```val df = dataFrameOf("id", "props")( 1, map...

# datascience

[Marian Schubert]

val df = dataFrameOf("id", "props")(
    1, mapOf("a" to 10, "b" to 20),
    2, mapOf("a" to 30, "b" to 40),
)

is it possible to convert this to dataframe with columns id, a, b?

[Jolan Rensen [JB]]

probably the easiest way would be something like:

df.add {
    "a" from { props["a"] }
    "b" from { props["b"] }
}.remove { props }

(if you don't have the compiler plugin or you're outside notebooks, this would look like:

df.add {
    "a" from { "props"<Map<String, Int>>()["a"] }
    "b" from { "props"<Map<String, Int>>()["b"] }
}.remove("props")

)

[Jolan Rensen [JB]]

(this may become easier in the future. We discovered Maps are treated differently in some places, so in the future you might be able to simply unfold { props }, but as of now that works unexpectedly)

[Jolan Rensen [JB]]

a more generic solution could be:

df.add {
    // or however you fetch all keys :)
    val keys = props.values().flatMap { it.keys }.toSet()
    
    for (key in keys) {
        key from { props[key] }
    }
}.remove { props }

[Jolan Rensen [JB]]

I came up with another neat solution 🙂

fun Map<String, Any?>.toDataRow() = mapValues { listOf(it.value) }.toDataFrame().single()

df.convert { props }.with { it.toDataRow() }.ungroup { props }

this Map.toDataRow() function will be in the 1.0 release, but not yet in Beta3

[Marian Schubert]

toDataRow works perfectly. Thank you! 🙂