What is the difference of Ruby's Array#to_a method - ruby

For example:
a = [1,2,3,4]
b = a
c = a.to_a
a.insert(0,0) #=> [0,1,2,3,4]
b #=> [0,1,2,3,4]
c #=> [0,1,2,3,4]
Why the output of array b and c is the same? If I want to get a copy of array a, not a reference one, which method should I use?

Why the output of array b and c is the same?
Because all three local variables referencing the same objects,as below:
a = [1,2,3,4]
b = a
c = a.to_a
a.object_id
# => 72187200
b.object_id
# => 72187200
c.object_id
# => 72187200
If i want to get a copy of array a, not a reference one , which method should i use?
Then use a.dup.Here documented Object#dup
a = [1,2,3,4]
b = a.dup
c = a.dup
a.object_id
# => 82139270
b.object_id
# => 82139210
c.object_id
# => 82134600
a.insert(0,0) # => [0, 1, 2, 3, 4]
b # => [1, 2, 3, 4]
c # => [1, 2, 3, 4]
Array#to_a says : Returns self.If called on a subclass of Array, converts the receiver to an Array object.
So it will not be helpful as per your need.

This is because Array#to_a returns self, so both variables contain a reference to the same Array object. In order to get a new Array with the same contents, you can use either dup or clone (read about the differences between dup and clone):
a = [1, 2, 3]
b = a.dup
a << 4
a #=> [1, 2, 3, 4]
b #=> [1, 2, 3]
Note, however, that the same object references are stored in the new array. This means if you mutate the objects themselves they will still change in both arrays:
a = ['foo', 'bar']
b = a.dup
a[0] << 'baz'
a #=> ["foobaz", "bar"]
b #=> ["foobaz", "bar"]
This is because dup and clone are shallow-copies.

Array#to_a returns the receiver. That is why b and c refer to the same thing. Regarding why to_a returns the original array, in principle, it could be defined one or the other way, but I guess one use case for to_a is to apply it to a variable that is potentially nil to ensure it becomes an array.
some_value.to_a # => `[]` if `some_value` is `nil`
In such use case, you don't need to replace the array with another one in case the receiver is already an array. That would be performantly more preferable.

The reason for that is that variables are merely references to data. Variables are stored in memory; variables keep address of where they are located in a memory. So when you do:
a = b
Those two variables point to the same memory location, hence if you alter a, b is altered as well because it is the same object.
There are a few ways to force Ruby to create another copy of the object. The most popular one is the dup method mentioned by LBg. Note however that it is only creating a shallow copy. If you run:
a = ['foo','bar', []]
b = a.dup
a << 'blah'
b #=> ['foo', 'bar', []] as expected but
b[3] << blah
a #=> ['foobar', 'bar', ['blah]]
The reason for that is that array is in fact an array of references and nested array has not been duplicated when performing dup, so they are the same object.
To create a deep copy of an object, you can use the Marshall module:
b = Marshal.load(Marshal.dump(a))
However, usually you don't really need to do this. Also, some objects cannot be duplicated (e.g. symbols).

You can just do
b = a.dup
OLD POST
You can try this if there is no easier way
b = a.map {|x| x}
It works
1.9.3-p448 :001 > a = [1,2,3] => [1, 2, 3]
1.9.3-p448 :002 > b = a => [1, 2, 3]
1.9.3-p448 :003 > c = a.map{|x| x} => [1, 2, 3]
1.9.3-p448 :004 > a<<0 => [1, 2, 3, 0]
1.9.3-p448 :005 > b => [1, 2, 3, 0]
1.9.3-p448 :006 > c => [1, 2, 3]
But it is a shallow copy though.
According to this post, a.dup is the easier way.

Related

How to stop shift from affecting other instances of an array

I need to shift through an array and keep a copy of the original array for future.
I tried creating another variable using a = b, but both are affected when I shift a.
rb(main):001:0> a = [1,2,3,4,5]
# => [1, 2, 3, 4, 5]
irb(main):002:0> b = a
# => [1, 2, 3, 4, 5]
irb(main):003:0> c = a.shift
# => 1
irb(main):004:0> a
# => [2, 3, 4, 5]
irb(main):005:0> b
# => [2, 3, 4, 5]
irb(main):006:0> c
# => 1
Is there a way to keep this from happening?
In Ruby it's important to remember variables are object references which behave a lot like pointers, so b = a does not make a copy, it is another reference to the same object.
To make a copy you must be explicit and use dup or clone to achieve this:
b = a.dup
If you're ever confused by Ruby's behaviour, stop and look at the objects you're dealing with:
a = [ 1 ]
b = a
a.object_id == b.object_id
# => true
They're exactly the same object, but when cloned:
b = a.dup
a.object_id == b.object_id
# => false
Now they're independent, at least on the top-level.
Note that this comes with some caveats, as this is only a shallow copy:
a = [ [ 1 ] ]
b = a.dup
b[0].object_id == a[0].object_id
# => true
This is where deep_clone tools come in handy if you need a complete clone, something available from various gems but most popularly ActiveSupport from Rails.
One thing you'll find in Ruby is it tends to steer towards a more functional style, as in if you wanted to strip an element from a and avoid mangling b:
a = [ 1, 2, 3, 4, 5 ]
b = a
a = a.drop(1)
# => [2, 3, 4, 5]
Where drop skips over the first N entries and returns the rest as a copy:
b
# => [1, 2, 3, 4, 5]

