if j.job_type_name
j.type = j.job_type_name
elsif j.type.length > 1
# jty = Jobtype.find_by_id(j.type)
# We don't want to relentlessly hit the db for JobTypes, Colors, Priorities, etc. so we use the stored 'all' variable
jty = allJobTypes.find { |h| h['_id'] == BSON::ObjectId(j.type) }
j.type = jty && jty.job_type ? jty.job_type : 'N/A'
end
if j.priority_name
j.priority = j.priority_name
elsif j.priority
pri = Jobpriority.find_by_id(j.priority)
j.priority = !pri.blank? && !pri.job_priority.blank? ? pri.job_priority : 'N/A'
end
I have these two if else blocks and I think they are better off as Procs, but I'm still wrapping my head around what a Proc is. It's just a piece of code that I want to call often, and put in one place so that I only have to update it in that one place. It's not a method on an object per-se, but just some repetitious code.
So what's the syntax for the above?
The code that you don't see (where 'j' is set) is an .each loop over a database cursor.
So Job.where(stuff).each do |j|
etc.
The syntax for making a proc is just:
my_proc = proc do |arglist|
# code here
end
You can also use Proc.new instead of proc if you want.
You should be able to make a proc and then pass it in as the block to .each, (but I don't see what this would get you):
def process_jobs(stuff)
process_job = proc do |j|
if j.job_type_name
#...
end
end
Job.where(stuff).each(&process_job)
end
Related
When using an accumulator, does the accumulator exist only within the reduce block or does it exist within the function?
I have a method that looks like:
def my_useless_function(str)
crazy_letters = ['a','s','d','f','g','h']
str.split.reduce([]) do |new_array, letter|
for a in 0..crazy_letters.length-1
if letter == crazy_letters[a]
new_array << letter
end
end
end
return true if (new_array == new_array.sort)
end
When I execute this code I get the error
"undefined variable new_array in line 11 (the return statement)"
I also tried assigning the new_array value to another variable as an else statement inside my reduce block but that gave me the same results.
Can someone explain to me why this is happening?
The problem is that new_array is created during the call to reduce, and then the reference is lost afterwards. Local variables in Ruby are scoped to the block they are in. The array can be returned from reduce in your case, so you could use it there. However, you need to fix a couple things:
str.split does not break a string into characters in Ruby 2+. You should use str.chars, or str.split('').
The object retained for each new iteration of reduce must be retained by returning it from the block each time. The simplest way to do this is to put new_array as the last expression in your block.
Thus:
def my_useless_function(str)
crazy_letters = ['a','s','d','f','g','h']
crazy_only = str.split('').reduce([]) do |new_array, letter|
for a in 0..crazy_letters.length-1
if letter == crazy_letters[a]
new_array << letter
end
end
new_array
end
return true if (crazy_only == crazy_only.sort)
end
Note that your function is not very efficient, and not very idiomatic. Here's a shorter version of the function that is more idiomatic, but not much more efficient:
def my_useless_function(str)
crazy_letters = %w[a s d f g h]
crazy_only = str.chars.select{ |c| crazy_letters.include?(c) }
crazy_only == crazy_only.sort # evaluates to true or false
end
And here's a version that's more efficient:
def efficient_useless(str)
crazy_only = str.scan(/[asdfgh]/) # use regex to search for the letters you want
crazy_only == crazy_only.sort
end
Block local variables
new_array doesn't exist outside the block of your reduce call. It's a "block local variable".
reduce does return an object, though, and you should use it inside your method.
sum = [1, 2, 3].reduce(0){ |acc, elem| acc + elem }
puts sum
# 6
puts acc
# undefined local variable or method `acc' for main:Object (NameError)
Your code
Here's the least amount of change for your method :
def my_useless_function(str)
crazy_letters = ['a','s','d','f','g','h']
new_array = str.split(//).reduce([]) do |new_array, letter|
for a in 0..crazy_letters.length-1
if letter == crazy_letters[a]
new_array << letter
end
end
new_array
end
return true if (new_array == new_array.sort)
end
Notes:
return isn't needed at the end.
true if ... isn't needed either
for loop should never be used in Ruby
reduce returns the result of the last expression inside the block. It was for in your code.
If you always need to return the same object in reduce, it might be a sign you could use each_with_object.
