add infer types parameter in DSL functions

[koperagen]

I believe tests that i added speak for themselves. Type can often be erased to Any and we need an easy way to get a meaningful result

[Jolanrensen]

I actually expected this to be done by default. If you need to manually specify to infer types, I think people will just not do it. And then, it will be Any?

[koperagen]

It's consistent with add itself (default infer = Infer.Nulls) and most other usages of this enum. I think it makes sense. Looking at values to deduce type is a reflective operation and, while i didn't test it myself and decision wasn't made by me, i suppose it can be quite heavy + produce unexpected results (more precise type than needed or something). So i'd say keeping it as an option is a reasonable compromise

[Jolanrensen]

same here, it's good it exists, but I think it should be done by default when using expr

[Jolanrensen]

I think this requires more explanation, as expr already has the ability to generate a type. However, this type is only known if it is supplied to expr as reified parameter. If a column needs to be added using a normal generic function, this type is not known and the newly added column will always be Any/Any?.
This PR adds the option to infer the (schema) type from the data even if the reified type is too broad.

[koperagen]

This summarizes the issue how i initially found it. There was no way to create anything other than Any column from my list of Any that i got from a library. Changes in AddDsl were made for consistency of expr functions

 val data: List<Any> = listOf(1, 2, 3)
val res = data.toDataFrame {
      "d" from { it } // Any :( 
      "e" from inferType { it } // Int
       expr(infer = Infer.Type) { it } into "d" // Int
}