added test and small fix to verify that primary constructor arguments keep their order as dataframe columns, even when they have @ColumnName annotations.

Author: Jolanrensen

core/src/main/kotlin/org/jetbrains/kotlinx/dataframe/impl/schema/Utils.kt

@@ -192,8 +192,8 @@ internal fun <T> Iterable<KCallable<T>>.sortWithConstructor(klass: KClass<*>): L
     // else sort the ones in the primary constructor first according to the order in there
     // leave the rest at the end in lexicographical order
     val (propsInConstructor, propsNotInConstructor) =
-        lexicographicalColumns.partition { it.columnName in primaryConstructorOrder.keys }
+        lexicographicalColumns.partition { it.name in primaryConstructorOrder.keys }
 
     return propsInConstructor
-        .sortedBy { primaryConstructorOrder[it.columnName] } + propsNotInConstructor
+        .sortedBy { primaryConstructorOrder[it.name] } + propsNotInConstructor
 }

core/src/test/kotlin/org/jetbrains/kotlinx/dataframe/api/toDataFrame.kt

@@ -5,6 +5,7 @@ import io.kotest.matchers.shouldBe
 import org.jetbrains.kotlinx.dataframe.DataColumn
 import org.jetbrains.kotlinx.dataframe.DataFrame
 import org.jetbrains.kotlinx.dataframe.DataRow
+import org.jetbrains.kotlinx.dataframe.annotations.ColumnName
 import org.jetbrains.kotlinx.dataframe.annotations.DataSchema
 import org.jetbrains.kotlinx.dataframe.columns.ColumnKind
 import org.jetbrains.kotlinx.dataframe.kind
@@ -79,14 +80,28 @@ class CreateDataFrameTests {
 
     @Test
     fun `preserve fields order`() {
+        // constructor properties will keep order, so x, c
         class B(val x: Int, val c: String, d: Double) {
+            // then child properties will be sorted lexicographically by column name, so a, b
             val b: Int = x
             val a: Double = d
         }
 
         listOf(B(1, "a", 2.0)).toDataFrame().columnNames() shouldBe listOf("x", "c", "a", "b")
     }
 
+    @Test
+    fun `preserve fields order with @ColumnName`() {
+        // constructor properties will keep order, so z, y
+        class B(@ColumnName("z") val x: Int, @ColumnName("y") val c: String, d: Double) {
+            // then child properties will be sorted lexicographically by column name, so w, x
+            @ColumnName("x") val a: Double = d
+            @ColumnName("w") val b: Int = x
+        }
+
+        listOf(B(1, "a", 2.0)).toDataFrame().columnNames() shouldBe listOf("z", "y", "w", "x")
+    }
+
     @DataSchema
     data class A(val v: Int)