A single method to delete! elements from Array by predicate and return deleted

I couldn't find a method in docs to do the following:
a = [1, 2, 3]
b = a.remove! {|x| x > 1}
puts a # [1]
puts b # [2, 3]
There is a select! method that does similar thing but it doesn't accept a predicate.
To my disappointment, delete_if, keep_if, reject! and select! mutate the array but also return the same array.
Currently, I achieve a desired in 2 steps like this but maybe there are better/smarter options?
a = [1, 2, 3]
b = a.reject {|x| x > 1}
a = a - b
puts a # [1]
puts b # [2, 3]
I don't know a way to accomplish that in 1 step, however if as in your 2nd example you'd accept to not mutate a from the ruby method, you could use Enumerable#partition
a = [1, 2, 3]
# note that b is first because first array is for elements returning true
b, a = a.partition { |x| x > 1 }
puts a # [1]
puts b # [2, 3]

Ruby enumerator with chaining

I am a newbie of Ruby and I don't quite understand the following code.
n = [1,2,3]
n.map!.select{|x| x+=1}
puts n.inspect
will give me
[nil, nil, nil]
I know that n.map! will give me an enumerator but I don't get it why calling select on it will make n becomes nil. I know that map! will modify the original object, I just don't understand how that's achieved.
Update:
I tried n.map!.select {|x| true}, n is still [nil, nil, nil]
I appreciate your help! Thanks!
It's because the select method is used for selection. It returns the element that satisfies the condition you placed in your block. In your case, you didn't put any condition so it returned nil
e.g.
You want to do something like this:
n = [1,2,3]
n.select { |num| num < 3 }
#=> This should return
#=> [1,2]
In the comments, it says:
It's to do with the in-place modification of map!
that's true. Let me make it clearer here:
In the definition of Array#map!, it says:
map! {|item| block } → ary click to toggle source
map! → Enumerator
Invokes the given block once for each element of self, replacing the element with the value returned by the block
So:
Example 1
[58] pry(main)> a = [1,2,3]
=> [1, 2, 3]
[60] pry(main)> e = a.map!
=> #<Enumerator: ...>
[61] pry(main)> e.each {|n| n + 1}
=> [2, 3, 4]
[62] pry(main)> a
=> [2, 3, 4]
However:
Example 2
[71] pry(main)> a = [1,2,3]
=> [1, 2, 3]
[72] pry(main)> e = a.map!
=> #<Enumerator: ...>
[73] pry(main)> e.next
=> 1
[74] pry(main)> a
=> [1, 2, 3]
[75] pry(main)> e.peek
=> 2
[76] pry(main)> a
=> [nil, 2, 3]
I think this due to Enumerator#each call takes a block, and that would be the return value from example 1; but when Enumerator#next is called (same goes to Enumerator.to_a), it cannot evaluate the block, therefore, returning nil.

ruby set a = b, perform operation on b, but a changes value too

I am seeing some weird behavior (or maybe I just don't understand it).
When I run the following code
a = [[1,2],[3,4]]
b = a
b.each do |r|
r[0] = r[0] + 1
end
the values for both a and b have changed. Why is this? Also, how does one make it so that there is a copy of the original array as well as one of the new array? Thanks
Mutable Objects with Same Object ID
When you say b = a you are assigning the same mutable object to two variables. Consider the following:
a = [[1,2],[3,4]]
b = a
a.object_id == b.object_id
#=> true
Since both variables point to the same mutable object, making changes to the object stored by either variable changes the underlying object. Now consider the same operation with the Enumerator#dup method inherited by the Array class:
a = [[1,2],[3,4]]
b = a.dup
a.object_id == b.object_id
#=> false
Because b and a point to the same object, which in turn means that their elements are the same objects.
To make a copy, use dup. To make a deep copy, apply dup to the elements as well.
That behavior is not weird, when you do "b = a" this simply means that now b points to the same object as a. a changes the value because it's the same memory location.
You can understand it by "object_id" method.
object_id: Returns an integer identifier for obj.
2.1.2 :101 > a= [[1,2],[3,4]]
=> [[1, 2], [3, 4]]
2.1.2 :102 > a.object_id
=> 70360382470860
2.1.2 :103 > b = a
=> [[1, 2], [3, 4]]
2.1.2 :104 > b.object_id
=> 70360382470860
2.1.2 :105 >
You can use "b = a.clone"

How to remember swapped array elements in another array?

