foldl versus foldr behavior with infinite lists - Stack Overflow
It stops processing an infinite list when the predicate is satisfied. I rewrote it using foldl: myAny :: (a -> Bool) -> [a] -> Bool myAny p list = foldl step False list where step acc item = p item || acc (Note that the arguments to the step function are correctly reversed.) However, it no longer stops processing infinite lists.
Haskell + Infinite lists = Hang - Stack Overflow
2018年6月25日 · However because my syntax isn't perfect I often mix up parts and while I want to ask Haskell to define (lazy) an infinite list and pick the first 5 elements (or whatever) I end up actually asking it to do something with the full infinite list …
How to filter an infinite list in Haskell - Stack Overflow
2012年11月24日 · In other words, you are giving to ghci an infinite list and it will do x<1000 for every element of the list. If you want to take elements until a predicate is True, use takeWhile. It doesn't matter that Haskell is lazy, the filtered list is evaluated because ghci print it.
Haskell - List comprehension with infinite lists - Stack Overflow
2014年11月13日 · The lists being merged are infinite, but that doesn't matter. What matters is that you only have a finite number of lists being merged, and so to compute the next element of the merge you only need to perform a finite number of comparisons. To compute the head of merge xs ys you only need to compute the head of xs and the head of ys.
In Haskell, how can you sort a list of infinite lists of strings?
2010年5月15日 · In general, merging an infinite list of sorted lists is impossible. For same reason that sorting them is. Merging an infinite list of sorted lists, which are sorted by heads (forall i j. i < j => head (lists !! i) <= head (lists !! j)), is possible. So I'm guessing that what you really want is to merge a sorted infinite list of sorted lists.
How to define infinite lists? - Mathematics Stack Exchange
$\omega$ is standard notation for the first infinite ordinal, denoting the order type of the natural numbers under their usual ordering. With the usual Von Neumann ordinals, this is represented by the set $\{0,1,2,3,\ldots\}$ -- that is, $\omega$ is just a fancy name for $\mathbb N$.
Haskell how to generate this infinite list? - Stack Overflow
2013年11月3日 · Can a similar styled code be written to generate the infinite list [1..]? I saw this link on cyclic structures on the Haskell website. There an example is given. cyclic = let x = 0 : y y = 1 : x in x I tried to define a list for my problem in a cyclic manner, but could not succeed.
python - Generating an infinite list? - Stack Overflow
2018年8月15日 · However, some languages support infinite lists by being "lazy", meaning they will only calculate the part of the list which they need at that certain point. Then you could define an infinite list and ask it to print out e.g. the first 100 values in it.
infinite lists, lazy evaluation and length - Stack Overflow
And if you're feeling adventurous, try re-implementing splitAt and length in terms of Peano numbers instead of Int and see if you can make your chunks well-behaved on infinite lists – jberryman Commented Mar 12, 2016 at 16:43
haskell - Why is lazy evaluation useful? - Stack Overflow
2008年11月5日 · Among other things, lazy languages allow multidimensional infinite data structures. While scheme, python, etc allow single dimensional infinite data structures with streams, you can only traverse along one dimension. Laziness is useful for the same fringe problem, but it's worth noting the coroutines connection mentioned in that link.