Hello, On 10/17/2011 05:22 PM, Brent Yorgey wrote:
On Mon, Oct 17, 2011 at 04:18:05PM +0100, Hugo Ferreira wrote:
Hello,
I came across the following code:
ngrams'' :: Int -> [a] -> [[a]] ngrams'' n l = do t<- Data.List.tails l l<- [take n t] Control.Monad.guard (length l == n) return l
and tried to use the ">>=" operator in order to figure out how Monads work. I came up with:
test l = (Data.List.tails l) >>= (\t -> [take 2 t]) >>= (\l -> if (length l == 2) then [l] else [])
Questions: 1. How can I use Control.Monad.guard directly in "test l"
test l = (Data.List.tails l) >>= \t -> [take 2 t] >>= \l -> Control.Monad.guard (length l == 2) >> return l
The rule is that
x<- foo
desugars to
foo>>= \x -> ...
and
blah
desugars to
blah>> ...
Ok, I was not aware of the >>.
One thing that might have been tripping you up is your extra parentheses around the lambda expressions. If you have
= (\l -> ...) foo...
the l does not scope over foo... so you cannot mention it. Instead what you want is
= \l -> ... foo...
so the lambda expression is actually \l -> ...>> foo..., that is, it includes *everything* after the \l -> ... and not just the stuff on that line.
Hmmm. Still cannot wrap my mind around this B-(. [[1],[2],[3]] >>= \l -> func1 l >>= \m -> func2 m \l will hold each of the 3 elements of initial list these are concatenated with the results of func1 results in a new list \m will have each element in the new list these are concatenated with the results of func2 results in a last list is equal to ? (([[1],[2],[3]] >>= \l -> func1 l) >>= \m -> func2 m) Hmmm.. going to lookup more info on this.
2. Related issue: how can create a List monad so that I can input "l" argument of "test l" directly?
I don't understand this question. Can you give an example of what you are trying to do?
Weird. I was trying to execute: test1' [[1,2,3],[2,3],[3]] in a version that does not use Data.List.tails but could not. Now it works. Must have done something wrong. Hugo F.
-Brent
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