Defining my_times Enumerator based on my_each Enumerator? - ruby

Say that I have my own implementation of each:
class Array
def my_each
c = 0
until c == size
yield(self[c])
c += 1
end
self
end
end
How would I do my own implementation of times by using my_each? Here is my approach:
class Integer
def my_times
Array.new(self) { |i| i }.my_each do |el|
yield el
end
end
end
But I don't particularly like it because I am creating an Array. But, is there any other way I could accomplish this?

You could do it like this:
class Integer
def my_times
return (0...self).to_enum unless block_given?
(0...self).each { |i| yield i }
end
end
5.my_times { |i| puts i*i }
0
1
4
9
16
5.my_times #=> #<Enumerator: 0...5:each>
I have used Range#each. To use Array#my_each we'd have to convert the range to an array:
[*(0...self)].my_each { |i| yield i }
Recall that Integer#times returns an enumerator if no block is given. The same is true of Array#each; you need to fix my_each for it to be equivalent to Array#each.
You don't need each or my_each, however:
class Integer
def my_times
return (0...self).to_enum unless block_given?
i = 0
while(i < self) do
yield i
i += 1
end
end
end
5.my_times { |i| puts i*i }
0
1
4
9
16
5.my_times #=> #<Enumerator: 0...5:each>

Related

Ruby Program to solve Circular Primes below number x

I'm working on project Euler #35. I am getting the wrong number returned and I can't find where I have done wrong!
def is_prime?(num)
(2..Math.sqrt(num)).each { |i| return false if num % i == 0}
true
end
def is_circular?(num)
len = num.to_s.length
return true if len == 1
(len - 1).times do
new_n = cycle(num)
break unless is_prime?(new_n)
end
end
def cycle(num)
ary = num.to_s.split("")
return ary.rotate!.join.to_i
end
def how_many
circulars = []
(2..1000000).each do |num|
if is_prime?(num) && is_circular?(num)
circulars << num
end
end
p circulars.count
end
how_many #=> 14426
The returned number is '14426'. I am only returning the circular primes, supposedly the correct answer is '55'
I have edited your code with few fixes in Ruby way. Your mistake was including corect set of [a, b, c] three times to count, instead of counting them as a one circular prime number. Your answer was correct, while 55 is the number of unique sets.
require 'prime'
class Euler35
def is_circular?(num)
circulars_for(num).all?{ |el| ::Prime.instance.prime?(el) }
end
def circulars_for(a)
a.to_s.split("").length.times.map{|el| a.to_s.split("").rotate(el).join.to_i }
end
def how_many
circulars = []
::Prime.each(1_000_000) do |num|
continue if circulars.include?(num)
if is_circular?(num)
circulars << circulars_for(num)
end
end
circulars.count
end
end
puts Euler35.new.how_many # => 55