"test".split is just ["test"]
String and Enumerable have methods that could help you. Using them, you could write a much cleaner and more efficient method, as in #Phrogz answer.
Here's some code:
i = 0
collection = []
loop do
i += 1
break if complicated_predicate_of(i)
collection << i
end
collection
I don't know in advance how many times I'll need to iterate; that depends on complicated_predicate_of(i). I could do something like 0.upto(Float::INFINITY).times.collect do ... end but that's pretty ugly.
I'd like to do this:
i = 0
collection = loop.collect do
i += 1
break if complicated_predicate_of(i)
i
end
But, though it's not a syntax error for some reason, loop.collect doesn't seem to collect anything. (Neither does loop.reduce). collection is nil at the end of the statement.
In other words, I want to collect the values of a loop statement without an explicit iterator. Is there some way to achieve this?
You could write
collection = 1.step.take_while do |i|
i <= 3 # this block needs to return *false* to stop the taking
end
Whatever solution you choose in the end, remember that you can always opt to introduce a helper method with a self-explanatory name. Especially if you need to collect numbers like this in many places in your source code.
Say you wanted to hide the intricate bowels of your solution above, then this could be your helper method:
def numbers_until(&block)
i = 0
collection = []
loop do
i += 1
break if yield i
collection << i
end
collection
end
collection = numbers_until do |i|
i > 3 # this block needs to return *true* to stop the taking
end
You could write
def complicated_predicate_of(i)
i > 3
end
1.step.with_object([]) { |i,collection| complicated_predicate_of(i) ?
(break collection) : collection << i }
#=> [1, 2, 3]
I have the following code
# colours a random cell with a correct colour
def colour_random!
while true do
col, row = rand(columns), rand(rows)
cell = self[row,col]
if cell.empty? then
cell.should_be_filled? ? cell.colour!(1) : cell.colour!(0)
break
end
end
end
it's not that important what's doing, although it should pretty obvious. The point is that Rubocop gives me a warning
Never use 'do' with multi-line 'while
Why should I not do that? How should I do it then?
while is a keyword,so you don't need to pass a block. Without do..end it will work fine. The below is fine
def colour_random!
while true
col, row = rand(columns), rand(rows)
cell = self[row,col]
if cell.empty? then
cell.should_be_filled? ? cell.colour!(1) : cell.colour!(0)
break
end
end
end
while is a keyword, and if you pass a block to it, like do..end, it still works as you asked it to do, by not throwing any error, rather just a warning. But it could be dangerous if you try to pass a Proc or Method object to it, and dynamically try to convert it to a block using & keyword, as we do generally. That means
# below code will work as expected just throwing an warning.
x = 2
while x < 2 do
#code
end
But if you try to do by mistake like below
while &block # booom!! error
The reason is while is a keyword, which don't support any to_proc method to satisfy your need. So it can be dangerous.
Ruby style guide also suggested that Never use while/until condition do for multi-line while/until
I think the reason is as Nobuyoshi Nakada said in the mailing list
loop is a kernel method which takes a block. A block introduces new local variable scope.
loop do
a = 1
break
end
p a #=> causes NameError
while doesn't.
while 1
a = 1
break
end
p a #=> 1
Ruby actually has a shortcut for while true: the loop statement.
def colour_random!
loop do
col, row = rand(columns), rand(rows)
cell = self[row,col]
if cell.empty? then
cell.should_be_filled? ? cell.colour!(1) : cell.colour!(0)
break
end
end
end
Can anyone help me to figure out the the use of yield and return in Ruby. I'm a Ruby beginner, so simple examples are highly appreciated.
Thank you in advance!
The return statement works the same way that it works on other similar programming languages, it just returns from the method it is used on.
You can skip the call to return, since all methods in ruby always return the last statement. So you might find method like this:
def method
"hey there"
end
That's actually the same as doing something like:
def method
return "hey there"
end
The yield on the other hand, excecutes the block given as a parameter to the method. So you can have a method like this:
def method
puts "do somthing..."
yield
end
And then use it like this:
method do
puts "doing something"
end
The result of that, would be printing on screen the following 2 lines:
"do somthing..."
