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I cannot understand this ruby behavior, the code explains better what I mean:
class DoNotUnderstand
def initialize
#tiny_array = [3,4]
test
end
def messing(ary)
return [ary[0]+=700, ary[1]+=999]
end
def test
puts #tiny_array.join(",") # before => 3,4
messing(#tiny_array)
puts #tiny_array.join(",") # after => 703,1003
end
end
question = DoNotUnderstand.new
#tiny_array was [3,4] and became [703,1003]
if I don't use a class, that happens:
#tiny = [1,2]
def messing(ary)
return [ary[0]+693,ary[1]+999]
end
puts #tiny.join(",") # before => 1,2
messing(#tiny)
puts #tiny.join(",") # after => 1,2
the array simply remains [1,2]
why?
The class is a red herring, and completely irrelevant to the issue.
In the first case, where the array was modified, you defined messing as:
def messing(ary)
return [ary[0]+=700, ary[1]+=999]
end
Whereas in the second case, where the array was not modified, you defined messing as:
def messing(ary)
return [ary[0]+693,ary[1]+999]
end
In one case you used +=, and in the other, you used merely +.
ary[0] += 700 is exactly equivalent to ary[0] = ary[0] + 700. In other words you are changing the value stored in the 0th index of ary.
In the second case you merely add to the values stored in the array and return the result, but in the first case you not only return the result, you also store it back in the array.
For an explanation of why modifying ary modifies #tiny_array, see this answer to the question Is Ruby pass by reference or by value?.
You're second code example (the one from outside the class) is missing the two characters in the first that make it work the way it does. In the first example, the += operator is used, modifying the array in place:
return [ary[0]+=700, ary[1]+=999]
In your second example, the + operator is used, leaving the array as is:
return [ary[0]+693,ary[1]+999]
If you change it use the += operator, it works the same way as the first code snippet.
I'm working on a module to format print output for the console. The problem I'm running into is when I call .each on an array that I have, it returns the array in the console. I need to control what gets returned in the console.
How can I iterate through an array without having the array returned to the console?
#values.each do |value| # The end result of this is being returned, do not want.
printf(#format,
value[0], value[1], value[2], value[3], value[4], value[5],
value[6]
)
end
Why not create a method for your desired behavior?
def print_each_value(values, format)
values.each do |value|
printf(format, value[0], value[1], value[2], value[3], value[4], value[5], value[6])
end
nil # Here you set what you would like to return to the console.
end
print_each_value(#values, #format)
Edit: Removed annotative variable.
If you're only interested in the output and not the intermediate array which the interactive Ruby will always display you have two options. The first is to pass in the value you want returned:
#values.each_with_object(nil) do |value, x|
# ...
end
Whatever you supply as the argument there will be what is returned.
The second is to return the strings and print those:
puts(
#values.collect do |value|
#format % values
end.join("\n")
)
This has the advantage of dramatically simplifying your call.
As a note, if you have an array and you want to pass it through as arguments then use the splat operator:
printf(#format, *values)
Add ; to the end.
#values.each do |value| # The end result of this is being returned, do not want.
printf(#format,
value[0], value[1], value[2], value[3], value[4], value[5],
value[6])
end;
You can see the explanation in this article.
If you chain multiple statements together in the interactive shell, only the output of the last command that was executed will be displayed to the screen
And basicaly ; in Ruby is used for chaining.
Try #values.cycle(1) do |value|
I wish there'd be a dedicated method for iterating for side-effects only that returns nothing tho.
[1,2,3,4,5,6,7].delete_if{|i| i < 4 }
For example, in the above, why do you need to put |i| before i < 4?
I'm new to Ruby programming and the purpose of this element escapes me.
This is very basic Ruby syntax for a block. A block can sometimes take parameters which are given between the bars |. In your case:
[1,2,3,4,5,6,7].delete_if { |i| i < 4 }
The delete_if method for the type Array accepts a block as a parameter. When the bock is given, the block accepts the array element as a parameter. So it iterates i over each value within the array in this case. More specifically, an element will be deleted from the array if that element is < 4. The result will be:
[4,5,6,7]
You'll often see documentation for methods for Ruby types which say, for example:
delete_if { |item| block } -> Array
Which means that the method accepts a block with a parameter, the block being some code that uses the parameter, and the output being another array. The method's description explains more detail in the documentation (e.g., Ruby Array).
I recommend reading some Ruby getting started information online or a good introductory book which will explain this in more detail.
You have to put i there for the same reason you would put i in the first line here:
def plus_one(i)
return i + 1
end
You have to name your method argument, which you later use as a local variable in the method.
Ruby blocks are similar to methods, they can also receive arguments, and syntax for declaring them is slightly different: enclosing them in | |.
I've redone my answer, even though the OP's question has already been answered, because I thought of a new way to explain this that may help future SO users with the same question.
From high school algebra, you should remember functions like this: f(x) = x + 1.
Imagine putting curly braces around the x + 1: f(x) = { x + 1 }
Then move the (x) to inside the curly braces: f = {(x) x + 1 }
And then get rid of the name f: {(x) x + 1 }. This makes it an "anonymous function," i.e. a "lambda."
Here's the problem: The braces could contain arbitrary statements, which may themselves use parentheses: (x + 1) * 4. So how would Ruby know that the (x) is supposed to be an argument to the function, and not an expression to execute? Some other syntax had to be used. Hence the vertical bars: |x|. (At least I assume that was the thought process).
So {|i| i > 4 } is just like f(i) = i > 4, except that it has no name and is not defined in advance, so the parameter has to be defined "inside" the function itself, rather than being outside attached to the name.