Turning a method into an enumerable method

I rewrote the map method:
def my_map(input, &block)
mod_input = []
x = -1
while x < input.length - 1
x = x + 1
if block == nil
return input
break
end
mod_input.push(block.call(input[x]))
end
return mod_input
end
I need to call this code as I would call map or reverse. Does anyone know the syntax for that?
Are you asking how you put a method into a module? That's trivial:
module Enumerable
def my_map(&block)
mod_input = []
x = -1
while x < length - 1
x = x + 1
if block == nil
return self
break
end
mod_input.push(block.call(self[x]))
end
return mod_input
end
end
[1, 2, 3, 4, 5].my_map(&2.method(:*))
# => [2, 4, 6, 8, 10]
Or are you asking how to make your method an Enumerable method? That's more involved: your method currently uses many methods that are not part of the Enumerable API. So, even if you make it a member of the Enumerable module, it won't be an Enumerable method. Enumerable methods can only use each or other Enumerable methods. You use length and [] both of which are not part of the Enumerable interface, for example, Set doesn't respond to [].
This would be a possible implementation, using the Enumerable#inject method:
module Enumerable
def my_map
return enum_for(__method__) unless block_given?
inject([]) {|res, el| res << yield(el) }
end
end
[1, 2, 3, 4, 5].my_map(&2.method(:*))
# => [2, 4, 6, 8, 10]
A less elegant implementation using each
module Enumerable
def my_map
return enum_for(__method__) unless block_given?
[].tap {|res| each {|el| res << yield(el) }}
end
end
[1, 2, 3, 4, 5].my_map(&2.method(:*))
# => [2, 4, 6, 8, 10]
Note that apart from being simply wrong, your code is very un-idiomatic. There is also dead code in there.
the break is dead code: the method returns in the line just before it, therefore the break will never be executed. You can just get rid of it.
def my_map(&block)
mod_input = []
x = -1
while x < length - 1
x = x + 1
if block == nil
return self
end
mod_input.push(block.call(self[x]))
end
return mod_input
end
Now that we have gotten rid of the break, we can convert the conditional into a guard-style statement modifier conditional.
def my_map(&block)
mod_input = []
x = -1
while x < length - 1
x = x + 1
return self if block == nil
mod_input.push(block.call(self[x]))
end
return mod_input
end
It also doesn't make sense that it is in the middle of the loop. It should be at the beginning of the method.
def my_map(&block)
return self if block == nil
mod_input = []
x = -1
while x < length - 1
x = x + 1
mod_input.push(block.call(self[x]))
end
return mod_input
end
Instead of comparing an object against nil, you should just ask it whether it is nil?: block.nil?
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x = x + 1
mod_input.push(block.call(self[x]))
end
return mod_input
end
Ruby is an expression-oriented language, the value of the last expression that is evaluated in a method body is the return value of that method body, there is no need for an explicit return.
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x = x + 1
mod_input.push(block.call(self[x]))
end
mod_input
end
x = x + 1 is more idiomatically written x += 1.
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x += 1
mod_input.push(block.call(self[x]))
end
mod_input
end
Instead of Array#push with a single argument it is more idiomatic to use Array#<<.
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x += 1
mod_input << block.call(self[x])
end
mod_input
end
Instead of Proc#call, you can use the .() syntactic sugar.
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x += 1
mod_input << block.(self[x])
end
mod_input
end
If you don't want to store, pass on or otherwise manipulate the block as an object, there is no need to capture it as a Proc. Just use block_given? and yield instead.
def my_map
return self unless block_given?
mod_input = []
x = -1
while x < length - 1
x += 1
mod_input << yield(self[x])
end
mod_input
end
This one is opinionated. You could move incrementing the counter into the condition.
def my_map
return self unless block_given?
mod_input = []
x = -1
while (x += 1) < length
mod_input << yield(self[x])
end
mod_input
end
And then use the statement modifier form.
def my_map
return self unless block_given?
mod_input = []
x = -1
mod_input << yield(self[x]) while (x += 1) < length
mod_input
end
Also, your variable names could use some improvements. For example, what does mod_input even mean? All I can see is that it is what you output, so why does it even have "input" in its name? And what is x?
def my_map
return self unless block_given?
result = []
index = -1
result << yield(self[index]) while (index += 1) < length
result
end
This whole sequence of initializing a variable, then mutating the object assigned to that variable and lastly returning the object can be simplified by using the K Combinator, which is available in Ruby as Object#tap.
def my_map
return self unless block_given?
[].tap {|result|
index = -1
result << yield(self[index]) while (index += 1) < length
}
end
The entire while loop is useless. It's just re-implementing Array#each, which is a) unnecessary because Array#each already exists, and b) means that your my_map method will only work with Arrays but not other Enumerables (for example Set or Enumerator). So, let's just use each instead.
def my_map
return self unless block_given?
[].tap {|result|
each {|element|
result << yield(element)
}
}
end
Now it starts to look like Ruby code! What you had before was more like BASIC written in Ruby syntax.
This pattern of first creating a result object, then modifying that result object based on each element of a collection and in the end returning the result is very common, and it even has a fancy mathematical name: Catamorphism, although in the programming world, we usually call it fold or reduce. In Ruby, it is called Enumerable#inject.
def my_map
return self unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
That return self is strange. map is supposed to return a new object! You don't return a new object, you return the same object. Let's fix that.
def my_map
return dup unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
And actually, map is also supposed to return an Array, but you return whatever it is that you called map on.
def my_map
return to_a unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
But really, if you look at the documentation of Enumerable#map, you will find that it returns an Enumerator and not an Array when called without a block.
def my_map
return enum_for(:my_map) unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
And lastly, we can get rid of the hardcoded method name using the Kernel#__method__ method.
def my_map
return enum_for(__method__) unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
Now, that looks a lot better!
class Array
def my_map(&block)
# your code, replacing `input` with `self`
end
end
The code itself is not really idiomatic Ruby - while is very rarely used for iteration over collections, and if you don't need to pass a block somewhere else, it is generally cleaner to use block_given? instead of block.nil? (let alone block == nil), and yield input[x] instead of block.call(input[x]).

