Closed dnxie12 closed 1 year ago
Hi @dnxie12 , I can actually reproduce the same error against snowflake-snowpark-python==0.12.0
. Could you please double check and confirm whether this is a regression?
Meanwhile, you could work around this by doing:
df2 = copy(df1.filter(col("a") != 5))
Hi @dnxie12 , I can actually reproduce the same error against
snowflake-snowpark-python==0.12.0
. Could you please double check and confirm whether this is a regression?
Ah that's odd, I re-tried on 0.12.0 and it works as expected, with the following generated SQL:
SELECT * FROM ( SELECT "l_nund_A" AS "A_1", "r_ifbk_A" AS "A_2", "l_nund_B" FROM ( SELECT * FROM (( SELECT "A" AS "l_nund_A", "B" AS "l_nund_B" FROM ( SELECT "A", "B" FROM ( SELECT $1 AS "A", $2 AS "B" FROM VALUES (1 :: INT, 2 :: INT), (3 :: INT, 4 :: INT), (5 :: INT, 6 :: INT)))) AS SNOWPARK_TEMP_TABLE_LX74PXI7IR INNER JOIN ( SELECT "A" AS "r_ifbk_A", "B" AS "r_ifbk_B" FROM ( SELECT * FROM ( SELECT "A", "B" FROM ( SELECT $1 AS "A", $2 AS "B" FROM VALUES (1 :: INT, 2 :: INT), (3 :: INT, 4 :: INT), (5 :: INT, 6 :: INT))) WHERE ("A" != 5 :: INT))) AS SNOWPARK_TEMP_TABLE_UXJOQWFJGO ON ("l_nund_A" = "r_ifbk_A")))) LIMIT 10 OFFSET 0
Thanks for the workaround!
Please answer these questions before submitting your issue. Thanks!
pip freeze
)?Self-join on a dataframe created through
filter
leads to aninvalid_identifier
error. The same code does not cause such errors on Snowpark 0.12.0 (last version I was on). Here's an example to reproduce this error:The join should happen without errors.