How to get the R value in GHCI ?

[Stéphane Laurent]

Hello,

 For example, how can we get 2 from [r|1+1] ? 

I've seen some documentations about evaluating R expressions in H, but how to do in ordinary GHCI ?

[Boespflug, Mathieu]

H is ordinary GHCi. It's just GHCi initialized in such a way that a handful of modules are in scope and a few language extensions are enabled by default. See here for the initialization script we use:

https://github.com/tweag/HaskellR/blob/master/H/H.ghci

So getting a Haskell value from a quasiquote works exactly the same way as in H:

> p [r| 1 + 1 |]

[1] 2

The p function above is a shorthand for calling R's printing function to display the value of the quasiquote on the terminal.

If you want an actual Haskell value, you need a conversion:

> x <- [r| 1 + 1 |]

> fromSomeSEXP x :: Double

2.0

Note that what you'll get back is a floating point number, not an integer. That's because the "1" literal in R actual stands for a a floating point number. You can tell R you really want a (32-bit) integer though:

> import Data.Int

> y <- [r| 1L + 1L |]

> fromSomeSEXP x :: Int32

2

HTH,

--
Mathieu Boespflug
Founder at http://tweag.io.

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[Stéphane Laurent]

Thank you !

[Stéphane Laurent]

Now to use it in a module, I'm doing :

{-# OPTIONS_GHC -fno-ghci-sandbox #-}
{-# LANGUAGE QuasiQuotes       #-}

{-# LANGUAGE TemplateHaskell   #-}

{-# LANGUAGE ViewPatterns      #-}

{-# LANGUAGE DataKinds #-}

{-# LANGUAGE GADTs #-}
{-# LANGUAGE PartialTypeSignatures #-}
{-# LANGUAGE OverloadedLists #-}
{-# LANGUAGE ScopedTypeVariables #-}
import H.Prelude.Interactive as H.Prelude

But I don't know how to deal with 

Language.R.Instance.initialize Language.R.Instance.defaultConfig

?

[Boespflug, Mathieu]

Call that function early on from your main function. Better yet, use Language.R.Instance.withEmbeddedR.

See http://tweag.github.io/HaskellR/docs/differences-repl-source.html

--
Mathieu Boespflug
Founder at http://tweag.io.

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[Stéphane Laurent]

This code works: 

:set -XQuasiQuotes
import qualified Foreign.R as R
import Foreign.R (SEXP, SEXPTYPE)
import Language.R.Instance as R
import Language.R.QQ
import H.Prelude (fromSomeSEXP)

z <-  R.withEmbeddedR R.defaultConfig $  R.unsafeRunRegion [r|1+1|]
fromSomeSEXP z :: Double

But if I do "z <- ..." a second time,  GHCI crashes after returning the message "R already initialized". 

[Stéphane Laurent]

Call that function early on from your main function.

Ok, I've managed by using this way.

[Stéphane Laurent]

My goal was to try to do a web app which calls R, and it works: http://stla.github.io/stlapblog/posts/RunRInYesod.html

Very cool.

[Boespflug, Mathieu]

Very cool indeed!

--
Mathieu Boespflug
Founder at http://tweag.io.

On 1 September 2016 at 11:37, 'Stéphane Laurent' via haskellr <hask...@googlegroups.com> wrote:

My goal was to try to do a web app which calls R, and it works: http://stla.github.io/stlapblog/posts/RunRInYesod.html

Very cool.

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[Stéphane Laurent]

fromSomeSEXP x :: Double

How to do when the result is an integer ? I can't get it:

> fromSomeSEXP y :: Int
<interactive>:35:1:
    No instance for (Literal Int form0)
      arising from a use of ‘fromSomeSEXP’
    In the expression: fromSomeSEXP y :: Int
    In an equation for ‘it’: it = fromSomeSEXP y :: Int

> fromSomeSEXP y :: Int
<interactive>:35:1:
    No instance for (Literal Int form0)
      arising from a use of ‘fromSomeSEXP’
    In the expression: fromSomeSEXP y :: Int
    In an equation for ‘it’: it = fromSomeSEXP y :: Int
 




[Stéphane Laurent]

I get it !

> import Data.Int (Int32)
> fromSomeSEXP y :: Int32
2

[Boespflug, Mathieu]

See my earlier email - you'll want to cast to Data.Int.Int32, not Int. We use explicit explicitly sized integer types to avoid overflow problems as you pass values from one language to another.

> import Data.Int

> fromSomeSEXP x :: Int32

--
Mathieu Boespflug
Founder at http://tweag.io.

On 1 September 2016 at 18:21, 'Stéphane Laurent' via haskellr <hask...@googlegroups.com> wrote:

fromSomeSEXP x :: Double

How to do when the result is an integer ? I can't get it:

> fromSomeSEXP x :: Int

<interactive>:35:1:
    No instance for (Literal Int form0)
      arising from a use of ‘fromSomeSEXP’
    In the expression: fromSomeSEXP y :: Int
    In an equation for ‘it’: it = fromSomeSEXP y :: Int

> fromSomeSEXP x :: Int

<interactive>:35:1:
    No instance for (Literal Int form0)
      arising from a use of ‘fromSomeSEXP’
    In the expression: fromSomeSEXP y :: Int
    In an equation for ‘it’: it = fromSomeSEXP y :: Int
 

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[Stéphane Laurent]

Another blog post, thanks to your help.