1 to 5 foldRight (0)(x:Int, y:Int => x+y)
I am trying to add all the values from right to left with 0 as the initial parameter. I get the following error
Int(0) does not take parameters
Can anyone explain to me what this error even means?
Thanks.
It's just the parser getting "confused", so it's trying to apply (x: Int, y: Int ...) as argument of (0).
Specifically, what you're using is a scala syntactic feature that allows to use
a.f(b)
as
a f b
This works with any method that has a single parameter. However when your method has multiple parameter lists (like foldRight), you have to use extra care.
This is what the parser sees
1 to 5 foldRight (0)(x: Int, y: Int => x + y)
|__a__| |___f___| |____________b_____________|
So when evaluating b, it treats 0 as a function with (x: Int, ...) as an argument.
Clearly this can't work, because "Int(0) does not take parameters".
Instead, you can use
(1 to 5).foldRight(0)((x,y) => x + y)
or even
(1 to 5).foldRight(0)(_ + _)
There are two things wrong with your code. The first is that you need to use brackets around 1 to 5, the second is the syntax of your anonymous function. You need brackets around the parameters there.
(1 to 5).foldRight(0)((x,y) => x + y)
Related
Folding list in scala using /: and :\ operator
I tried to to look at different sites and they only talk about foldRight and foldLeft functions.
def sum(xs: List[Int]): Int = (0 /: xs) (_ + _)
sum(List(1,2,3))
res0: 6
The code segment works as described. But I am not able to completely understand the method definition. What I understand is that the one inside the first parenthesis -> 0 /: xs where /: is a right associate operator. The object is xs and the parameter is 0. I am not sure about the return type of the operation (most probably it would be another list?). The second part is a functional piece which sums its two parameters. But I don't understand what object invokes it ? and the name of function. Can someone please help me to understand.
The signature of :/ is
/:[B](z: B)(op: (B, A) ⇒ B): B
It is a method with multiple argument lists, so when it is just invoked with on argument (i.e. 0 /: xs in your case) the return type is (op: (B, A) ⇒ B): B. So you have to pass it a method with 2 parameters ( _ + _ ) that is used to combine the elements of the list starting from z.
This method is usually called foldLeft:
(0 /: xs)(_ + _) is the same as xs.foldLeft(0)(_ + _)
You can find more details here: https://www.scala-lang.org/api/2.12.3/scala/collection/immutable/List.html
Thanks #HaraldGliebe & #LuisMiguelMejíaSuárez for your great responses. I am enlightened now!. I am just summarisig the answer here which may benefit others who read this thread.
"/:" is actually the name of the function which is defined inside the List class. The signature of the function is: /:[B](z: B)(op: (B, A) ⇒ B): B --> where B is the type parameter, z is the first parameter; op is the second parameter which is of functional type.
The function follows curried version --> which means we can pass less number of parameters than that of the actual number. If we do that,
the partially applied function is stored in a temporary variable; we can then use the temporary variable to pass the remaining parameters.
If supplied with all parameters, "/:" can be called as: x./:(0)(_+_) where x is val/var of List type. OR "/:" can be called in two steps which are given as:
step:1 val temp = x./:(0)(_) where we pass only the first parameter. This results in a partially applied function which is stored in the temp variable.
step:2 temp(_+_) here using the partially applied function temp is passed with the second (final) parameter.
If we decide to follow the first style ( x./:(0)(_+_) ), calling the first parameter can be written in operator notion which is: x /: 0
Since the method name ends with a colon, the object will be pulled from right side. So x /: 0 is invalid and it has to be written as 0 /: x which is correct.
This one is equivalent to the temp variable. On following 0 /: x, second parameter also needs to be passed. So the whole construct becomes: (0/:x)(_+_)
This is how the definition of the function sum in the question, is interpreted.
We have to note that when we use curried version of the function in operator notion, we have to supply all the parameters in a single go.
That is: (0 /: x) (_) OR (0 /: x) _ seems throwing syntax errors.
