21 февр. 2023 г. · How about the rows() method? df = pl.DataFrame( { "a": [1, 3, 5], "b": [2, 4, 6], } ) df.rows(). [(1, 2), (3, 4), (5, 6)]

4 мар. 2024 г. · You could do this. import polars as pl df = pl.DataFrame( { 'Letter': ['A', 'A', 'B', 'B', 'B', 'C', 'C', 'D','D','E'], 'Value': [1, 2, 3, ...

20 дек. 2023 г. · As long as your string column is valid JSON, you could use polars.Expr.str.json_decode as follows.

18 февр. 2024 г. · You can do the same, but columns= is replaced with schema= pl.DataFrame(table[1:], schema=table[0])

17 нояб. 2023 г. · Each column in Polars has a schema ("type"). If we take {"a": 1}, {"b": 2} as an example: df = pl.select(pl.concat_list(pl.struct(a=1), ...

3 июл. 2023 г. · Append items within a list to Polars Dataframe · python-3.x · python-polars · Share.

11 мая 2024 г. · Performance · Elapsed time: 0.45117 seconds. For comparison, if we perform the first list.eval step in your example: · Elapsed time: 1.99926 ...

1 сент. 2023 г. · So you can explode a column in a select context but you can't do that when other columns are present since then the dimension of the columns ...

25 мар. 2024 г. · For a fixed column of type pl.List(...) , you could simply use an pl.when().then() construct as follows. import polars as pl df = pl.

12 июн. 2024 г. · As already mentioned in the comments, there is pl.Expr.list.join in polars' native expression API to join all string items in a sublist with ...