polars.DataFrame.apply#

DataFrame.apply(f: Callable[[tuple[Any, ...]], Any], return_dtype: Optional[Union[Type[DataType], DataType]] = None, inference_size: int = 256) DF[source]#

Apply a custom/user-defined function (UDF) over the rows of the DataFrame.

The UDF will receive each row as a tuple of values: udf(row).

Implementing logic using a Python function is almost always _significantly_ slower and more memory intensive than implementing the same logic using the native expression API because:

  • The native expression engine runs in Rust; UDFs run in Python.

  • Use of Python UDFs forces the DataFrame to be materialized in memory.

  • Polars-native expressions can be parallelised (UDFs cannot).

  • Polars-native expressions can be logically optimised (UDFs cannot).

Wherever possible you should strongly prefer the native expression API to achieve the best performance.

Parameters:
f

Custom function/ lambda function.

return_dtype

Output type of the operation. If none given, Polars tries to infer the type.

inference_size

Only used in the case when the custom function returns rows. This uses the first n rows to determine the output schema

Notes

The frame-level apply cannot track column names (as the UDF is a black-box that may arbitrarily drop, rearrange, transform, or add new columns); if you want to apply a UDF such that column names are preserved, you should use the expression-level apply syntax instead.

Examples

>>> df = pl.DataFrame({"foo": [1, 2, 3], "bar": [-1, 5, 8]})

Return a DataFrame by mapping each row to a tuple:

>>> df.apply(lambda t: (t[0] * 2, t[1] * 3))
shape: (3, 2)
┌──────────┬──────────┐
│ column_0 ┆ column_1 │
│ ---      ┆ ---      │
│ i64      ┆ i64      │
╞══════════╪══════════╡
│ 2        ┆ -3       │
├╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌┤
│ 4        ┆ 15       │
├╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌┤
│ 6        ┆ 24       │
└──────────┴──────────┘

It is better to implement this with an expression:

>>> (
...     df.select([pl.col("foo") * 2, pl.col("bar") * 3])
... )  

Return a Series by mapping each row to a scalar:

>>> df.apply(lambda t: (t[0] * 2 + t[1]))
shape: (3, 1)
┌───────┐
│ apply │
│ ---   │
│ i64   │
╞═══════╡
│ 1     │
├╌╌╌╌╌╌╌┤
│ 9     │
├╌╌╌╌╌╌╌┤
│ 14    │
└───────┘

In this case it is better to use the following expression:

>>> df.select(pl.col("foo") * 2 + pl.col("bar"))