How to create an unique autogenerated Id column in a spark dataframe

Or if you don't want to restrict your program into one only partition with df.coalesce(1) you can use zipWithIndex which starts with index = 0 as next:

lines = [["a1", "a2", "a3"],
            ["b1", "b2", "b3"],
            ["c1", "c2", "c3"]]

    cols = ["c1", "c2", "c3"]

    df = spark.createDataFrame(lines, cols)

    start_indx = 10
    df = df.rdd.zipWithIndex() \
           .map(lambda (r, indx): (indx + start_indx, r[0], r[1], r[2])) \
           .toDF(["id", "c1", "c2", "c3"])

    df.show(10, False)

In this case I set the start_index = 10. And this will be the output:

+---+---+---+---+
|id |c1 |c2 |c3 |
+---+---+---+---+
|10 |a1 |a2 |a3 |
|11 |b1 |b2 |b3 |
|12 |c1 |c2 |c3 |
+---+---+---+---+

You can simply add to it to provide a minimum value for the id. Note that it is not guaranteed the values will start from the minimum value

.withColumn("id", monotonically_increasing_id + 123)

Explanation: Operator + is overloaded for columns https://github.com/apache/spark/blob/master/sql/core/src/main/scala/org/apache/spark/sql/Column.scala#L642