I try to programm snakes in Ruby. In order to get myself more familiar with Ruby. I define the position of every part of the snake through saving its X and Y value in two 1D arrays one for a X value and one for a Y value.
$x = [2,...]
$y = [2,...]
(What I forgot to tell is that the head of the Snake moves through user input while the rest just inherits its position from the part before like this.)
def length(f)
if $length >= f
$y[f] = $y[f-1]
$x[f] = $x[f-1]
end
end
In order to get a field for the Snake to move around I programmed this.
for a in (1..20)
for b in (1..20)
print " X "
end
puts" "
end
Which gives me a 20*20 field.
I then tried to display every part of the snake like on the field like this.(While also drawing a boarder around the field.)
for a in (1..20)
for b in (1..20)
if a == 1 || a == 20
if b == 1 || b == 20
print " + "
else
print " - "
end
elsif b == 1 || b == 20
print " | "
elsif a == $x[0] && b == $y[0]
body
elsif a == $x[1] && b == $y[1]
body
elsif a == $x[2] && b == $y[2]
body
elsif a == $x[3] && b == $y[3]
body
elsif a == $x[4] && b == $y[4]
body
else
print " "
end
end
puts""
end
This works but if the user is really good/ has a lot of spare time I need to make allot of elsif for every one represents a part of the snake if the snake should have as a limit a length of 100 I would need to make 100 elsif statements.(The body is just:
def body
print " # ".green
end
)
I tried fixing it with a for loop like this:
for c in (1..100)
if a == $x[c] && b == $y[c]
body
end
end
and this
loop do
$x.size.times do |index|
if $x[index] == a && $y[index] == b
body
end
end
break
end
But sadly this didn't gave the desired outcome for this interfered with the ifthat draws the boarders of the field.
Is there a way to combine these multiple elsif statements?
Every help would be highly appreciated. ( Sorry for being to vague in the first draft.)
Recommended Refactorings
NB: You included no sample data in your original post, so your mileage with answers will vary.
You have a number of issues, not just one. Besides not being DRY, your code is also not very testable because it's not broken out into discrete operations. There are a number of things you can (and probably should) do:
Break your "body" stuff into discrete methods.
Use Array or Enumerator methods to simplify the data.
Use dynamic methods to loop over your arrays, rather than fixed ranges or for-loops.
Use case/when statements inside your loop to handle multiple conditionals for the same variable.
In short, you need to refactor your code to be more modular, and to leverage the language to iterate over your data objects rather than using one conditional per element as you're currently doing.
Simplify Your Data Set and Handle Procedurally
As an example, consider the following:
def handle_matched_values array
end
def handle_mismatched_values array
end
paired_array = a.zip b
matched_pairs = paired_array.select { |subarray| subarray[0] == subarray[1] }
unmatched_pairs = paired_array.reject { |subarray| subarray[0] == subarray[1] }
matched_pairs.each { |pair| handle_matched_values pair }
matched_pairs.each { |pair| handle_mismatched_values pair }
In this example, you may not even need an if statement. Instead, you could use Array#select or Array#reject to find indices that match whatever criteria you want, and then call the relevant handler for the results. This has the advantage of being very procedural, and makes it quite clear what data set and handler are being paired. It's also quite readable, which is extremely important.
Dynamic Looping and Case Statements
If you truly need to handle your data within a single loop, use a case statement to clean up your conditions. For example:
# Extract methods to handle each case.
def do_something_with data; end
def do_something_else_with data; end
def handle_borders data; end
# Construct your data any way you want.
paired_array = a.zip b
# Loop over your data to evaluate each pair of values.
paired_array.each do |pair|
case pair
when a == b
do_something_with pair
when a == paired_array.first || paired_array.last
handle_borders pair
else
do_something_else_with pair
end
end
There are certainly plenty of other ways to work pairwise with a large data set. The goal is to give you a basic structure for refactoring your code. The rest is up to you!
I would start with something like this:
(1..20).each do |a|
(1..20).each do |b|
if [1, 20].include?(a)
print([1, 20].include?(b) ? ' + ' : ' - ')
elsif (1..100).any? { |i| a == $x[i] && b == $y[i] }
body
else
print(' ')
end
puts('')
end
end
I suppose this would work as a solution even if it is not that advanced?
loop do
$x.size.times do |index|
if $x[index] == a && $y[index] == b
body
end
end
break
end
This question already has answers here:
Ruby method for +=
(2 answers)
Closed 9 years ago.