I have two functions
val mul3 = 3*(_: Double)
val pow2 = (x: Double) => x*x
What I don't understand how it works at all is this:
println((pow2.andThen[Double] _ )(mul3)(5))
1) I thought andThen operates with results of the function to the left, but here it is [Double] _ - what is this? (I was expecting something like pow2 andThen mul3)
2) Why is mul3 passed to pow2 if pow2 expects Double? (mul3(pow2) gives an error)
3) What is being passed to pow2? Is it mul3 or 15?
4) What does . mean?
scala fiddle
Lets go step by step.
val mul3 = 3*(_: Double)
Is a Function from a Double to another Double. Thus, is type is Function1[Double, Double].
The same applies to:
val pow2 = (x: Double) => x*x
Now, remember that in Scala, everything is an object. And that there are not operators, only methods. So:
pow2.andThen[Double]
Is calling the andThen on method on the Function1 class.
As you can see on the scaldoc, that method receives another function. Also, it is parametric in the return type of the second function, which determines the return type of the composed function that is returned.
So the [Double] part is just specifying that type.
pow2.andThen[Double] _
Is using the underscore syntax to create another function.
The above line is equivalent to:
f => pow2.andThen[Double](f)
So, it is creating a function that takes another function as input, and returns another function. The resulting function will call pow2 first and then call the function passed as the argument.
Thus, it is of type Function1[Function1[Double, Double], Function[Double, Double]].
Then
(pow2.andThen[Double] _ )(mul3)
Is passing mul3 as the argument of that function, returning a final function (Function1[Double, Doule]) that calls pow2 first and pass the result to mul3.
Finally
(pow2.andThen[Double] _ )(mul3)(5)
Is calling the resulting function with 5 as is input.
After substitution, the code is equivalent to:
x => mul3(pow2(x))
x => 3 * (x * x)
5 => 3 * (x * x)
3 * (5 * 5)
75.0
Side note, IMHO, that code is very cryptic and far for what I would call idiomatic in Scala.
This question already has answers here:
Understanding Currying in Scala
(3 answers)
Closed 4 years ago.
def sum(f: Int => Int): (Int, Int) => Int = {
def sumF(a: Int, b: Int): Int =
if (a > b) 0
else f(a) + sumF(a + 1, b)
sumF
}
def sumCubes = sum(a => a * a * a)
sumCubes // output Function2
sumCubes(1,10) // output 3025.
sumCubes() // doesn't work
I feel I dont understand currying well enough. In the first statement , we are calling sumCubes without parameters , hence sum gets called with the anonymous function as parameter and returns a function2.
Whats really happening in 2nd and 3rd invocation ,
Why are we able to do
sum(a => a * a * a)(1,10)
but not
sumCubes()(1,10)
My understanding is that in sum(a => a * a * a)(1,10) , we are partially applying sum to the anonymous function, which returns a Function2 ,which is applied to the second pair of parameters (1,10) and hence we are getting 3025,
However the same should happen in case of sumCubes()(1,10) , invoking sumCubes without parameters first , would inturn invoke sum with the anonymous function and the Function2 returned would be applied to (1,10)
Why does sumCubes(1,10) work but not sumCubes()(1,10) , shouldn't sumCubes and sumCubes() mean the same thing , invocation of function sumCubes. Also if a simple reference sumCubes is invoking it , how can I pass it around. I feel like I am not understanding something fundamental about Scala.
Scala's methods can have multiple argument lists.
For example, here is a method foo that has ten argument lists, of which the first seven are empty:
def foo()()()()()()()(a: Int)(b: Int)(c: Int): Int = a + b + c
You can invoke it as follows:
println(foo()()()()()()()(1)(20)(300))
and it will print 321.
Note that when you invoke a method, the number of argument lists, as well as the number of arguments in each list (and also their types) have to match the declaration.
Another example. The following method has two argument lists:
def bar(i: Int)(f: Int => Int) = f(i)
you can invoke it as follows:
bar(42)(x => x * x)
but not as
bar()(x => x * x)
bar()(42)(x => x * x)
bar(42)()
or anything like it.
