I have a model, FacebookProfile, that has three columns, a, b and c.
I invoke this object by doing:
#facebook_profile = FacebookProfile.find_by_facebook_id(1)
Then I can access each of the values for the columns a, b and c
#facebook_profile.a # result: 1
#facebook_profile.b # result: 2
#facebook_profile.c # result: 3
If I have an array of items a, b and c, how can I iterate through calling these on #facebook_profile?
My best guess:
my_array = ["a", "b", "c"]
for i in my_array
puts #facebook_profile.i
end
Expected output: 1, 2, 3
I have seen a way to iterate through all columns, but I have other columns besides a, b and c, so I don't think that will work.
I think you could use the sendoperator to do something along the lines of
my_array = ["a","b","c"]
for i in my_array
puts #facebook_profile.send(i.to_sym)
end
Related
After about a year of Ruby, I just saw this somewhere and my mind is blown. Why in the world does this work?
>> words = ['uno', 'dos']
=> ["uno", "dos"]
>> first, second = words
=> ["uno", "dos"]
>> first
=> "uno"
>> second
=> "dos"
Specifically, how does this work:
>> first, second = ['uno', 'dos']
Why can I do this? It makes no syntactical sense!
It makes no syntactical sense
But this is part of Ruby's syntax! In the Ruby docs it is known as array decomposition:
Like Array decomposition in method arguments you can decompose an
Array during assignment using parenthesis:
(a, b) = [1, 2]
p a: a, b: b # prints {:a=>1, :b=>2}
You can decompose an Array as part of a larger multiple assignment:
a, (b, c) = 1, [2, 3]
p a: a, b: b, c: c # prints {:a=>1, :b=>2, :c=>3}
Since each decomposition is considered its own multiple assignment you
can use * to gather arguments in the decomposition:
a, (b, *c), *d = 1, [2, 3, 4], 5, 6
p a: a, b: b, c: c, d: d
# prints {:a=>1, :b=>2, :c=>[3, 4], :d=>[5, 6]}
Edit
as Stefan points out in the comments, the docs don't mention that array decomposition also occurs implicitly (i.e. without parenthesis) if there is only one value on the right-hand side:
a, b = [1, 2] works like (a, b) = [1, 2]
Why can I do this? It makes no syntactical sense!
It makes a perfect sense. It is an example of parallel assignment.
When you use = what is happening is each of the list of variables on the left of = are assigned to each of the list of expressions on the right of =.
first, second = ['uno', 'dos']
# is equivalent to
first, second = 'uno', 'dos'
If there are more variables on the left, than expressions on the right, those left variables are assigned with nil:
first, second = 'uno'
first #=> 'uno'
second #=> nil
As to
words = ['uno', 'dos']
first, second = words
first #=> 'uno'
second #=> 'dos'
It is not assigning the whole words array to first leaving second with nil, because while parallel assignment Ruby tries to decompose the right side expression, and does so if it is an instance of Array.
[TIL] Moreover, it attempts to call to_ary on the right side expression, and if it responds to the method, decomposes accordingly to that object's to_ary implementation (credits to #Stefan):
string = 'hello world'
def string.to_ary; split end
first, second = string
first #=> 'hello'
second #=> 'world'
This is called multiple assignment, handy to assign multiple variables at once.
example
one, two = 1,2
puts one #=>1
puts two #=>2
one, two = [1,2] # this makes sense
one, two = 1 # obviously this doesn't it will assign nil to two
Hope its bit clear now
Whenever I swap values in an array, I make sure I stored one of the values in a reference variable. But I found that Ruby can return two values as well as automatically swap two values. For example,
array = [1, 3, 5 , 6 ,7]
array[0], array[1] = array[1] , array[0] #=> [3, 1]
I was wondering how Ruby does this.
Unlike other languages, the return value of any method call in Ruby is always an object. This is possible because, like everything in Ruby, nil itself is an object.
There's three basic patterns you'll see. Returning no particular value:
def nothing
end
nothing
# => nil
Returning a singular value:
def single
1
end
x = single
# => 1
This is in line with what you'd expect from other programming languages.
