英文:
Make a new column with list of unique values grouped by, or over, another column in Polars
问题
在 Polars 0.18.0 之前，我能够通过类型创建一个包含所有独特宝可梦的列。我看到 Expr.list() 已经重构为 implode()，但是我在使用新语法时遇到了困难，无法复制以下操作：
df.with_columns(lst_of_pokemon = pl.col('name').unique().list().over('Type 1'))
英文:
Before Polars 0.18.0, I was able to create a column with a list of all unique pokemon by type. I see Expr.list() has been refactored to implode(), but I'm having trouble replicating the following using the new syntax:
df.with_columns(lst_of_pokemon = pl.col('name').unique().list().over('Type 1'))
答案1
得分: 2
mapping_strategy= 参数已添加到 .over 中。
df = pl.from_repr("""┌──────┬──────┐│ name ┆ type ││ --- ┆ --- ││ str ┆ i64 │╞══════╪══════╡│ a ┆ 1 ││ a ┆ 1 ││ a ┆ 2 ││ b ┆ 2 ││ c ┆ 3 ││ c ┆ 3 │└──────┴──────┘""")df.with_columns(lst_of_pokemon =pl.col('name').unique().over('type', mapping_strategy='join'))
shape: (6, 3)┌──────┬──────┬────────────────┐│ name ┆ type ┆ lst_of_pokemon ││ --- ┆ --- ┆ --- ││ str ┆ i64 ┆ list[str] │╞══════╪══════╪════════════════╡│ a ┆ 1 ┆ ["a"] ││ a ┆ 1 ┆ ["a"] ││ a ┆ 2 ┆ ["a", "b"] ││ b ┆ 2 ┆ ["a", "b"] ││ c ┆ 3 ┆ ["c"] ││ c ┆ 3 ┆ ["c"] │└──────┴──────┴────────────────┘
英文:
The mapping_strategy= argument for .over was added.
df = pl.from_repr("""┌──────┬──────┐│ name ┆ type ││ --- ┆ --- ││ str ┆ i64 │╞══════╪══════╡│ a ┆ 1 ││ a ┆ 1 ││ a ┆ 2 ││ b ┆ 2 ││ c ┆ 3 ││ c ┆ 3 │└──────┴──────┘""")df.with_columns(lst_of_pokemon =pl.col('name').unique().over('type', mapping_strategy='join'))
shape: (6, 3)┌──────┬──────┬────────────────┐│ name ┆ type ┆ lst_of_pokemon ││ --- ┆ --- ┆ --- ││ str ┆ i64 ┆ list[str] │╞══════╪══════╪════════════════╡│ a ┆ 1 ┆ ["a"] ││ a ┆ 1 ┆ ["a"] ││ a ┆ 2 ┆ ["a", "b"] ││ b ┆ 2 ┆ ["a", "b"] ││ c ┆ 3 ┆ ["c"] ││ c ┆ 3 ┆ ["c"] │└──────┴──────┴────────────────┘
答案2
得分: 1
编辑：jqurios发布的答案更加优雅
我认为以下代码可以生成所需的结果：
df.join(df.groupby("Type 1", maintain_order=True).agg(pl.col("Name").alias("combined_names")),on="Type 1")
使用groupby和agg，您可以创建按'Type 1'分组的所有宝可梦名称的列表，然后将其与原始数据框连接起来。
英文:
Edit: the answer posted by jqurios is way more elegant
In think the following produces the required result:
df.join(df.groupby("Type 1", maintain_order=True).agg(pl.col("Name").alias("combined_names")),on="Type 1")
With the groupby and agg you can create the lists with all the pokemon names per 'Type 1', which is joined to the original dataframe.
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