Consider
class Container
def initialize(value = 0)
#value = value
end
def + (other)
return #value + other
end
def - (other)
return #value - other
end
def * (other)
return #value * other
end
def / (other)
return #value / other
end
def get
return #value
end
end
I would like to use += to increase the value in the container, like this:
c = Container.new(100)
c += 100
print c.get # Expecting 200
The above won't work, as 100 will overwrite c.
I know I could use something like an attr_accessor to generate a getter and setter for the value, but I'm curious if I could do this in a prettier way such as using +=.
Since c += 100 is just a sugar for c = c + 100, you can't escape overwriting c. BUT you can overwrite it with a similar object (and not with fixnum, as in your question).
class Container
def initialize(value = 0)
#value = value
end
def + (other)
Container.new(#value + other)
end
def get
#value
end
end
c = Container.new(100)
c += 100
c.get # => 200
x += y is just syntactic sugar for x = x + y. So you only have to implement + in your class and you get += for free.
No, you can't overload +=. See list of ruby operators that can be overridden/implemented for the full list of overloadable operators.
In Ruby x += y always means x = x + y. The only way to change the meaning of += for a given x is overriding + in x.class. However, + has a different semantics, and the user most probably expects that + returns a new object. If you make + return the original x, that may confuse some of your users. If you make + return a different object, then x would point to that other object in your example, and as far as I understand your question you don't want that.
Hey I have a problem with my simulation.
I am a Ruby-Starter and don't know what's wrong in my code. That is only the part with the simulation:
def mean
mean = self.reduce(:+)/self.length.to_f
return mean
end
def randn
begin
rg1 = (rand*2)-1
rg2 = (rand*2)-1
q = rg1**2 + rg2**2
end while (q == 0 || q > 1)
p = Math.sqrt((-2*Math.log(q))/q)
rn1 = rg1 * p
rn2 = rg2 * p
return rn1, rn2
end
monte_carlo = 10
ren1_sim = Array.new
ren2_sim = Array.new
monte_carlo.times {
(1..20).each{ |i|
(1..250).each { |j|
r = randn()
ren1= * Math.exp(mu_ren1 + sigma_ren1 * r[0])
# ren1 is an array with prices, mu_ren1 and sigma_ren1 are individual values
ren2= * Math.exp(mu_ren2 + chol_21 * r[0] + chol_22 * r[1])
# chol_21 and chol_22 are also individual values
ren1_sim.push(ren1)
ren2_sim.push(ren2)
}
}
}
puts ren1_sim.mean
puts ren2_sim.mean
I don't get an error without the last two puts, but when I want to calculate the average of the arrays ren1_sim and rent_sim I get the error:
undefined method 'mean' for #<Array:0x2acf318> (NoMethodError)
Do you know how to fix that?
You're trying to invoke mean on an Array, which is not a method of Array. Perhaps you meant to use Statsample::Vector, which is Statsample's extension of Array, and does have mean?
ren1_sim = Statsample::Vector.new
ren2_sim = Statsample::Vector.new
You can also call to_vector on an Array instance to get a Statsample::Vector.
You've defined a mean method at the top of your file, but that just creates a method on the top level object, and you're trying to call it on an individual array. You could either change that code to
def mean(array)
array.reduce(:+)/array.length.to_f
end
and then change your usage of it later on to mean(ren1_sim)
or change your code so that you are adding the method to array, i.e.
class Array
def mean
self.reduce(:+)/self.length.to_f
end
end
have a look at this post to calculate the average of a array
How do I create an average from a Ruby array?
What does the following code mean in Ruby?
||=
Does it have any meaning or reason for the syntax?
a ||= b is a conditional assignment operator. It means:
if a is undefined or falsey, then evaluate b and set a to the result.
Otherwise (if a is defined and evaluates to truthy), then b is not evaluated, and no assignment takes place.
For example:
a ||= nil # => nil
a ||= 0 # => 0
a ||= 2 # => 0
foo = false # => false
foo ||= true # => true
foo ||= false # => true
Confusingly, it looks similar to other assignment operators (such as +=), but behaves differently.
a += b translates to a = a + b
a ||= b roughly translates to a || a = b
It is a near-shorthand for a || a = b. The difference is that, when a is undefined, a || a = b would raise NameError, whereas a ||= b sets a to b. This distinction is unimportant if a and b are both local variables, but is significant if either is a getter/setter method of a class.
Further reading:
http://www.rubyinside.com/what-rubys-double-pipe-or-equals-really-does-5488.html
This question has been discussed so often on the Ruby mailing-lists and Ruby blogs that there are now even threads on the Ruby mailing-list whose only purpose is to collect links to all the other threads on the Ruby mailing-list that discuss this issue.
