Hi,
I'll try to explain the meaning of different parts of the comprehension::
unique xs = [x | *(x,y) <- zip xs [0..]*, x `notElem` (take y xs)]
*this* part means: consider the couples (x,y) that are output of zip xs
[0..]
for example if xs = [10,20,10,30,30], you are taking
[(10,0),(20,1),(10,2),(30,3),(30,4)]
unique xs = [x | (x,y) <- zip xs [0..], *x `notElem` (take y xs)*]
*this* part means: among the things you considered before (i.e. all the
couples obtained before), consider only those that satisfy the property
that the first element of (x,y) is not an alement of a certain list (take y
xs, i.e. the first y elements of xs).
So in the example above:
is 10 an element of (take 0 xs)=[ ] ? No ----> we consider (10,0)
is 20 an element of (take 1 xs)=[10]? No ----> we consider (20,1)
is 10 an element of (take 2 xs)=[10,20]? Yes ----> we DON'T consider
(10,2)
is 30 an element of (take 3 xs)=[10,20,10]? No ----> we consider
(30,3)
is 30 an element of (take 4 xs)=[10,20,10,30]? Yes ----> we DON'T
consider (30,4)
So we are considering only [(10,0),(20,1),(30,3)]
So as you said this is a refinement of the elements generated by the first
generator. A refinement means that some of the elements generated before
are (possibly) discarded, so you don't obtain a [Bool], but something of
the same type of what was generated before, i.e. another [(Eq,Int)]
possibly shorter than the previous one.
Finally
unique xs =* [x |* (x,y) <- zip xs [0..], x `notElem` (take y xs)]
*this* part says that, of each couple produced and refined before, you take
the first element (that was called x).
So in the example you get [10,20,30]
Hope this is clear,
Ut
Il giorno dom 26 apr 2020 alle ore 14:51 Ken Overton
Hello all,
I recently came across this function which made me realize I don't understand list comprehensions well. I hope someone can help me understand them better by understanding this example better. The function takes a list of Eq and returns the list of unique elements from it:
unique :: Eq a => [a] -> [a] unique xs = [x | (x,y) <- zip xs [0..], x `notElem` (take y xs)]
It's using a list comprehension with multiple 'generators' (hope I have the term correctly). My understanding of multiple generators in a list comprehension is that they refine the results of the previous generator.
So the first generator should produce [(Eq,Int)] as input to the second generator? And the second generator should produce [Bool]?
My understanding must be wrong though; how do we end up with just the items where the second generator produced True?
Thanks,
-- Ken Overton (917) 863-3937 ken.overton@gmail.com
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