Pyspark add sequential and deterministic index to dataframe

Question

I need to add an index column to a dataframe with three very simple constraints:

• start from 0

• be sequential

• be deterministic

I'm sure I'm missing something obvious because the examples I'm finding look very convoluted for such a simple task, or use non-sequential, non deterministic increasingly monotonic id's. I don't want to zip with index and then have to separate the previously separated columns that are now in a single column because my dataframes are in the terabytes and it just seems unnecessary. I don't need to partition by anything, nor order by anything, and the examples I'm finding do this (using window functions and row_number). All I need is a simple 0 to df.count sequence of integers. What am I missing here?

1, 2, 3, 4, 5

Answer 1

What I mean is: how can I add a column with an ordered, monotonically increasing by 1 sequence 0:df.count? (from comments)

You can use row_number() here, but for that you'd need to specify an orderBy(). Since you don't have an ordering column, just use monotonically_increasing_id().

from pyspark.sql.functions import row_number, monotonically_increasing_id
from pyspark.sql import Window

df = df.withColumn(
    "index",
    row_number().over(Window.orderBy(monotonically_increasing_id()))-1
)

Also, row_number() starts at 1, so you'd have to subtract 1 to have it start from 0. The last value will be df.count - 1.

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I don't want to zip with index and then have to separate the previously separated columns that are now in a single column

You can use zipWithIndex if you follow it with a call to map, to avoid having all of the separated columns turn into a single column:

cols = df.columns
df = df.rdd.zipWithIndex().map(lambda row: (row[1],) + tuple(row[0])).toDF(["index"] + cols