoked with an iteration variable (x), it begins at 0 and increments on each iteration through the loop. You never fed it a value, however it manages to successfully solve the problem with the iteration variable x being passed in as the parameter to fib.
The second part of your code says:
"From the instance 10 of the class Integer, call the method times with the given block" (The method "recive" a block implicitly).
What is a block? A small piece of code between {braces} or a do-end (like you did).
The method times is called, "iterator". And it will yield 0,1,2,..,9 (in your case). An iterator and the yield statement are always together. Think that yield is like return with memory, when you look for more information.
So, your code could be re-writing like:
10.times { |x| fib_array << fib(x) }
And it will call, the block you pass, on every yield that the method times
does. Calling the method << (append) to the result of fib(x) on your array.
We have:
def fib (n)
return n if n <= 1
fib(n - 1) + fib(n - 2)
end
which you understand. First let's see what are the first 5 Fibonacci numbers::
fib_array = []
5.times do |x|
fib_array << fib(x)
end
fib_array
#=> [0, 1, 1, 2, 3]
Now let's break this down and see what happening. First look at the docs for the method times. To find them, we need to know what class or module the method is from, because that's how the docs are organized. As 5.class #=> Fixnum, we might look at the docs for Fixnum. Hmmm. times is not there. Evidentally, it was inherited from another class. Let's check:
Fixnum.ancestors
#=> [Fixnum, Integer, Numeric, Comparable, Object, Kernel, BasicObject]
Integer includes Fixnum and BigNum, Numeric includes Integer and Float. (3.14).times doesn't make sense, so it appears times is defined in Integer, and so it is: Integer#times. By defining it there, it is inherited by both Fixnum and Bignum.
Here's a direct way to determine where the method came from:
5.method(:times).owner #=> Integer
It's not important understand this now, but you'll find it handy as you gain experience with Ruby.
OK, the docs say that times return a value if given a block, or an enumerator if not. Let's forget about the block a moment and look at the enumerator that is returned:
enum = 5.times #=> #<Enumerator: 5:times>
The method Enumerator#each passes the elements of the enumerator enum to its block:
do |x|
fib_array << fib(x)
end
assigning them to the block variable x. To see the contents of the enumerator, convert it to an array:
enum.to_a #=> [0, 1, 2, 3, 4]
The result is:
fib_array = []
enum.each do |x|
fib_array << fib(x)
end
fib_array
#=> [0, 1, 1, 2, 3]
which of course is the same result that we obtained previously. Now let's see what's happening step-by-by, by using the method Enumerator#next to extract each element of the enumerator:
x = enum.next #=> 0
fib(x) #=> 0
fib_array << fib(x) #=> [0]
x = enum.next #=> 1
fib(x) #=> 1
fib_array << fib(x) #=> [0, 1]
x = enum.next #=> 2
fib(x) #=> 1
fib_array << fib(x) #=> [0, 1, 1]
x = enum.next #=> 3
fib(x) #=> 2
fib_array << fib(x) #=> [0, 1, 1, 2]
x = enum.next #=> 4
fib(x) #=> 3
fib_array << fib(x) #=> [0, 1, 1, 2, 3]
print fib_array # [0, 1, 1, 2, 3]
That's all there is to it.
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