如何在 pyspark 中按有序分类变量创建和排序
how to create & sort by an ordered categorical variable in pyspark
我正在将一些代码从 pandas 迁移到 pyspark。我的源数据框如下所示:
a b c
0 1 insert 1
1 2 update 1
2 3 seed 1
3 4 insert 2
4 5 update 2
5 6 delete 2
6 7 snapshot 1
我正在应用的操作(在 python / pandas 中)是:
df.b = pd.Categorical(df.b, ordered=True, categories=['insert', 'seed', 'update', 'snapshot', 'delete'])
df.sort_values(['c', 'b'])
导致输出数据帧:
a b c
0 1 insert 1
2 3 seed 1
1 2 update 1
6 7 snapshot 1
3 4 insert 2
4 5 update 2
5 6 delete 2
我不确定如何最好地使用 pyspark 设置有序分类,我最初的方法是使用 case-when 创建一个新列并随后尝试使用它:
df = df.withColumn(
"_precedence",
when(col("b") == "insert", 1)
.when(col("b") == "seed", 2)
.when(col("b") == "update", 3)
.when(col("b") == "snapshot", 4)
.when(col("b") == "delete", 5)
)
您可以使用地图:
from pyspark.sql.functions import create_map, lit, col
categories=['insert', 'seed', 'update', 'snapshot', 'delete']
# per @HaleemurAli, adjusted the below list comprehension to create map
map1 = create_map([val for (i, c) in enumerate(categories) for val in (c, lit(i))])
#Column<b'map(insert, 0, seed, 1, update, 2, snapshot, 3, delete, 4)'>
df.orderBy('c', map1[col('b')]).show()
+---+---+--------+---+
| id| a| b| c|
+---+---+--------+---+
| 0| 1| insert| 1|
| 2| 3| seed| 1|
| 1| 2| update| 1|
| 6| 7|snapshot| 1|
| 3| 4| insert| 2|
| 4| 5| update| 2|
| 5| 6| delete| 2|
+---+---+--------+---+
反转 b 列的顺序:df.orderBy('c', map1[col('b')].desc()).show()
你也可以使用 coalesce
和你的 when statements
.
from pyspark.sql import functions as F
categories=['insert', 'seed', 'update', 'snapshot', 'delete']
cols=[(F.when(F.col("b")==x,F.lit(y))) for x,y in zip(categories,[x for x in (range(1, len(categories)+1))])]
df.orderBy("c",F.coalesce(*cols)).show()
#+---+--------+---+
#| a| b| c|
#+---+--------+---+
#| 1| insert| 1|
#| 3| seed| 1|
#| 2| update| 1|
#| 7|snapshot| 1|
#| 4| insert| 2|
#| 5| update| 2|
#| 6| delete| 2|
#+---+--------+---+
我正在将一些代码从 pandas 迁移到 pyspark。我的源数据框如下所示:
a b c
0 1 insert 1
1 2 update 1
2 3 seed 1
3 4 insert 2
4 5 update 2
5 6 delete 2
6 7 snapshot 1
我正在应用的操作(在 python / pandas 中)是:
df.b = pd.Categorical(df.b, ordered=True, categories=['insert', 'seed', 'update', 'snapshot', 'delete'])
df.sort_values(['c', 'b'])
导致输出数据帧:
a b c
0 1 insert 1
2 3 seed 1
1 2 update 1
6 7 snapshot 1
3 4 insert 2
4 5 update 2
5 6 delete 2
我不确定如何最好地使用 pyspark 设置有序分类,我最初的方法是使用 case-when 创建一个新列并随后尝试使用它:
df = df.withColumn(
"_precedence",
when(col("b") == "insert", 1)
.when(col("b") == "seed", 2)
.when(col("b") == "update", 3)
.when(col("b") == "snapshot", 4)
.when(col("b") == "delete", 5)
)
您可以使用地图:
from pyspark.sql.functions import create_map, lit, col
categories=['insert', 'seed', 'update', 'snapshot', 'delete']
# per @HaleemurAli, adjusted the below list comprehension to create map
map1 = create_map([val for (i, c) in enumerate(categories) for val in (c, lit(i))])
#Column<b'map(insert, 0, seed, 1, update, 2, snapshot, 3, delete, 4)'>
df.orderBy('c', map1[col('b')]).show()
+---+---+--------+---+
| id| a| b| c|
+---+---+--------+---+
| 0| 1| insert| 1|
| 2| 3| seed| 1|
| 1| 2| update| 1|
| 6| 7|snapshot| 1|
| 3| 4| insert| 2|
| 4| 5| update| 2|
| 5| 6| delete| 2|
+---+---+--------+---+
反转 b 列的顺序:df.orderBy('c', map1[col('b')].desc()).show()
你也可以使用 coalesce
和你的 when statements
.
from pyspark.sql import functions as F
categories=['insert', 'seed', 'update', 'snapshot', 'delete']
cols=[(F.when(F.col("b")==x,F.lit(y))) for x,y in zip(categories,[x for x in (range(1, len(categories)+1))])]
df.orderBy("c",F.coalesce(*cols)).show()
#+---+--------+---+
#| a| b| c|
#+---+--------+---+
#| 1| insert| 1|
#| 3| seed| 1|
#| 2| update| 1|
#| 7|snapshot| 1|
#| 4| insert| 2|
#| 5| update| 2|
#| 6| delete| 2|
#+---+--------+---+