Here's one: The definitive list of ||= (OR Equal) threads and pages
If you really want to know what is going on, take a look at Section 11.4.2.3 "Abbreviated assignments" of the Ruby Language Draft Specification.
As a first approximation,
a ||= b
is equivalent to
a || a = b
and not equivalent to
a = a || b
However, that is only a first approximation, especially if a is undefined. The semantics also differ depending on whether it is a simple variable assignment, a method assignment or an indexing assignment:
a ||= b
a.c ||= b
a[c] ||= b
are all treated differently.
Concise and complete answer
a ||= b
evaluates the same way as each of the following lines
a || a = b
a ? a : a = b
if a then a else a = b end
-
On the other hand,
a = a || b
evaluates the same way as each of the following lines
a = a ? a : b
if a then a = a else a = b end
-
Edit: As AJedi32 pointed out in the comments, this only holds true if: 1. a is a defined variable. 2. Evaluating a one time and two times does not result in a difference in program or system state.
In short, a||=b means: If a is undefined, nil or false, assign b to a. Otherwise, keep a intact.
Basically,
x ||= y means
if x has any value leave it alone and do not change the value, otherwise
set x to y
It means or-equals to. It checks to see if the value on the left is defined, then use that. If it's not, use the value on the right. You can use it in Rails to cache instance variables in models.
A quick Rails-based example, where we create a function to fetch the currently logged in user:
class User > ActiveRecord::Base
def current_user
#current_user ||= User.find_by_id(session[:user_id])
end
end
It checks to see if the #current_user instance variable is set. If it is, it will return it, thereby saving a database call. If it's not set however, we make the call and then set the #current_user variable to that. It's a really simple caching technique but is great for when you're fetching the same instance variable across the application multiple times.
To be precise, a ||= b means "if a is undefined or falsy (false or nil), set a to b and evaluate to (i.e. return) b, otherwise evaluate to a".
Others often try to illustrate this by saying that a ||= b is equivalent to a || a = b or a = a || b. These equivalencies can be helpful for understanding the concept, but be aware that they are not accurate under all conditions. Allow me to explain:
a ||= b ⇔ a || a = b?
The behavior of these statements differs when a is an undefined local variable. In that case, a ||= b will set a to b (and evaluate to b), whereas a || a = b will raise NameError: undefined local variable or method 'a' for main:Object.
a ||= b ⇔ a = a || b?
The equivalency of these statements are often assumed, since a similar equivalence is true for other abbreviated assignment operators (i.e. +=,-=,*=,/=,%=,**=,&=,|=,^=,<<=, and >>=). However, for ||= the behavior of these statements may differ when a= is a method on an object and a is truthy. In that case, a ||= b will do nothing (other than evaluate to a), whereas a = a || b will call a=(a) on a's receiver. As others have pointed out, this can make a difference when calling a=a has side effects, such as adding keys to a hash.
a ||= b ⇔ a = b unless a??
The behavior of these statements differs only in what they evaluate to when a is truthy. In that case, a = b unless a will evaluate to nil (though a will still not be set, as expected), whereas a ||= b will evaluate to a.
a ||= b ⇔ defined?(a) ? (a || a = b) : (a = b)????
Still no. These statements can differ when a method_missing method exists which returns a truthy value for a. In this case, a ||= b will evaluate to whatever method_missing returns, and not attempt to set a, whereas defined?(a) ? (a || a = b) : (a = b) will set a to b and evaluate to b.
Okay, okay, so what is a ||= b equivalent to? Is there a way to express this in Ruby?
Well, assuming that I'm not overlooking anything, I believe a ||= b is functionally equivalent to... (drumroll)
begin
a = nil if false
a || a = b
end
Hold on! Isn't that just the first example with a noop before it? Well, not quite. Remember how I said before that a ||= b is only not equivalent to a || a = b when a is an undefined local variable? Well, a = nil if false ensures that a is never undefined, even though that line is never executed. Local variables in Ruby are lexically scoped.
If X does NOT have a value, it will be assigned the value of Y. Else, it will preserve it's original value, 5 in this example:
irb(main):020:0> x = 5
=> 5
irb(main):021:0> y = 10
=> 10
irb(main):022:0> x ||= y
=> 5
# Now set x to nil.
irb(main):025:0> x = nil
=> nil
irb(main):026:0> x ||= y
=> 10
x ||= y
is
x || x = y
"if x is false or undefined, then x point to y"
||= is a conditional assignment operator
x ||= y
is equivalent to
x = x || y
or alternatively
if defined?(x) and x
x = x
else
x = y
end
unless x
x = y
end
unless x has a value (it's not nil or false), set it equal to y
is equivalent to
x ||= y
Suppose a = 2 and b = 3
THEN, a ||= b will be resulted to a's value i.e. 2.
