[random-fu] sampleRVar with StdGen in RWS?
Dear Café, I would like to use random-fu to do some pseudo-random simulations for a given StdGen (so that I can run the same simulation multiple times, if needed). The following works: testState :: StdGen -> Int testState = evalState (sampleRVar $ uniform 1 10) The following doesn't: testRWS :: StdGen -> Int testRWS = fst . evalRWS (sampleRVar $ uniform 1 10) () I get <interactive>:2:26: error: • No instance for (MonadRandom (RWST () b0 StdGen Identity)) arising from a use of ‘sampleRVar’ • In the first argument of ‘evalRWS’, namely ‘(sampleRVar $ uniform 1 10)’ In the second argument of ‘(.)’, namely ‘evalRWS (sampleRVar $ uniform 1 10) ()’ In the expression: fst . evalRWS (sampleRVar $ uniform 1 10) () Indeed, I do see a MonadRandom instance for StateT, but none for RWST [0]. Is there a reason to not have a MonadRandom instance for RWST? Am I looking in the wrong place? - Sergiu [0] https://hackage.haskell.org/package/random-source-0.3.0.6/docs/Data-Random-S...
It seems like you need your write-state to be a monoid in order to infer a `MonadRandom` instance. Based on your code, I'm guessing the compiler can't infer that. Have a look at https://hackage.haskell.org/package/MonadRandom-0.5.1.1/docs/Control-Monad-R... On 08/03/2018 07:31 AM, Sergiu Ivanov wrote:
Dear Café,
I would like to use random-fu to do some pseudo-random simulations for a given StdGen (so that I can run the same simulation multiple times, if needed).
The following works:
testState :: StdGen -> Int testState = evalState (sampleRVar $ uniform 1 10)
The following doesn't:
testRWS :: StdGen -> Int testRWS = fst . evalRWS (sampleRVar $ uniform 1 10) ()
I get
<interactive>:2:26: error: • No instance for (MonadRandom (RWST () b0 StdGen Identity)) arising from a use of ‘sampleRVar’ • In the first argument of ‘evalRWS’, namely ‘(sampleRVar $ uniform 1 10)’ In the second argument of ‘(.)’, namely ‘evalRWS (sampleRVar $ uniform 1 10) ()’ In the expression: fst . evalRWS (sampleRVar $ uniform 1 10) ()
Indeed, I do see a MonadRandom instance for StateT, but none for RWST [0].
Is there a reason to not have a MonadRandom instance for RWST?
Am I looking in the wrong place?
- Sergiu
[0] https://hackage.haskell.org/package/random-source-0.3.0.6/docs/Data-Random-S...
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Hello Vanessa, MonadRandom from Control.Monad.Random does indeed have an RWST instance, but I'd like to stick with random-fu because of the long list of provided distributions, and the two seem incompatible : https://stackoverflow.com/a/13946470 I did try to force a very monoidal write-state: testRWS :: StdGen -> Int testRWS = fst . evalRWS work () where work = do tell "hello" sampleRVar $ uniform 1 10 I get <interactive>:2:25: error: • No instance for (MonadRandom (RWST () String StdGen Identity)) arising from a use of ‘work’ • In the first argument of ‘evalRWS’, namely ‘work’ In the second argument of ‘(.)’, namely ‘evalRWS work ()’ In the expression: fst . evalRWS work () which seems to be saying the same thing as before :-( - Sergiu Thus quoth Vanessa McHale on Fri Aug 03 2018 at 14:43 (+0200):
It seems like you need your write-state to be a monoid in order to infer a `MonadRandom` instance. Based on your code, I'm guessing the compiler can't infer that.
Have a look at https://hackage.haskell.org/package/MonadRandom-0.5.1.1/docs/Control-Monad-R...
On 08/03/2018 07:31 AM, Sergiu Ivanov wrote:
Dear Café,
I would like to use random-fu to do some pseudo-random simulations for a given StdGen (so that I can run the same simulation multiple times, if needed).
The following works:
testState :: StdGen -> Int testState = evalState (sampleRVar $ uniform 1 10)
The following doesn't:
testRWS :: StdGen -> Int testRWS = fst . evalRWS (sampleRVar $ uniform 1 10) ()
I get
<interactive>:2:26: error: • No instance for (MonadRandom (RWST () b0 StdGen Identity)) arising from a use of ‘sampleRVar’ • In the first argument of ‘evalRWS’, namely ‘(sampleRVar $ uniform 1 10)’ In the second argument of ‘(.)’, namely ‘evalRWS (sampleRVar $ uniform 1 10) ()’ In the expression: fst . evalRWS (sampleRVar $ uniform 1 10) ()
Indeed, I do see a MonadRandom instance for StateT, but none for RWST [0].
Is there a reason to not have a MonadRandom instance for RWST?
Am I looking in the wrong place?
- Sergiu
[0] https://hackage.haskell.org/package/random-source-0.3.0.6/docs/Data-Random-S...
