The code goes:
def stratified_subsampling(arr, n_strata, n_points, srand)
<<-DOC
The function samples "n_points" points from an array "arr".
It preserves the original distribution of the data. "n_strata"
is the number of strata to devide the values in. Have to find a
trade-off for this value, but I think that as a rule of thumb,
10 bins should be fair enough in most cases.
DOC
vs = arr.sort
...
Rubocop says:
W: Lint/Void: Literal <<-DOC used in void context.
<<-DOC
^^^^^^
I don't understand what that means and how should I fix it.
Thank you!
It means that you have a string literal here that you do absolutely nothing with. Does it need to be a string literal? A comment should serve just fine.
Related
Am a little confused about array behaviors in Ruby
Given this;
string = "abcac"
remainder = 1
What will the following function do?
string[0, remainder]
let's try to check documentation for class String. Just google rubydoc string
and https://ruby-doc.org/core-2.4.0/String.html
then look for square brackets:
str[start, length] → new_str or nil
have a nice day and good luck with coding.
Max
Thank you.
I kept researching and found that
The first operand specifies an index(which may be negative), and the second specifies the length(which must be non negative)
[index, length]
foo = 1
p +foo
This example code prints 1 just like if the + was not there. I know - in front of a variable gets the opposite of what the variable was (-29 becomes 29) but is there any case where a variable with a + in front of it ever does anything or can I safely remove it every time I see it? To clarify this a bit I am asking no specifically about numbers assigned to variables but any datatype in ruby.
+ is both a unary operator (one argument) and a binary operator (two arguments). It is defined on the Numeric class. Unary operators are defined using # suffix to differentiate from the binary operator.
Unary Plus—Returns the receiver.
This is the source for the method:
num_uplus(VALUE num)
{
return num;
}
So to answer your question,
Does + in front of a variable in ruby ever do anything?
NO, for Numeric values.
I just looked it up for strings and yes it does do something for frozen strings.
If the string is frozen, then return duplicated mutable string.
If the string is not frozen, then return the string itself.
static VALUE
str_uplus(VALUE str)
{
if (OBJ_FROZEN(str)) {
return rb_str_dup(str);
}
else {
return str;
}
}
is there any case where a variable with a + in front of it ever does anything
Yes. Every time. It calls the +# method.
or can I safely remove it every time I see it?
No, you can't. It will change the semantics of your code: before, it will call the +# method, after, it won't.
Whether or not that changes the outcome of your program, depends on what that method is doing. The default implementation for Numeric#+# simply returns self, but of course someone could have monkey-patched it to do something different.
Also, String#+# does something more interesting: if self is a frozen string, it will return an unfrozen, mutable copy; if self is already mutable, it returns self.
Other objects in the core library don't have a +# method, so they will usually raise a NoMethodError. If you remove the call, they won't; that is also a behavioral change.
I need to make a number out of the string. to do that i use well known maneuver looking something like this:
Float(string_var) rescue nil
This works nicely, however I do have a tiny, little problem. If a string is "2.50", variable I get is 2.5. Is it even possible to create Float object with 'unnecessary' 0 digit at the end? can I literally translate "2.50" into 2.50 ?
In short, the answer is no, given the question, as any Float, when examined, will use Float's to_s function, eliciting an answer without trailing zeroes.
Float will always give you a numeric value that can be interpreted any way you wish, though. In your example, you will get a float value (given a string that is a parsable float). What you are asking then, is how to display that value with trailing zeroes. To do that, you will be turning the float value back into a string.
Easiest way to accomplish that is to use the format given by one of your respondents, namely
string_var = "2.50"
float_value = Float(string_var) rescue nil # 2.5
with_trailing_zeroes = "%0.2f" % float_value # '2.50'
I was playing around in irb, and noticed one cannot do
5 * "Hello".
Error
String can't be coerced into Fixnum
However "Hello"*5 provided "HelloHelloHelloHelloHello" as expected.
What is the exact reason for this? I've been looking around in the doc's and could not find the exact reason for this behavior. Is this something the designers of ruby decided?
Basically, you are asking "why is multiplication not commutative"? There are two possible answers for this. Or rather one answer with two layers.
The basic principle of OO is that everything happens as the result of one object sending a message to another object and that object responding to that message. This "messaging" metaphor is very important, because it explains a lot of things in OO. For example, if you send someone a message, all you can observe is what their response is. You don't know, and have no idea of finding out, what they did to come up with that response. They could have just handed out a pre-recorded response (reference an instance variable). They could have worked hard to construct a response (execute a method). They could have handed the message off to someone else (delegation). Or, they just don't understand the message you are sending them (NoMethodError).
Note that this means that the receiver of the message is in total control. The receiver can respond in any way it wishes. This makes message sending inherently non-commutative. Sending message foo to a passing b as an argument is fundamentally different from sending message foo to b passing a as an argument. In one case, it is a and only a that decides how to respond to the message, in the other case it is b and only b.
Making this commutative requires explicit cooperation between a and b. They must agree on a common protocol and adhere to that protocol.
In Ruby, binary operators are simply message sends to the left operand. So, it is solely the left operand that decides what to do.
So, in
'Hello' * 5
the message * is sent to the receiver 'Hello' with the argument 5. In fact, you can alternately write it like this if you want, which makes this fact more obvious:
'Hello'.*(5)
'Hello' gets to decide how it responds to that message.
Whereas in
5 * 'Hello'
it is 5 which gets to decide.
So, the first layer of the answer is: Message sending in OO is inherently non-commutative, there is no expectation of commutativity anyway.
But, now the question becomes, why don't we design in some commutativity? For example, one possible way would be to interpret binary operators not as message sends to one of the operands but instead message sends to some third object. E.g., we could interpret
5 * 'Hello'
as
*(5, 'Hello')
and
'Hello' * 5
as
*('Hello', 5)
i.e. as message sends to self. Now, the receiver is the same in both cases and the receiver can arrange for itself to treat the two cases identically and thus make * commutative.
Another, similar possibility would be to use some sort of shared context object, e.g. make
5 * 'Hello'
equivalent to
Operators.*(5, 'Hello')
In fact, in mathematics, the meaning of a symbol is often dependent on context, e.g. in ℤ, 2 / 3 is undefined, in ℚ, it is 2/3, and in IEEE754, it is something close to, but not exactly identical to 0.333…. Or, in ℤ, 2 * 3 is 6, but in ℤ|5, 2 * 3 is 1.
So, it would certainly make sense to do this. Alas, it isn't done.
Another possibility would be to have the two operands cooperate using a standard protocol. In fact, for arithmetic operations on Numerics, there actually is such a protocol! If a receiver doesn't know what to do with an operand, it can ask that operand to coerce itself, the receiver, or both to something the receiver does know how to handle.
Basically, the protocol goes like this:
you call 5 * 'Hello'
5 doesn't know how to handle 'Hello', so it asks 'Hello' for a coercion. …
… 5 calls 'Hello'.coerce(5)
'Hello' responds with a pair of objects [a, b] (as an Array) such that a * b has the desired result
5 calls a * b
One common trick is to simply implement coerce to flip the operands, so that when 5 retries the operation, 'Hello' will be the receiver:
class String
def coerce(other)
[self, other]
end
end
5 * 'Hello'
#=> 'HelloHelloHelloHelloHello'
Okay, OO is inherently non-commutative, but we can make it commutative using cooperation, so why isn't it done? I must admit, I don't have a clear-cut answer to this question, but I can offer two educated guesses:
coerce is specifically intended for numeric coercion in arithmetic operations. (Note the protocol is defined in Numeric.) A string is not a number, nor is string concatenation an arithmetic operation.
We just don't expect * to be commutative with wildly different types such as Integer and String.
Of course, just for fun, we can actually observe that there is a certain symmetry between Integers and Strings. In fact, you can implement a common version of Integer#* for both String and Integer arguments, and you will see that the only difference is in what we choose as the "zero" element:
class Integer
def *(other)
zero = case other
when Integer then 0
when String then ''
when Array then []
end
times.inject(zero) {|acc, _| acc + other }
end
end
5 * 6
#=> 30
5 * 'six'
#=> 'sixsixsixsixsix'
5 * [:six]
#=> [:six, :six, :six, :six, :six, :six]
The reason for this is, of course, that the set of strings with the concatenation operation and the empty string as the identity element form a monoid, just like arrays with concatenation and the empty array and just like integers with addition and zero. Since all three are monoids, and our "multiplication as repeated addition" only requires monoid operations and laws, it will work for all monoids.