As when a evaluates to some value not resulted to false or nil.. That's why it ll not evaluate b's value.
Now Suppose a = nil and b = 3.
Then a ||= b will be resulted to 3 i.e. b's value.
As it first try to evaluates a's value which resulted to nil.. so it evaluated b's value.
The best example used in ror app is :
#To get currently logged in iser
def current_user
#current_user ||= User.find_by_id(session[:user_id])
end
# Make current_user available in templates as a helper
helper_method :current_user
Where, User.find_by_id(session[:user_id]) is fired if and only if #current_user is not initialized before.
||= is called a conditional assignment operator.
It basically works as = but with the exception that if a variable has already been assigned it will do nothing.
First example:
x ||= 10
Second example:
x = 20
x ||= 10
In the first example x is now equal to 10. However, in the second example x is already defined as 20. So the conditional operator has no effect. x is still 20 after running x ||= 10.
a ||= b
Signifies if any value is present in 'a' and you dont want to alter it the keep using that value, else if 'a' doesnt have any value, use value of 'b'.
Simple words, if left hand side if not null, point to existing value, else point to value at right side.
a ||= b
is equivalent to
a || a = b
and not
a = a || b
because of the situation where you define a hash with a default (the hash will return the default for any undefined keys)
a = Hash.new(true) #Which is: {}
if you use:
a[10] ||= 10 #same as a[10] || a[10] = 10
a is still:
{}
but when you write it like so:
a[10] = a[10] || 10
a becomes:
{10 => true}
because you've assigned the value of itself at key 10, which defaults to true, so now the hash is defined for the key 10, rather than never performing the assignment in the first place.
It's like lazy instantiation.
If the variable is already defined it will take that value instead of creating the value again.
Please also remember that ||= isn't an atomic operation and so, it isn't thread safe. As rule of thumb, don't use it for class methods.
This is the default assignment notation
for example: x ||= 1
this will check to see if x is nil or not. If x is indeed nil it will then assign it that new value (1 in our example)
more explicit:
if x == nil
x = 1
end
b = 5
a ||= b
This translates to:
a = a || b
which will be
a = nil || 5
so finally
a = 5
Now if you call this again:
a ||= b
a = a || b
a = 5 || 5
a = 5
b = 6
Now if you call this again:
a ||= b
a = a || b
a = 5 || 6
a = 5
If you observe, b value will not be assigned to a. a will still have 5.
Its a Memoization Pattern that is being used in Ruby to speed up accessors.
def users
#users ||= User.all
end
This basically translates to:
#users = #users || User.all
So you will make a call to database for the first time you call this method.
Future calls to this method will just return the value of #users instance variable.
As a common misconception, a ||= b is not equivalent to a = a || b, but it behaves like a || a = b.
But here comes a tricky case. If a is not defined, a || a = 42 raises NameError, while a ||= 42 returns 42. So, they don't seem to be equivalent expressions.
irb(main):001:0> a = 1
=> 1
irb(main):002:0> a ||= 2
=> 1
Because a was already set to 1
irb(main):003:0> a = nil
=> nil
irb(main):004:0> a ||= 2
=> 2
Because a was nil
This ruby-lang syntax. The correct answer is to check the ruby-lang documentation. All other explanations obfuscate.
Google
"ruby-lang docs Abbreviated Assignment".
Ruby-lang docs
https://docs.ruby-lang.org/en/2.4.0/syntax/assignment_rdoc.html#label-Abbreviated+Assignment
a ||= b is the same as saying a = b if a.nil? or a = b unless a
But do all 3 options show the same performance? With Ruby 2.5.1 this
1000000.times do
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
end
takes 0.099 Seconds on my PC, while
1000000.times do
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
end
takes 0.062 Seconds. That's almost 40% faster.
and then we also have:
1000000.times do
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
end
which takes 0.166 Seconds.
Not that this will make a significant performance impact in general, but if you do need that last bit of optimization, then consider this result.
By the way: a = 1 unless a is easier to read for the novice, it is self-explanatory.
Note 1: reason for repeating the assignment line multiple times is to reduce the overhead of the loop on the time measured.
Note 2: The results are similar if I do a=nil nil before each assignment.