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It seems like the instance comes from http://hackage.haskell.org/package/random-source-0.3.0.6/docs/Data-Random-In... then, no? I only see an instance for `StateT` unfortunately. I suppose you could write an orphan instance for RWST if you really wanted to. On 08/03/2018 07:54 AM, Sergiu Ivanov wrote:
Hello Vanessa,
MonadRandom from Control.Monad.Random does indeed have an RWST instance, but I'd like to stick with random-fu because of the long list of provided distributions, and the two seem incompatible :
https://stackoverflow.com/a/13946470
I did try to force a very monoidal write-state:
testRWS :: StdGen -> Int testRWS = fst . evalRWS work () where work = do tell "hello" sampleRVar $ uniform 1 10
I get
<interactive>:2:25: error: • No instance for (MonadRandom (RWST () String StdGen Identity)) arising from a use of ‘work’ • In the first argument of ‘evalRWS’, namely ‘work’ In the second argument of ‘(.)’, namely ‘evalRWS work ()’ In the expression: fst . evalRWS work ()
which seems to be saying the same thing as before :-(
- Sergiu
Thus quoth Vanessa McHale on Fri Aug 03 2018 at 14:43 (+0200):
It seems like you need your write-state to be a monoid in order to infer a `MonadRandom` instance. Based on your code, I'm guessing the compiler can't infer that.
Have a look at https://hackage.haskell.org/package/MonadRandom-0.5.1.1/docs/Control-Monad-R...
On 08/03/2018 07:31 AM, Sergiu Ivanov wrote:
Dear Café,
I would like to use random-fu to do some pseudo-random simulations for a given StdGen (so that I can run the same simulation multiple times, if needed).
The following works:
testState :: StdGen -> Int testState = evalState (sampleRVar $ uniform 1 10)
The following doesn't:
testRWS :: StdGen -> Int testRWS = fst . evalRWS (sampleRVar $ uniform 1 10) ()
I get
<interactive>:2:26: error: • No instance for (MonadRandom (RWST () b0 StdGen Identity)) arising from a use of ‘sampleRVar’ • In the first argument of ‘evalRWS’, namely ‘(sampleRVar $ uniform 1 10)’ In the second argument of ‘(.)’, namely ‘evalRWS (sampleRVar $ uniform 1 10) ()’ In the expression: fst . evalRWS (sampleRVar $ uniform 1 10) ()
Indeed, I do see a MonadRandom instance for StateT, but none for RWST [0].
Is there a reason to not have a MonadRandom instance for RWST?
Am I looking in the wrong place?
- Sergiu
[0] https://hackage.haskell.org/package/random-source-0.3.0.6/docs/Data-Random-S...
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Thus quoth Vanessa McHale on Fri Aug 03 2018 at 15:06 (+0200):
It seems like the instance comes from http://hackage.haskell.org/package/random-source-0.3.0.6/docs/Data-Random-In... then, no? I only see an instance for `StateT` unfortunately.
Yes, that's what I see as well.
I suppose you could write an orphan instance for RWST if you really wanted to.
That's what I'll most probably do for my small project, but I was wondering whether it made sense to make a pull request to random-fu. To me, having a MonadRandom for both StateT and RWST seems very natural. - Sergiu
On 08/03/2018 07:54 AM, Sergiu Ivanov wrote:
Hello Vanessa,
MonadRandom from Control.Monad.Random does indeed have an RWST instance, but I'd like to stick with random-fu because of the long list of provided distributions, and the two seem incompatible :
https://stackoverflow.com/a/13946470
I did try to force a very monoidal write-state:
testRWS :: StdGen -> Int testRWS = fst . evalRWS work () where work = do tell "hello" sampleRVar $ uniform 1 10
I get
<interactive>:2:25: error: • No instance for (MonadRandom (RWST () String StdGen Identity)) arising from a use of ‘work’ • In the first argument of ‘evalRWS’, namely ‘work’ In the second argument of ‘(.)’, namely ‘evalRWS work ()’ In the expression: fst . evalRWS work ()
which seems to be saying the same thing as before :-(
- Sergiu
Thus quoth Vanessa McHale on Fri Aug 03 2018 at 14:43 (+0200):
It seems like you need your write-state to be a monoid in order to infer a `MonadRandom` instance. Based on your code, I'm guessing the compiler can't infer that.
Have a look at https://hackage.haskell.org/package/MonadRandom-0.5.1.1/docs/Control-Monad-R...
On 08/03/2018 07:31 AM, Sergiu Ivanov wrote:
Dear Café,
I would like to use random-fu to do some pseudo-random simulations for a given StdGen (so that I can run the same simulation multiple times, if needed).
The following works:
testState :: StdGen -> Int testState = evalState (sampleRVar $ uniform 1 10)
The following doesn't:
testRWS :: StdGen -> Int testRWS = fst . evalRWS (sampleRVar $ uniform 1 10) ()
I get
<interactive>:2:26: error: • No instance for (MonadRandom (RWST () b0 StdGen Identity)) arising from a use of ‘sampleRVar’ • In the first argument of ‘evalRWS’, namely ‘(sampleRVar $ uniform 1 10)’ In the second argument of ‘(.)’, namely ‘evalRWS (sampleRVar $ uniform 1 10) ()’ In the expression: fst . evalRWS (sampleRVar $ uniform 1 10) ()
Indeed, I do see a MonadRandom instance for StateT, but none for RWST [0].
Is there a reason to not have a MonadRandom instance for RWST?
Am I looking in the wrong place?
- Sergiu
[0] https://hackage.haskell.org/package/random-source-0.3.0.6/docs/Data-Random-S...
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participants (2)
-
Sergiu Ivanov -
Vanessa McHale