How to remove last N chars from a string column in python-polars?

Question

Given this dataframe:

df = pl.DataFrame({"s": ["pear", None, "papaya", "dragonfruit"]})

I want to remove the last X chars, e.g. remove the last 2 chars from the column. This obviously doesn't do what I want:

df.with_columns(   
    pl.col("s").str.slice(2).alias("s_sliced"),

)

I'd like the result to be:

shape: (4, 2)
┌─────────────┬──────────┐
│ s           ┆ s_sliced │
│ ---         ┆ ---      │
│ str         ┆ str      │
╞═════════════╪══════════╡
│ pear        ┆ pe       │
│ null        ┆ null     │
│ papaya      ┆ papa     │
│ dragonfruit ┆ dragonfru|  

Answer 1

You could use a regex with .str.replace

• . matches a "single character"
• $ matches the "end"
• {N} matches exactly N times

Meaning we could use ..$ or .{2}$

df.with_columns(   
    pl.col("s").str.replace(r"..$", "").alias("s_sliced"),
)

shape: (4, 2)
┌─────────────┬───────────┐
│ s           ┆ s_sliced  │
│ ---         ┆ ---       │
│ str         ┆ str       │
╞═════════════╪═══════════╡
│ pear        ┆ pe        │
│ null        ┆ null      │
│ papaya      ┆ papa      │
│ dragonfruit ┆ dragonfru │
└─────────────┴───────────┘

Answer 2

A non-regex way to do this is pretty ugly but here it is...

df.with_columns(
    s_sliced=(pl.col('s').str.explode().implode().over('s')
                .list.take(pl.arange(0,(pl.col('s').str.n_chars()-2))))
            .list.eval(pl.element().str.concat("")).list.get(0))

The first thing to notice is that we're converting the character column into a list using str.explode.implode, this is because str.slice doesn't take an expression for the length but pl.arange does. Then we just use take the elements we want which is all over them except the last 2 Lastly, we have to convert our list back into a string.

** There was a really concise answer using a simply regex but its author deleted it and I'm reluctant to copy it. Maybe the post's author will undelete it.