polars.DataFrame.apply#

DataFrame.apply(
function: Callable[[tuple[Any, ...]], Any],
return_dtype: PolarsDataType | None = None,
*,
inference_size: int = 256,
) → DataFrame[source]#

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

Warning

This method is much slower than the native expressions API. Only use it if you cannot implement your logic otherwise.

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 typically 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:
function

Custom function or lambda.

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.

  • If your function is expensive and you don’t want it to be called more than once for a given input, consider applying an @lru_cache decorator to it. With suitable data you may achieve order-of-magnitude speedups (or more).

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       │
└──────────┴──────────┘

However, it is much better to implement this with a native expression:

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

Return a DataFrame with a single column 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 native expression:

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