Note: Python has an interesting twist on this double-dispatch idea. Just like in Ruby, if you write
a * b
Python will re-write that into a message send:
a.__mul__(b)
However, if a can't handle the operation, instead of cooperating with b, it cooperates with Python by returning NotImplemented. Now, Python will try with b, but with a slight twist: it will call
b.__rmul__(a)
This allows b to know that it was on the right side of the operator. It doesn't matter much for multiplication (because multiplication is (usually but not always, see e.g. matrix multiplication) commutative), but remember that operator symbols are distinct from their operations. So, the same operator symbol can be used for operations that are commutative and ones that are non-commutative. Example: + is used in Ruby for addition (2 + 3 == 3 + 2) and also for concatenation ('Hello' + 'World' != 'World' + 'Hello'). So, it is actually advantageous for an object to know whether it was the right or left operand.
This is because that operators are also methods(Well there are exceptions as Cary has listed in the comments which I wasn't aware of).
For example
array << 4 == array.<<4
array[2] == array.[](2)
array[2] ='x' == array.[] =(2,'x')
In your example:
5 * "Hello" => 5.*("Hello")
Meanwhile
"hello" *5 => 5.*("hello")
An integer cannot take that method with a string param
If you ever dabble around in python try 5*hello and hello*5, both work. Pretty interesting that ruby has this feature to be honest.
Well, as Muntasir Alam has already told that Fixnum does not has a method named * which takes a string as argument. So, 5*"Hello" produces that error.But, to have fun we can actually achieve 5*"Hello" this by adding that missing method to the Fixnum class.
class Fixnum # open the class
def * str # Override the *() method
if str.is_a? String # If argument is String
temp = ""
self.times do
temp << str
end
temp
else # If the argument is not String
mul = 0
self.times do
mul += str
end
mul
end
end
end
now
puts 5*"Hello" #=> HelloHelloHelloHelloHello
puts 4*5 #=> 20
puts 5*10.4 #=> 52.0
Well, that was just to show that the opposite is also possible. But that will bring a lot of overhead. I think we should avoid that at all cost.
In his book programming in scala (Chapter 5 Section 5.9 Pg 93)
Odersky mentioned this expression "bills !*&^%~ code!"
In the footnote on same page:
"By now you should be able to figure out that given this code,the Scala compiler would
invoke (bills.!*&^%~(code)).!()."
That's a bit to cryptic for me, could someone explain what's going on here?
What Odersky means to say is that it would be possible to have valid code looking like that. For instance, the code below:
class BadCode(whose: String, source: String) {
def ! = println(whose+", what the hell do you mean by '"+source+"'???")
}
class Programmer(who: String) {
def !*&^%~(source: String) = new BadCode(who, source)
}
val bills = new Programmer("Bill")
val code = "def !*&^%~(source: String) = new BadCode(who, source)"
bills !*&^%~ code!
Just copy&paste it on the REPL.
The period is optional for calling a method that takes a single parameter, or has an empty parameter list.
When this feature is utilized, the next chunk after the space following the method name is assumed to be the single parameter.
Therefore,
(bills.!*&^%~(code)).!().
is identical to
bills !*&^%~ code!
The second exclamation mark calls a method on the returned value from the first method call.
I'm not sure if the book provides method signatures but I assume it's just a comment on Scala's syntactic sugar so it assumes if you type:
bill add monkey
where there is an object bill which has a method add which takes a parameter then it automatically interprets it as:
bill.add(monkey)
Being a little Scala rusty, I'm not entirely sure how it splits code! into (code).!() except for a vague tickling of the grey cells that the ! operator is used to fire off an actor which in compiler terms might be interpretted as an implicit .!() method on the object.
The combination of the '.()' being optional with method calls (as Wysawyg explained above) and the ability to use (almost) whatever characters you like for naming methods, makes it possible to write methods in Scala that look like operator overloading. You can even invent your own operators.
For example, I have a program that deals with 3D computer graphics. I have my own class Vector for representing a 3D vector:
class Vector(val x: Double, val y: Double, val z: Double) {
def +(v: Vector) = new Vector(x + v.x, y + v.y, z + v.z)
// ...etc.
}
I've also defined a method ** (not shown above) to compute the cross product of two vectors. It's very convenient that you can create your own operators like that in Scala, not many other programming languages have this flexibility.