Looking to apply a string replace for all columns of a Polars DataFrame without specifying each column

Question

I’m trying to apply a string replace to a Polars DataFrame, similar to how you would apply it to a Pandas DataFrame.

I would like the equivalent to the following:

df = df.apply(lambda x: x.str.replace("A", "B")

I understand with Polars you can use the following:

`
df=df.with_columns(
    pl.col("col1").str.replace("A", "B"),
    pl.col("col2").str.replace("A", "B"))`

Is there a way to apply it to all columns? I have a very large DataFrame (100+ columns) and would prefer to not have to specify all columns.

I’ve tried using:

df = df.select([
    pl.all().map(lambda x: x.replace("A", "B"), df)
])

and I get an error: ValueError: Cannot infer dtype from 'shape: (150000, 150)

Is there something that I may be missing?

Asked By: EKCoote

||

Answer 1

You can select all columns of a certain type, or multiple columns. See Selectors for more.

df=df.with_columns(pl.col(pl.Utf8).str.replace("A", "B"))

pl.all().str … is fine too if the df is fully string columns.

Answered By: Wayoshi

Answer 2

Instead of your initial solution:

df = df.select([
    pl.all().map(lambda x: x.replace("A", "B"), df)
])

Try:

df = df.with_columns([
    pl.all().str.replace("A", "B")
])

Answered By: bilibraker

Question: