I'm following a tutorial. (Real World Haskell)
And I have one beginner question about head and tail called on empty lists: In GHCi it returns exception.
Intuitively I think I would say they both should return an empty list. Could you correct me ? Why not ? (as far as I remember in OzML left or right of an empty list returns nil)
I surely have not yet covered this topic in the tutorial, but isnt it a source of bugs (if providing no arguments)?
I mean if ever passing to a function a list of arguments which may be optionnal, reading them with head may lead to a bug ?
I just know the GHCi behaviour, I don't know what happens when compiled.
Intuitively I think would say they both should return an empty list. Could you correct me ? Why not ?
Well - head is [a] -> a. It returns the single, first element; no list.
And when there is no first element like in an empty list? Well what to return? You can't create a value of type a from nothing, so all that remains is undefined - an error.
And tail? Tail basically is a list without its first element - i.e. one item shorter than the original one. You can't uphold these laws when there is no first element.
When you take one apple out of a box, you can't have the same box (what happened when tail [] == []). The behaviour has to be undefined too.
This leads to the following conclusion:
I surely have not yet covered this topic in the tutorial, but isnt it a source of bugs ? I mean if ever passing to a function a list of arguments which may be optionnal, reading them with head may lead to a bug ?
Yes, it is a source of bugs, but because it allows to write flawed code. Code that's basically trying to read a value that doesn't exist. So: *Don't ever use head/tail** - Use pattern matching.
sum [] = 0
sum (x:xs) = x + sum xs
The compiler can guarantee that all possible cases are covered, values are always defined and it's much cleaner to read.
Related
What I want is roughly equivalent to
df.where(<condition>).count() != 0
But I'm pretty sure it's not quite smart enough to stop once it finds any such violation. I would expect some sort of aggregator to be able to do this, but I haven't found one? I could do it with a max and some sort of conversion, but again I don't think it would necessarily know to quit (not being specific to bool, I'm not sure if understands no value is larger than true).
More specifically, I want to check if a column contains only a single element. Right now my best idea is to do this is by grabbing the first value and comparing everything.
I would try this option, it should be much faster:
df.where(<condition>).head(1).isEmpty
You can also try to define your conditions on a row together with scala's exists (which stops at the first occurence of true):
df.mapPartitions(rows => if(rows.exists(row => <condition>)) Iterator(1) else Iterator.empty).isEmpty
At the end you should benchmark the alternatives
I'm currently working on my functional programming - I am fairly new to it. Am i using Options correctly here? I feel pretty insecure on my skills currently. I want my code to be as safe as possible - Can any one point out what am I doing wrong here or is it not that bad? My code is pretty straight forward here:
def main(args: Array[String]): Unit =
{
val file = "myFile.txt"
val myGame = Game(file) //I have my game that returns an Option here
if(myGame.isDefined) //Check if I indeed past a .txt file
{
val solutions = myGame.get.getAllSolutions() //This returns options as well
if(solutions.isDefined) //Is it possible to solve the puzzle(crossword)
{
for(i <- solutions.get){ //print all solutions to the crossword
i.solvedCrossword foreach println
}
}
}
}
-Thanks!! ^^
When using Option, it is recommended to use match case instead of calling 'isDefined' and 'get'
Instead of the java style for loop, use higher-order function:
myGame match {
case Some(allSolutions) =>
val solutions = allSolutions.getAllSolutions
solutions.foreach(_.solvedCrossword.foreach(println))
case None =>
}
As a rule of thumb, you can think of Option as a replacement for Java's null pointer. That is, in cases where you might want to use null in Java, it often makes sense to use Option in Scala.
Your Game() function uses None to represent errors. So you're not really using it as a replacement for null (at least I'd consider it poor practice for an equivalent Java method to return null there instead of throwing an exception), but as a replacement for exceptions. That's not a good use of Option because it loses error information: you can no longer differentiate between the file not existing, the file being in the wrong format or other types of errors.
