pyspark：将字典数组转换为新列

Question

我正在努力转换我的 pyspark 数据框，它看起来像这样：

df = spark.createDataFrame([('0018aad4',[300, 450], ['{"v1": "blue"}', '{"v2": "red"}']), ('0018aad5',[300], ['{"v1": "blue"}'])],[ "id","Tlist", 'Tstring'])
df.show(2, False)

+--------+----------+-------------------------------+
|id      |Tlist     |Tstring                        |
+--------+----------+-------------------------------+
|0018aad4|[300, 450]|[{"v1": "blue"}, {"v2": "red"}]|
|0018aad5|[300]     |[{"v1": "blue"}]               |
+--------+----------+-------------------------------+

对此：

df_result = spark.createDataFrame([('0018aad4',[300, 450], 'blue', 'red'), ('0018aad5',[300], 'blue', None)],[ "id","Tlist", 'v1', 'v2'])
df_result.show(2, False)

+--------+----------+----+----+
|id      |Tlist     |v1  |v2  |
+--------+----------+----+----+
|0018aad4|[300, 450]|blue|red |
|0018aad5|[300]     |blue|null|
+--------+----------+----+----+

我尝试了旋转和其他一些东西，但没有得到上面的结果。

请注意，我在 Tstring

列中没有确切的字典数

你知道我该怎么做吗？

Answer 1

使用transform函数你可以将数组的每个元素转换成一个映射类型。之后，您可以使用 aggregate 函数获取一张地图，将其分解然后旋转键以获得所需的输出：

from pyspark.sql import functions as F

df1 = df.withColumn(
    "Tstring",
    F.transform("Tstring", lambda x: F.from_json(x, "map<string,string>"))
).withColumn(
    "Tstring",
    F.aggregate(
        F.expr("slice(Tstring, 2, size(Tstring))"), 
        F.col("Tstring")[0], 
        lambda acc, x: F.map_concat(acc, x)
    )
).select(
    "id", "Tlist", F.explode("Tstring")
).groupby(
    "id", "Tlist"
).pivot("key").agg(F.first("value"))


df1.show()
#+--------+----------+----+----+
#|id      |Tlist     |v1  |v2  |
#+--------+----------+----+----+
#|0018aad4|[300, 450]|blue|red |
#|0018aad5|[300]     |blue|null|
#+--------+----------+----+----+

我使用的是 Spark 3.1+，因此 higher-order 函数（例如 transform 可以在数据帧 API 中使用，但您可以使用 expr 执行相同的操作火花 <3.1:

df1 = (df.withColumn("Tstring", F.expr("transform(Tstring, x-> from_json(x, 'map<string,string>'))"))
       .withColumn("Tstring", F.expr("aggregate(slice(Tstring, 2, size(Tstring)), Tstring[0], (acc, x) -> map_concat(acc, x))"))
       .select("id", "Tlist", F.explode("Tstring"))
       .groupby("id", "Tlist")
       .pivot("key")
       .agg(F.first("value"))
       )

Answer 2

from pyspark.sql import SparkSession
from pyspark.sql import functions as F

from pyspark.sql.types import *
from datetime import datetime
from pyspark.sql import *
from collections import *
from pyspark.sql.functions import udf,explode
from pyspark.sql.types import StringType
from pyspark.context import SparkContext
from pyspark.sql.session import SparkSession

spark = SparkSession(sc)

df= spark.createDataFrame(
    [
        ('0018aad4', [{"val1":"blue", "val2":"red"}],[300,500]), 
         ('0018aad', [{"val1":"blue", "val2":"null"}],[300])
       
        ],("ID","List","Tlist")
    )

df2 = df.select(df.ID,explode(df.List).alias("Dict"),df.Tlist )
df2.withColumn("Val1", F.col("Dict").getItem("val1")).withColumn("Val2", F.col("Dict").getItem("val2")).show(truncate=False)


+--------+----------------------------+----------+----+----+
|ID      |Dict                        |Tlist     |Val1|Val2|
+--------+----------------------------+----------+----+----+
|0018aad4|{val2 -> red, val1 -> blue} |[300, 500]|blue|red |
|0018aad |{val2 -> null, val1 -> blue}|[300]     |blue|null|

+--------+----------------------------+------ ----+----+----+

这就是您要找的。

Answer 3

稍微 over-fitting 这个例子（您可能需要调整它以进行任何概括），您可以使用它们的索引从 Tstring 列中获取元素：

partial_results = df.withColumn("v1", df.Tstring[0]).withColumn("v2", df.Tstring[1])

+--------+----------+--------------+-------------+
|      id|     Tlist|            v1|           v2|
+--------+----------+--------------+-------------+
|0018aad4|[300, 450]|{"v1": "blue"}|{"v2": "red"}|
|0018aad5|     [300]|{"v1": "blue"}|         null|
+--------+----------+--------------+-------------+

有了这个你可以做一些清洁来达到想要的结果

from pyspark.sql.functions import regexp_replace


maximum_color_length = 100
wanted_df = df.withColumn(
    "v1",
    regexp_replace(df.Tstring[0].substr(9, maximum_color_length), r"\"\}", "")
).withColumn(
    "v2",
    regexp_replace(df.Tstring[1].substr(9, maximum_color_length), r"\"\}", "")
).drop(
    "Tstring"
)

+--------+----------+----+----+
|      id|     Tlist|  v1|  v2|
+--------+----------+----+----+
|0018aad4|[300, 450]|blue| red|
|0018aad5|     [300]|blue|null|
+--------+----------+----+----+

pyspark：将字典数组转换为新列

pyspark: turn array of dict to new columns

python

apache-spark

apache-spark-sql

pyspark