Array#delete_if expects such a function (called a "block" when it's used like this) and knows what to do with it. It passes each member of the array [1,2,3,4,5,6,7] into the block as the argument i to see whether i > 4 is true for that member. It's equivalent to doing something like this:
def greater_than_four(x)
x > 4
end
arr = [1,2,3,4,5,6,7]
arr.each do |el|
arr.delete(el) if greater_than_four(el)
end
You could avoid defining the greater_than_four method in advance by defining a lambda on the fly like this:
arr = [1,2,3,4,5,6,7]
arr.each do |el|
arr.delete(el) if lambda{|i| i > 4}.call(el)
end
But since Array#delete_if already expects a block, and already knows to call it on each element, you can save yourself a whole lot of code:
[1,2,3,4,5,6,7].delete_if{|i| i < 4 }
The parameter which you are passing to the delete_if method is a block and the thing inside the parameter you pass to the block.
Think of the block as a method of sorts. The delete_if method iterates over the block and passes the current item as the parameter i to the block. If the condition evaluates to true then that element gets deleted.
could anyone tell me the return value of this function give the parameter listed blew:
def sequence(*enumerables)
enumerables.each do |enumerable|
print "#{enumerable},"
end
end
a,b,c = [1,2,3],4..6,'a'..'e'
value = sequence(a,b,c)
print value
why the value is evaluated to be:
[[1,2,3],4..6,"a".."e"]
Remember that the last thing left on the stack is the return value of your method. This is always the case. If the return value is important, you must pay close attention to how you exit from your method.
The each method returns what it has been iterating over. Since the each is the last statement in your method, stack-wise, that value gets returned.
You can fix this by returning nothing:
def sequence(*enumerables)
enumerables.each do |enumerable|
print "#{enumerable},"
end
return
end
This approach is generally frowned on as the return method seems out of place. If the caller of this method is not expecting any particular return value, then it's not necessary.
The alternative is to return something useful:
def sequence(*enumerables)
enumerables.join(",")
end
puts sequence(a,b,c)
It's often the case that methods which do not set an expectation for a particular return value may return an arbitrary one.
enumerables is an array. The splat (*) operator causes this.
The return value of sequence is the return value of enumerables.each which is enumerables
A simple example:
def foo(*args)
args
end
foo(1,2,3,4) == [1,2,3,4] # true
You're returning enumerables, which is an array containing all the arguments to the method.
What did you expect value to contain? You haven't made any explicit attempt to return anything, so whatever value the last statement in the method resolves to "falls off" the method to become its return value.
I am new to Ruby and I am trying to write a method that groups an array of words into anagram groups. Here is the code:
def combine_anagrams(words)
dict = words.inject(Hash.new(0)) do |list,ws|
key = sort_word(ws)
if !list.has_key?(key)
list[key] = []
end
list[key].push(ws)
list #What is this
end
return dict.values
end
My question is what the statement list is for. If I take it out list becomes an array instead of hash.
Every method/block/etc. in Ruby returns something, and unless there is an early return statement, whatever the last statement in the method/block/etc. is, is what is returned.
In your case, having list be the last line in the block passed to inject ensures that list is returned by the block. When you remove it, the return value of list[key].push(ws) is returned, which obviously isn't what you want.
Note that this behavior also makes using the return keyword when it is the last statement that would be executed otherwise is unnecessary (this includes the return you have at the end of your method). Though some prefer to be explicit that they intend to return something and use them even when not needed.
On an unrelated note: your if !list.has_key?(key) can be rewritten unless list.has_key?(key).
inject works like this:
final = enumerable.inject(initial_value) do |current_value, iteration|
# calculations, etc. here
value # next iteration, current_value will be whatever the block returns
end
So, in your case, initial_value is Hash.new(0), or an empty Hash with 0 as the default value for a key that doesn't exist instead of nil. This is passed into the inject block for the first element in enumerable.
Inside the inject block, you check to see if key already exists as a key on the hash. If it does not, set it equal to an empty array. In either case, take the current iteration of words (ws) and push it onto the array.
Finally, the block returns the current version of list; it becomes current_value (the first parameter to the inject block) the next time the loop processes an element from enumerable.
As a more simple example, check out this sample:
numbers = [1, 2, 3, 4]
sum = inject(0) do |total, number| # first time, total will be 0
total + number # set total next time to be whatever total is now plus the current number
end
Take a look at http://ruby-doc.org/core-1.9.3/Enumerable.html#method-i-inject
In the inject method if you pass two arguments into it (in your case list and ws) the first one - list - is so-called accumulator value. The value which is returned by the inject block at each iteration step is assigned to the list variable. So the line with the only word "list" which you commented as "#What is this" is used for assigning the value of the list in the block to the "list" accumulator variable.
the statement "list" is the return value of the whole block. The line: "list[key] = []" has a return value of "list", therefore it doesnt need another line to set the return value of the if condition to 'list', but the return value of list[key].push(ws) is list[key]. we want to get the updated value of list in the end, therefore we need to return that value from the block each time, so that further processing acts of the updated list, and not something else.
As a background, each ruby line also has a return value, so if that were the last line of a block, or a function, it automatically becomes the return value of the whole block or the function respectively.
To understand this further, try some code like this in irb:
a = [1,2,3,4,5]
b = a.inject(0) {|sum, val| puts sum; puts val; sum + val}
the inner block comprises of three statememts; the last statement returns the value of sum+val to the block, which get stored in sum, to be used in next iterations.
Also, try some code like this:
h = {:a => []}
b = h[:a].push 6
See what b evaluates to; in your code, you need 'b' to be the accumulated hash, and not the array that is stored in h[:a]