Thats a good question. In this particular example, it is really only
internal implementation details that make it ambiguous. However, fixing
this was a very large change so we have defered it to Spark 1.4 and instead
print a warning now when you construct trivially equal expressions. I can
try t
Hi,
Thanks for your response. I am not clear about why the query is ambiguous.
val both = df_2.join(df_1, df_2("country")===df_1("country"), "left_outer")
I thought df_2("country")===df_1("country") indicates that the country
field in the 2 dataframes should match and df_2("country") is the
equ
Unfortunately you are now hitting a bug (that is fixed in master and will
be released in 1.3.1 hopefully next week). However, even with that your
query is still ambiguous and you will need to use aliases:
val df_1 = df.filter( df("event") === 0)
. select("country", "cnt").as("a"
Hi,
Thanks for your response. I modified my code as per your suggestion, but
now I am getting a runtime error. Here's my code:
val df_1 = df.filter( df("event") === 0)
. select("country", "cnt")
val df_2 = df.filter( df("event") === 3)
. select("country", "cnt
You need to use `===`, so that you are constructing a column expression
instead of evaluating the standard scala equality method. Calling methods
to access columns (i.e. df.county is only supported in python).
val join_df = df1.join( df2, df1("country") === df2("country"),
"left_outer")
On Tue,
