How do you express the identity expression?

Question

How do you express the identity expression in Polars?

By this I mean the expression idexpr that when you do lf.filter(idexpr) you get the entirety of lf.

Similar to SELECT(*) in SQL.

I'm resorting to a logical expression like

idexpr = (pl.col("a") == 0) | (pl.col("a") != 0)

Answer 1

According to Documentation, what's passed to filter as a predicate needs to be an "Expression(s) that evaluates to a boolean Series."

This you already know, since you are passing a logical expression to circumvent it. Easily enough, there's a very simple expression that always evaluates to true: pl.lit(True) or just True.

import polars as pl

df = pl.DataFrame({
    "id": [1, 2, 3],
    "name": ["Alice", "Bob", "Charlie"],
    "age": [24, 28, 23],
    "bool": [True, False, True]
})

print(df.filter(pl.lit(True)))

This gives the output of:

shape: (3, 4)
┌─────┬─────────┬─────┬───────┐
│ id  ┆ name    ┆ age ┆ bool  │
│ --- ┆ ---     ┆ --- ┆ ---   │
│ i64 ┆ str     ┆ i64 ┆ bool  │
╞═════╪═════════╪═════╪═══════╡
│ 1   ┆ Alice   ┆ 24  ┆ true  │
│ 2   ┆ Bob     ┆ 28  ┆ false │
│ 3   ┆ Charlie ┆ 23  ┆ true  │
└─────┴─────────┴─────┴───────┘

Edit:

Careful though, I found at least two cases where this does not work.

Series: This does not seem to work for them at all (polars 1.15):

>>> series = pl.Series("A", ["test", "test"])
>>> series.filter(pl.lit(True))
...
AttributeError: 'Expr' object has no attribute '_s'

I assume this is the case because pl.Series.filter works only with a mask anyway. This works however:

series.filter([True])

it also works for a dataframe it seems, so if you want to use the same solution for both series and dataframes, this is the one.

Null-column: When there's a column of type null (polars 1.12, seems to be fixed in 1.15 though):

>>> null_frame = pl.DataFrame({"A": [None, None], "B": ["test", "test"})
>>> null_frame.filter(pl.lit(True))
...
polars.exceptions.ShapeError: filter's length: 1 differs from that of the series: 2