Things get a bit different when dealing with multiple return values. These need to be specified explicitly:
def multiple
return 1, 2
end
x = multiple
# => [ 1, 2 ]
x
# => [ 1, 2 ]
When making a call that returns multiple values, you can break them out into independent variables:
x, y = multiple
# => [ 1, 2 ]
x
# => 1
y
# => 2
This strategy also works for the sorts of substitution you're talking about:
a, b = 1, 2
# => [1, 2]
a, b = b, a
# => [2, 1]
a
# => 2
b
# => 1
No, Ruby doesn't actually support returning two objects. (BTW: you return objects, not variables. More precisely, you return pointers to objects.)
It does, however, support parallel assignment. If you have more than one object on the right-hand side of an assignment, the objects are collected into an Array:
foo = 1, 2, 3
# is the same as
foo = [1, 2, 3]
If you have more than one "target" (variable or setter method) on the left-hand side of an assignment, the variables get bound to elements of an Array on the right-hand side:
a, b, c = ary
# is the same as
a = ary[0]
b = ary[1]
c = ary[2]
If the right-hand side is not an Array, it will be converted to one using the to_ary method
a, b, c = not_an_ary
# is the same as
ary = not_an_ary.to_ary
a = ary[0]
b = ary[1]
c = ary[2]
And if we put the two together, we get that
a, b, c = d, e, f
# is the same as
ary = [d, e, f]
a = ary[0]
b = ary[1]
c = ary[2]
Related to this is the splat operator on the left-hand side of an assignment. It means "take all the left-over elements of the Array on the right-hand side":
a, b, *c = ary
# is the same as
a = ary[0]
b = ary[1]
c = ary.drop(2) # i.e. the rest of the Array
And last but not least, parallel assignments can be nested using parentheses:
a, (b, c), d = ary
# is the same as
a = ary[0]
b, c = ary[1]
d = ary[2]
# which is the same as
a = ary[0]
b = ary[1][0]
c = ary[1][1]
d = ary[2]
When you return from a method or next or break from a block, Ruby will treat this kind-of like the right-hand side of an assignment, so
return 1, 2
next 1, 2
break 1, 2
# is the same as
return [1, 2]
next [1, 2]
break [1, 2]
By the way, this also works in parameter lists of methods and blocks (with methods being more strict and blocks less strict):
def foo(a, (b, c), d) p a, b, c, d end
bar {|a, (b, c), d| p a, b, c, d }
Blocks being "less strict" is for example what makes Hash#each work. It actually yields a single two-element Array of key and value to the block, but we usually write
some_hash.each {|k, v| }
instead of
some_hash.each {|(k, v)| }
tadman and Jörg W Mittag know Ruby better than me, and their answers are not wrong, but I don't think they are answering what OP wanted to know. I think that the question was not clear though. In my understanding, what OP wanted to ask has nothing to do with returning multiple values.
The real question is, when you want to switch the values of two variables a and b (or two positions in an array as in the original question), why is it not necessary to use a temporal variable temp like:
a, b = :foo, :bar
temp = a
a = b
b = temp
but can be done directly like:
a, b = :foo, :bar
a, b = b, a
The answer is that in multiple assignment, the whole right hand side is evaluated prior to assignment of the whole left hand side, and it is not done one by one. So a, b = b, a is not equivalent to a = b; b = a.
First evaluating the whole right hand side before assignment is a necessity that follows from adjustment when the both sides of = have different numbers of terms, and Jörg W Mittag's description may be indirectly related to that, but that is not the main issue.
Arrays are a good option if you have only a few values. If you want multiple return values without having to know (and be confused by) the order of results, an alternative would be to return a Hash that contains whatever named values you want.
e.g.
def make_hash
x = 1
y = 2
{x: x, y: y}
end
hash = make_hash
# => {:x=>1, :y=>2}
hash[:x]
# => 1
hash[:y]
# => 2
Creating a hash as suggested by some is definitely better than array as array indexing can be confusing. When an additional attribute needs to be returned at a certain index, we'll need to make changes to all the places where the return value is used with array.