How to recreate .select()?

I'm trying to recreate the #select method. So far, I have array portion working, but when I try to call #my_select on a Hash, I get an empty Hash.
FYI, for starters, I had to recreate my own `#each' method. Here's that.
module Enumerable
def my_each
i = 0
while i < self.size
yield(self[i])
i += 1
end
self
end
Now, here's the #my_select method I created:
def my_select
if self.instance_of?(Array)
ret = []
self.my_each do |item|
ret << item if yield(item)
end
ret
elsif self.instance_of?(Hash)
ret = {}
self.my_each do |key, value|
if yield(key,value)
ret[key] = value
end
end
ret
end
end
end
...my input/output...
> h = { "a" => 100, "b" => 200, "c" => 300 }
> h.select { |k,v| k == "a"}
=> {"a"=>100}
> h.my_select { |k,v| k == "a"}
=> {}
Maybe you could change my_each to handle Hash as well?
def my_each
if self.instance_of?(Array)
i = 0
while i < self.size
yield(self[i])
i += 1
end
self
elsif self.instance_of?(Hash)
i = 0
arr = self.to_a
while i < arr.size
yield(arr[i][0], arr[i][1])
i += 1
end
self
end
end

Combining multiple methods into one

I am wondering if there was a way to combine multiple methods into one using Proc.new, lambda, or another method.
Also there is something wrong with my code below.
def multiples(n,p)
(n..p).each do |e|
puts e if e%2 == 0 or e%3 == 0
end
end
def summing(n)
sum = 0
n.each { |x| puts sum+=x }
end
It provides the correct answer when I test them individually like multiples(2,43) or summing([2,6,5,4]), but when I combine the two like summing(multiples(0,10)), I get the wrong answer.
If somebody could help me, that would be great.
The below should work..
def multiples(n,p)
(n..p).map do |e|
e if e%2 == 0 or e%3 == 0
end.compact
end
def summing(n)
sum = 0
n.each { |x| p sum+=x }
end
summing(multiples(0,10))
# >> 0
# >> 2
# >> 5
# >> 9
# >> 15
# >> 23
# >> 32
# >> 42
Array#each will not help you,for your task,as it returns the receiver when finished.
Jamaal,
You want summing to return sum (rather than n), so you need sum at the end:
def summing(n)
sum = 0
n.each { |x| puts sum += x}
sum
end

infinite enumerator rewind

I have a function that generates an enumerator in the following manner:
def create_example_enumerator(starting_value)
current = starting_value
e = Enumerator.new do |y|
loop do
current += 1
y << current
end
end
end
The current behavior is pretty straightforward.
> e = create_example_enumerator(0)
#<Enumerator: details>
> e.next
1
> e.next
2
> e.rewind
#<Enumerator: details>
> e.next
3
I would like e.rewind to reset the enumerator back to it's starting value.
Is there a nice way to do that while still using an infinite enumerator?
This should work:
n = Enumerator.new do |y|
number = 1
loop do
y.yield number
number += 1
end
end
n.next #=> 1
n.next #=> 2
n.next #=> 3
n.rewind
n.next #=> 1

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