"doing something"
Hope that clears it up a bit. For more info on blocks, you can check out this link.
yield is used to call the block associated with the method. You do this by placing the block (basically just code in curly braces) after the method and its parameters, like so:
[1, 2, 3].each {|elem| puts elem}
return exits from the current method, and uses its "argument" as the return value, like so:
def hello
return :hello if some_test
puts "If it some_test returns false, then this message will be printed."
end
But note that you don't have to use the return keyword in any methods; Ruby will return the last statement evaluated if it encounters no returns. Thus these two are equivelent:
def explicit_return
# ...
return true
end
def implicit_return
# ...
true
end
Here's an example for yield:
# A simple iterator that operates on an array
def each_in(ary)
i = 0
until i >= ary.size
# Calls the block associated with this method and sends the arguments as block parameters.
# Automatically raises LocalJumpError if there is no block, so to make it safe, you can use block_given?
yield(ary[i])
i += 1
end
end
# Reverses an array
result = [] # This block is "tied" to the method
# | | |
# v v v
each_in([:duck, :duck, :duck, :GOOSE]) {|elem| result.insert(0, elem)}
result # => [:GOOSE, :duck, :duck, :duck]
And an example for return, which I will use to implement a method to see if a number is happy:
class Numeric
# Not the real meat of the program
def sum_of_squares
(to_s.split("").collect {|s| s.to_i ** 2}).inject(0) {|sum, i| sum + i}
end
def happy?(cache=[])
# If the number reaches 1, then it is happy.
return true if self == 1
# Can't be happy because we're starting to loop
return false if cache.include?(self)
# Ask the next number if it's happy, with self added to the list of seen numbers
# You don't actually need the return (it works without it); I just add it for symmetry
return sum_of_squares.happy?(cache << self)
end
end
24.happy? # => false
19.happy? # => true
2.happy? # => false
1.happy? # => true
# ... and so on ...
Hope this helps! :)
def cool
return yield
end
p cool {"yes!"}
The yield keyword instructs Ruby to execute the code in the block. In this example, the block returns the string "yes!". An explicit return statement was used in the cool() method, but this could have been implicit as well.
Let's say I have a function
def odd_or_even n
if n%2 == 0
return :even
else
return :odd
end
end
And I had a simple enumerable array
simple = [1,2,3,4,5]
And I ran it through map, with my function, using a do-end block:
simple.map do
|n| odd_or_even(n)
end
# => [:odd,:even,:odd,:even,:odd]
How could I do this without, say, defining the function in the first place? For example,
# does not work
simple.map do |n|
if n%2 == 0
return :even
else
return :odd
end
end
# Desired result:
# => [:odd,:even,:odd,:even,:odd]
is not valid ruby, and the compiler gets mad at me for even thinking about it. But how would I implement an equivalent sort of thing, that works?
edit
In reality, the solution to my problem matters to me a lot less than the motivation/reasoning behind it, to help me understand more how ruby blocks work :)
You're so close. Just remove the returns and you're golden.
This is because the block passed to map is a proc (i.e. created with Proc.new), and not a lambda. A return within a proc doesn't just jump out of the proc- it jumps out of the method that executed (i.e. called call on) the proc. A return within a lambda, on the other hand, jumps out of only the lambda.
The proc method returns a lambda in Ruby 1.8, and a Proc in Ruby 1.9. It's probably best to just not use this method and be explicit with which construct you want to use.
I'm guessing you were either in IRB or a plain ruby script when you were trying this out.
a = Proc.new { return }
a.call # fails. Nothing to return from.
def foobar
a = Proc.new { return }
a.call
puts 'hello' # not reached. The return within the proc causes execution to jump out of the foobar method.
end
foobar # succeeds, but does not print 'hello'. The return within the proc jumps out of the foobar method.
b = lambda { return }
b.call # succeeds. The return only returns from the lambda itself.
def bazquux
b = lambda { return }
b.call
puts 'hello' # this is reached. The lambda only returned from itself.
end
bazquux # succeeds, and prints 'hello'
The lesson to learn from this is to use implicit returns unless you can't, I guess.
I suspect this may be a duplicate question, but to give a value out of a block, use next
simple.map do |n|
if n%2 == 0
next :even
else
next :odd
end
end
Shortest variant using Andrew's answer:
simple.map { |n| next :even if n % 2 == 0; :odd }