Another better way to do this is by using OpenStruct. Its advantage over using a hash is its ease of accessibility.
Example: computer = OpenStruct.new(ram: '4GB')
there are multiple ways to access the value of ram
as a symbol key: computer[:ram]
as a string key: computer['ram']
as an attribute(accessor method): computer.ram
Reference Article: https://medium.com/rubycademy/openstruct-in-ruby-ab6ba3aff9a4
I want to know how I can declare multiple variables. I typed a,b=1 expecting to get a=1,b=1, but I got:
a,b=1
a #=> 1
b #=> nil
How am I able to do this?
After this code, I did:
a="Hello "
b=a
c="World~"
b << c
b #=> "Hello World"
Why is b the same as a's value?
To declare multiple vars on the same line, you can do that:
a = b = "foo"
puts a # return "foo"
puts b # return "foo" too
About your second question, when doing b << c, you are assigning c's value to b. Then, you are overriding previous value stored in b. Meanwhile, a keeps the same value because Ruby does not user pointers.
What you are doing is called destructuring assignment. Basically, you take what is on the right side of the equals sign, and destructure it, or break it apart, and then assign each section to each corresponding variable on the left.
Ruby is super friendly, and is providing some syntactic sugar that might be confusing.
When you type this:
a, b = 1
You are really saying something closer to this:
[a, b] = [1, nil]
A good example of destructuring assignment can be found here. It's for JavaScript, but I like it because the syntax is very explicit about what is happen when you do such an assigment.
I suppose, in the case of
a, b, c = 1, 2
the runtime system works the following way:
a, b, c = [1, 2]
_result = ( a, b, c = (_values = [1, 2]) )
a = _values[0] # => 1
b = _values[1] # => 2
c = _values[2] # => nil
_result = _values # => [1, 2]
However, in the case of a single value on the right hand side: a, b = 1, the computation process looks a bit different:
_result = ( a, b = ( _value = (_values = [1]).first ) )
a = _values[0] # => 1
b = _values[1] # => nil
_result = _value # => 1
Can someone approve or disprove my assumption?
Given this array: a = ["a", "b", "c"]
If I run a.cycle { |x| puts x } I get print, a, b, c, a, b, c,.. forever.
Is there a way to set the starting point so that it would begin with "b" or the second index like this: print, b, c, a, b, c, a,.. forever. ?
Use Array#rotate
a.rotate.cycle {|x| puts x}
Pass an argument to rotate to shift to whichever index you want.
Is there a Ruby method similar to Haskell's cycle? Haskell's cycle takes a list and returns that list infinitely appended to itself. It's commonly used with take which grabs a certain number of elements off the top of an array. Is there a Ruby method that takes an array and returns the array appended to itself some n number of times?
Yes, it's called cycle. From the documentation:
Array.cycle
(from ruby core)
------------------------------------------------------------------------------
ary.cycle(n=nil) {|obj| block } -> nil
ary.cycle(n=nil) -> an_enumerator
------------------------------------------------------------------------------
Calls block for each element repeatedly n times or forever if none
or nil is given. If a non-positive number is given or the array is empty, does
nothing. Returns nil if the loop has finished without getting interrupted.
If no block is given, an enumerator is returned instead.
a = ["a", "b", "c"]
a.cycle {|x| puts x } # print, a, b, c, a, b, c,.. forever.
a.cycle(2) {|x| puts x } # print, a, b, c, a, b, c.
Edit:
It seems like whats inside the block is basically a "Lambda", and as far as I know, I can't make a lambda concat each element onto an existing array.
b = [1, 2, 3]
z = []
b.cycle(2) { |i| z << i }
z # => [1, 2, 3, 1, 2, 3]
You can multiply an array by an integer using Array#*:
ary * int → new_ary
[...] Otherwise, returns a new array built by concatenating the int copies of self.
So you can do things like this:
>> [1, 2] * 3
=> [1, 2, 1, 2, 1, 2]