英文:
Sort list of datetime in Python polars dataframe
问题
I have problems working with datetime
我的问题出在处理日期时间数据上。
my dataset is like this:
我的数据集如下所示：
And I wish to sort the dates inside the list on every row, and get the first and second datetime, because some rows has more than 2 values. But I can't do it on Polars.
我希望对每一行中的日期进行排序，并获取第一个和第二个日期，因为有些行包含多于2个值。但我无法在Polars上实现这个。
I'm trying with :
我正在尝试使用：
filter_df.filter(pl.col("order_purchase_timestamp").arr.sort())
but I got this error:
但我收到了以下错误信息：
ComputeError: Filter predicate must be of type Boolean, got: SchemaMisMatch(Owned("Series of dtype: List(Datetime(Microseconds, None)) != Boolean"))
英文:
I have problems working with datetime
my dataset is like this:
And I wish to sort the dates inside the list on every row, and get the first and second datetime, because some rows has more than 2 values. But I can't do it on Polars.
I'm trying with :
filter_df.filter(pl.col("order_purchase_timestamp").arr.sort())
but I got this error:
> ComputeError: Filter predicate must be of type Boolean, got: SchemaMisMatch(Owned("Series of dtype: List(Datetime(Microseconds, None)) != Boolean"))
答案1
得分: 3
你可以使用.with_columns()上下文来选择和修改列。
col = pl.col("order_purchase_timestamp")df.with_columns(col.arr.sort(reverse=True).arr.slice(0, 2)) # / /# 按日期排序 取片段 [0:2]
从评论讨论中得出的解决方案
要根据条件在列下执行某些操作，您可以使用pl.when -> then -> otherwise结构。pl.when()接受某个布尔Series（条件）。了解更多信息这里。
col = pl.col("order_purchase_timestamp")# 2个日期之间的持续时间（微秒）diff = pl.duration(microseconds=(col.arr.get(0) - col.arr.get(1)))df.filter(col.arr.lengths() >= 2).with_columns(col.arr.sort(reverse=True).arr.slice(0, 2)).with_columns(less_30 = pl.when(diff <= pl.duration(days=30))\.then(True).otherwise(False))
英文:
You can use .with_columns() context to select & modify columns.
col = pl.col("order_purchase_timestamp")df.with_columns(col.arr.sort(reverse=True).arr.slice(0, 2)) # / /# sort by date take slice [0:2]
Solution that follows from the discussion in the comments
To perform some action under column based on condition, you can use pl.when -> then -> otherwise construct. pl.when() takes some boolean Series (condition). Read more here.
col = pl.col("order_purchase_timestamp")# duration (difference) between 2 dates (in microseconds)diff = pl.duration(microseconds=(col.arr.get(0) - col.arr.get(1)))df.filter(col.arr.lengths() >= 2).with_columns(col.arr.sort(reverse=True).arr.slice(0, 2)).with_columns(less_30 = pl.when(diff <= pl.duration(days=30))\.then(True).otherwise(False))
答案2
得分: 0
以下是代码部分的翻译：
-
Instead of operating on lists - you could
.explode()the column.
你可以使用.explode()列代替操作列表。 -
.with_row_count()can be used as a "group id":
.with_row_count()可以用作“group id”（分组标识）： -
You could then filter based on the group count ("size"):
之后，你可以基于分组计数进行筛选（"size"）： -
.groupby().agg()can be used to recreate the lists.
.groupby().agg()可用于重新创建列表。 -
.head/.limitto keep only the firstNresults.
使用.head/.limit来保留前N个结果。
英文:
Instead of operating on lists - you could .explode() the column.
df = pl.DataFrame({"letters": [["f", "a", "m"], ["b"], ["e", "d", "c"]]})
shape: (3, 1)┌─────────────────┐│ letters ││ --- ││ list[str] │╞═════════════════╡│ ["f", "a", "m"] ││ ["b"] ││ ["e", "d", "c"] │└─────────────────┘
.with_row_count() can be used as a "group id":
df.with_row_count("group").explode("letters").sort("letters")
shape: (7, 2)┌───────┬─────────┐│ group | letters ││ --- | --- ││ u32 | str │╞═══════╪═════════╡│ 0 | a ││ 1 | b ││ 2 | c ││ 2 | d ││ 2 | e ││ 0 | f ││ 0 | m │└───────┴─────────┘
You could then filter based on the group count ("size"):
(df.with_row_count("group").explode("letters").sort("letters").filter(pl.count().over("group") > 1))
shape: (6, 2)┌───────┬─────────┐│ group | letters ││ --- | --- ││ u32 | str │╞═══════╪═════════╡│ 0 | a ││ 2 | c ││ 2 | d ││ 2 | e ││ 0 | f ││ 0 | m │└───────┴─────────┘
.groupby().agg() can be used to recreate the lists.
.head / .limit to keep only the first N results.
(df.with_row_count("group").explode("letters").sort("letters").filter(pl.count().over("group") > 1).groupby("group").agg(pl.col("letters").head(2)))
shape: (2, 2)┌───────┬────────────┐│ group | letters ││ --- | --- ││ u32 | list[str] │╞═══════╪════════════╡│ 0 | ["a", "f"] ││ 2 | ["c", "d"] │└───────┴────────────┘
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