Instead you should use Either. Either consists of the cases Left and Right where Right is like Option's Some, but Left differs from None in that it also takes an argument. Here that argument can be used to store information about the error. So you can create a case class containing the possible types of errors and use that as an argument to Left. Or, if you never need to handle the errors differently, but just present them to the user, you can use a string with the error message as the argument to Left instead of case classes.
In getAllSolutions you're just using None as a replacement for the empty list. That's unnecessary because the empty list needs no replacement. It's perfectly fine to just return an empty list when there are no solutions.
When it comes to interacting with the Options, you're using isDefined + get, which is a bit of an anti pattern. get can be used as a shortcut if you know that the option you have is never None, but should generally be avoided. isDefined should generally only be used in situations where you need to know whether an option contains a value, but don't need to know the value.
In cases where you need to know both whether there is a value and what that value is, you should either use pattern matching or one of Option's higher-order functions, such as map, flatMap, getOrElse (which is kind of a higher-order function if you squint a bit and consider by-name arguments as kind-of like functions). For cases where you want to do something with the value if there is one and do nothing otherwise, you can use foreach (or equivalently a for loop), but note that you really shouldn't do nothing in the error case here. You should tell the user about the error instead.
If all you need here is to print it in case all is good, you can use for-comprehension which is considered quite idiomatic Scala way
for {
myGame <- Game("mFile.txt")
solutions <- myGame.getAllSolutions()
solution <- solutions
crossword <- solution.solvedCrossword
} println(crossword)
I was following a Scala video tutorial and he mentioned prepend :: takes constant time and append :+ time increases with length of list. And, also he mentioned most of the time reversing the list prepending and re-reversing the list gives better performance than appending.
Question 1
Why prepend :: takes constant time and append :+ time increases with length of list?
But reason for that is not mentioned in the tutorial and I tried in google. I didn’t find the answer but I found another surprising thing.
Question 2
ListBuffer takes constant time for both append and prepend. If possible why it wasnt implemented in List?
Obvious there would be reason behind! Appreciate if someone could explain.
Answer 1:
List is implemented as Linked list. The reference you hold is to it's head.
e.g. if you have a list of 4 elements (1 to 4) it will be:
[1]->[2]->[3]->[4]->//
Prepending meaning adding new element to the head and return the new head:
[5]->[1]->[2]->[3]->[4]->//
The reference to the old head [1] still valid and from it's point of view there are still 4 elements.
On the other hand, appending meaning adding element to the end of the list.
Since List is immutable, we can't just add it to the end, but we need to clone the entire List:
[1']->[2']->[3']->[4']->[5]->//
Since clone mean copy the entire list in the same order, we need to iterate over each element and append it.
Answer 2:
ListBuffer is mutable collection, changing it will change all the references.
Ad. 1. The list in Scala is defined (simplifying) as a head and a tail. The tail is also a list. Adding an element to the head means creation a new list with a new head and the existing list as a new tail. The existing list is not changed. This is why it is a constant time operation.
Appending to a list needs rebuilding the existing list, which cannot be done in constant time.
Ad. 2. ListBuffer is a mutable collection. It may be more efficient in some applications, but on the other hand immutable collections are thread-safe and easily scalable.
I am tutoring someone in basic search and sorts. In insertion sort I iterate negatively when I have a value that is greater than the one previous to it in numerical terms. Now of course this approach can cause issues because there is a check which calls for array[-1] which does not exist.
As underlined in bold below, adding the and x > 0 boolean prevents the index issue.
My question is how is this the case? Wouldn't the call for array[-1] still be made to ensure the validity of both booleans?
the_list = [10,2,4,3,5,7,8,9,6]
for x in range(1,len(the_list)):
value = the_list[x]
while value < the_list[x-1] **and x > 0**:
the_list[x] = the_list[x-1]
x=x-1
the_list[x] = value
print the_list
I'm not sure I completely understand the question, and I don't know what programming language this is, but most modern programming languages use so-called short-circuit Boolean evaluation by default so that the logical expression isn't evaluated further once the outcome is known.
You can use that to guard against range overflow, like this:
while x > 0 and value < the_list[x-1]
but the check of x's range here must come before the use.
