如何在极坐标中给定任意步长（一步）将行转换为列？

Question

我已经设法分两步解决这个问题。

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  │
└───────┴────────┴───────┘
# 

我的印象是，这应该可以通过单个表达式链轻松解决，但我还没有做到。有什么建议吗？

Answer 1

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


(df.with_columns(
    (pl.int_range(pl.len()) // step).alias("step")
).group_by("step", maintain_order=True)
 .agg(
     pl.col("a").get(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  │
└──────┴───────┴────────┴───────┘