This is driving me crazy! I've been trying to write a Ruby method to find all permutations, to solve Project Euler's problem 24. When I swap the elements of an array, they are swapped properly. But when I try to STORE this swapped array in a DIFFERENT array, this new array only remembers the latest copy of my swapped array! It won't remember the older version.
When I print out a during the loop, it shows all permutations properly. But when I print out perm (which I use to store all different permutations of a), it only shows 1 version of a repeated several times. How do I fix this?
a = [0, 1, 2, 3]
perms = []
p "a = #{a}" # output: "a = [0, 1, 2, 3]"
perms << a # add a to perms array
p "perms = #{perms}" # output: "perms = [[0, 1, 2, 3]]"
a[0], a[1] = a[1], a[0] # swap 1st 2 elements of a
p "a = #{a}" # output: "a = [1, 0, 2, 3]"
perms << a # add a to perms array
p "perms = #{perms}" # "perms = [[1, 0, 2, 3], [1, 0, 2, 3]]"
a[1], a[2] = a[2], a[1] # swap 2nd 2 elements of a
p "a = #{a}" # "a = [1, 2, 0, 3]"
perms << a # add a to perms array
p "perms = #{perms}" # "perms = [[1, 2, 0, 3], [1, 2, 0, 3], [1, 2, 0, 3]]"
Thanks to Sawa below, both "dup" and "clone" methods solved my problem! Why doesn't my original way work? When would I use "dup" vs. "clone"? Please give me some code examples.
a[0], a[1] = a[1], a[0] # swap 1st 2 elements of a
p "a = #{a}" # output: "a = [1, 0, 2, 3]"
b = a.dup (or a.clone)
perms << b
p "perms = #{perms}" # "perms = [[0, 1, 2, 3], [1, 0, 2, 3]]" *** it remembers!
a[1], a[2] = a[2], a[1] # swap 2nd 2 elements of a
p "a = #{a}" # "a = [1, 2, 0, 3]"
b = a.dup (or a.clone)
perms << b
p "perms = #{perms}" # "perms = [[0, 1, 2, 3], [1, 0, 2, 3], [1, 2, 0, 3]]"
Variables in Ruby (with some exceptions, such as variables bound to integers) contain references to objects, not values. Here's an example from running "irb":
1.9.3p374 :021 > str1="hi"
=> "hi"
1.9.3p374 :022 > str2=str1
=> "hi"
1.9.3p374 :023 > str1.replace("world")
=> "world"
1.9.3p374 :024 > str2
=> "world"
You'll notice that once I replace the value for str1, str2's "value" changes as well. That's because it contains a reference to the str1 object. I know one difference between dup and clone has to do with the "freeze" method. If I had called str1.freeze, then it would prevent the object str1 references from being modified, e.g.,
1.9.3p374 :055 > str1.freeze
=> "hi"
1.9.3p374 :056 > str1[0]="b"
RuntimeError: can't modify frozen String
from (irb):56:in `[]='
from (irb):56
from /.rvm/rubies/ruby-1.9.3-p374/bin/irb:13:in `<main>
"Dup"-ing a frozen object doesn't create a frozen object whereas cloning does.
EDIT: just a slight update....When assigning an object on the right to a variable on the left (e.g., str = Object.new), the variable receives an object reference. When assigning one variable to another, the left-hand side variable receives a copy of the reference that the variable on the right contains. In either case, you are still storing object references in the left-hand side variable.
Your original didn't work because you kept modifying the same array instance a.
Take a dup of the original array each time before you modify it into a different array. Or, create a new instance of Array by not relying on a destructive method.
a = original_array
b = a.dup
... # do some modifications to `b`
perms << b
c = a.dup
... # do some modifications to `c`
perms << c
...
If you don't like reinventing the wheel, you can use the facets gem.
gem install facets
https://github.com/rubyworks/facets/blob/d96ec0d700d1d7180ccbb5452e0a926386ec0b32/lib/backport/facets/array/permutation.rb
require 'facets'
[1, 2, 3].permutation
#=> [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

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