AND operation returns true if and only if both arguments are true, so if one of arguments is false there's no point of checking others as the final value is already known at that point. As for your example, usually evaluation goes from left to right but it is not a principle and it looks the language you used is not following that rule (othewise it still should crash on array lookup). But ut may be, this particular implementation optimizes this somehow (which IMHO is not good idea) and evaluates "simpler" things first (like checking if x > 0) before it look up the array. check the specs why this exact order works for you as in most popular languages you would still crash if test x > 0 wouldn't be evaluated before lookup
If I have an ArrayList<Double> dblList and a Predicate<Double> IS_EVEN I am able to remove all even elements from dblList using:
Collections2.filter(dblList, IS_EVEN).clear()
if dblList however is a result of a transformation like
dblList = Lists.transform(intList, TO_DOUBLE)
this does not work any more as the transformed list is immutable :-)
Any solution?
Lists.transform() accepts a List and helpfully returns a result that is RandomAccess list. Iterables.transform() only accepts an Iterable, and the result is not RandomAccess. Finally, Iterables.removeIf (and as far as I see, this is the only one in Iterables) has an optimization in case that the given argument is RandomAccess, the point of which is to make the algorithm linear instead of quadratic, e.g. think what would happen if you had a big ArrayList (and not an ArrayDeque - that should be more popular) and kept removing elements from its start till its empty.
But the optimization depends not on iterator remove(), but on List.set(), which is cannot be possibly supported in a transformed list. If this were to be fixed, we would need another marker interface, to denote that "the optional set() actually works".
So the options you have are:
Call Iterables.removeIf() version, and run a quadratic algorithm (it won't matter if your list is small or you remove few elements)
Copy the List into another List that supports all optional operations, then call Iterables.removeIf().
The following approach should work, though I haven't tried it yet.
Collection<Double> dblCollection =
Collections.checkedCollection(dblList, Double.class);
Collections2.filter(dblCollection, IS_EVEN).clear();
The checkCollection() method generates a view of the list that doesn't implement List. [It would be cleaner, but more verbose, to create a ForwardingCollection instead.] Then Collections2.filter() won't call the unsupported set() method.
The library code could be made more robust. Iterables.removeIf() could generate a composed Predicate, as Michael D suggested, when passed a transformed list. However, we previously decided not to complicate the code by adding special-case logic of that sort.
Maybe:
Collection<Double> odds = Collections2.filter(dblList, Predicates.not(IS_EVEN));
or
dblList = Lists.newArrayList(Lists.transform(intList, TO_DOUBLE));
Collections2.filter(dblList, IS_EVEN).clear();
As long as you have no need for the intermediate collection, then you can just use Predicates.compose() to create a predicate that first transforms the item, then evaluates a predicate on the transformed item.
For example, suppose I have a List<Double> from which I want to remove all items where the Integer part is even. I already have a Function<Double,Integer> that gives me the Integer part, and a Predicate<Integer> that tells me if it is even.
I can use these to get a new predicate, INTEGER_PART_IS_EVEN
Predicate<Double> INTEGER_PART_IS_EVEN = Predicates.compose(IS_EVEN, DOUBLE_TO_INTEGER);
Collections2.filter(dblList, INTEGER_PART_IS_EVEN).clear();
After some tries, I think I've found it :)
final ArrayList<Integer> ints = Lists.newArrayList(1, 2, 3, 4, 5);
Iterables.removeIf(Iterables.transform(ints, intoDouble()), even());
System.out.println(ints);
[1,3,5]
I don't have a solution, instead I found some kind of a problem with Iterables.removeIf() in combination with Lists.TransformingRandomAccessList.
The transformed list implements RandomAccess, thus Iterables.removeIf() delegates to Iterables.removeIfFromRandomAccessList() which depends on an unsupported List.set() operation.
Calling Iterators.removeIf() however would be successful, as the remove() operation IS supported by Lists.TransformingRandomAccessList.
see: Iterables: 147
Conclusion: instanceof RandomAccess does not guarantee List.set().
Addition:
In special situations calling removeIfFromRandomAccessList() even works:
if and only if the elements to erase form a compact group at the tail of the List or all elements are covered by the Predicate.