Add stable Expr.top_k

[MarcoGorelli]

Description

In #10054, there was a request for a way to answer the query:

For each group 'd', find the rows corresponding to the top k values from column 'b'

One possible API could have been: df.top_k(k=k, by='b', group_by='d'), or, as the OP suggested, df.group_by('d').top_k(k=k, by='b').

The response was that

df.group_by('d').agg(pl.all().top_k(k=1, by='b'))

is enough, and that's what's currently suggested in the top_k docs

However, as there's no ordering guarantees, then if there's ties in the by column, then the risk is that this solution produces a result with rows which never appeared in the original dataframe: #10054 (comment)

This was discussed in #15238, and the suggestion is now to introduce a stable Expr.top_k. This would solve the original issue