Completely analogously, if you have defined a method with zero argument lists
def baz = (x: Int) => x * x
then you must invoke it with zero argument lists too:
baz
Since it returns an Int => Int function, you can of course apply the result to an Int:
baz(42)
which is the same as (baz)(42), but you cannot do the following:
baz()(42)
because baz itself has no argument lists, and () does not contain a single integer argument.
Note that all of the above is actually a simplification: under certain circumstances, methods that have an empty argument list can be called without parentheses, i.e. def foo(): Unit = ... can be invoked as foo, without (). This is a somewhat strange feature, and I can't say exactly why it's there. My best guess would be: it has something to do with java-interop, where you really want to omit parentheses on zero-ary getters.
w.r.t Currying in scala, partly I understood below sample code.
def product1(f:Int => Int )(a:Int, b:Int):Int = {
println()
if(a > b ) 1
else
f(a) * product1(f)(a+1, b)
}
product(x => x * x) (3, 4)
Out of it., I am bit confused with
product1(f)
in
product1(f)(a+1, b)
Just need explanation, what goes on here.... :( and how to pronounce verbally while explaining...
Thanks in advance..
product1 has two parameter lists. product1(f) is the application of f which is a function of kind Int => Int. If you were only to call product1(f) like so:
product1(f)
without the second parameter list, you'd get a so-called partially-applied function, i.e. a function which doesn't have all of its parameters bound (you'd still need to provide it with a and b)
Look at the parameter declaration:
f:Int => Int
f is a function that maps an Int to an Int -- it takes an Int as an argument and returns an Int. An example is given:
x => x * x
returns the square of its argument.
product1(x => x * x) (3, 4)
returns the product of f(3) .. f(4) = 3*3 * 4*4
BTW, this isn't really an example of currying, since all the arguments are given. Currying would be something like
val prodSquare = product1(x => x * x)
Then,
prodSquare(1, 5)
yields 1*1 * 2*2 * 3*3 * 4*4 * 5*5
For most idiomatic purposes I've seen, you may as well think of your function as having just one parameter list. Multiple parameter lists are used mostly for type inference purposes in generic functions, since type inference is done one parameter list at a time, rather than using Hindley-Milner/Algorithm W to infer the type of everything at once. Some other language features work on individual parameter lists, such as implicit parameters and implicit use of braces in place of parentheses for single-parameter parameter lists, etc.
Functions with multiple parameter lists are called in a similar way to a curried functions from a syntactic perspective, but by default, the intermediate functions aren't created. In fully curried style, each function takes only at most one argument and returns one result (which might happen to be another function, which expects on argument, and so forth). Technically, currying the function would be:
def product2(f: Int => Int): Int => Int => Int = {
a: Int => {
b: Int => {
if(a > b ) 1
else f(a) * product2(f)(a+1)(b)
}
}
}
For completeness, you can treat a function with multiple parameter lists as a curried function by using an underscore after a complete parameter list. In your original example, you'd do product1(f)_, which would return a function of type (Int, Int) => Int.
In researching this question, I came upon another SO question worth checking out to understand this aspect of the language better.
Are the parenthesis around the final tuple really needed? It doesn't compile without them and the compiler tries to add only the Sort("time") and complains that it expects a tuple instead.
val maxSortCounts: Map[Sort, Int] =
sorts.map(s => s -> usedPredicates.map(pred => pred.signature.count(_ == s)).max)
.toMap + ((Sort("time"), 1))
I've tried to reproduce this behaviour inside the REPL with a shorter example, but there it behaves as intended. The variable sorts is a Seq[Sort].
error: type mismatch;
found : <snip>.Sort
required: (<snip>.Sort, Int)
.toMap + (Sort("time"), 1)
Yes, they are needed. Otherwise the compiler will interpret the code as
x.+(y, z) instead of x.+((y, z)).
Instead, you can use ArrowAssoc again: x + (y -> z). Notice, the parentheses are also needed because + and - have the same precedence (only the first sign of a method defines its precedence).
Yes, they're needed. They make the expression a tuple. Parentheses surrounding a comma-separated list create tuple objects. For example, (1, 2, 3) is a 3-tuple of numbers.
Map's + method accepts a pair - in other words a tuple of two elements. Map represents entries in the map as (key,value) tuples.