How to convert rows into columns given an arbitrary step size in polars (in a single step)?

user_489010 0
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I've managed to solve this problem in two steps.

import polars as pl
text = "a brown fox jumps over a lazy dog's head"
step = 3
df = pl.DataFrame({"a":text.split(" ")})


first =  df.filter(pl.arange(0, pl.count())%step==0)
second = df.filter(pl.arange(0, pl.count())%step==1)
third=   df.filter(pl.arange(0, pl.count())%step==2)

dff = (
    pl.DataFrame({
        'first':first['a'], 
        'second':second['a'], 
        'third':third['a']})
)
print(dff)
shape: (3, 3)
┌───────┬────────┬───────┐
│ first ┆ second ┆ third │
│ ---   ┆ ---    ┆ ---   │
│ str   ┆ str    ┆ str   │
╞═══════╪════════╪═══════╡
│ a     ┆ brown  ┆ fox   │
├╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┤
│ jumps ┆ over   ┆ a     │
├╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┤
│ lazy  ┆ dog's  ┆ head  │
└───────┴────────┴───────┘
#

I have the impression that this should be easily solved in a single chain of expressions but I haven't managed to do so. Any suggestions?

python-polars rust-polars
1 Answers
text = "a brown fox jumps over a lazy dog's head"
step = 3
df = pl.DataFrame({"a":text.split(" ")})


(df.with_column(
    (pl.arange(0, pl.count()) // step).alias("step")
).groupby("step", maintain_order=True)
 .agg([
     pl.col("a").take(i).alias(name) for i, name in enumerate(["first", "second", "third"])
 ]))

shape: (3, 4)
┌──────┬───────┬────────┬───────┐
│ step ┆ first ┆ second ┆ third │
│ ---  ┆ ---   ┆ ---    ┆ ---   │
│ i64  ┆ str   ┆ str    ┆ str   │
╞══════╪═══════╪════════╪═══════╡
│ 0    ┆ a     ┆ brown  ┆ fox   │
├╌╌╌╌╌╌┼╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┤
│ 1    ┆ jumps ┆ over   ┆ a     │
├╌╌╌╌╌╌┼╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┤
│ 2    ┆ lazy  ┆ dog's  ┆ head  │
└──────┴───────┴────────┴───────┘
user_67170540
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