经过一些处理后,我得到一个数据帧,我在数据帧列中有一个字典。现在我想更改列中字典的键。从“_ 1”到“product_id”和“_ 2”到“timestamp”。
以下是处理代码:
df1 = data.select("user_id","product_id","timestamp_gmt").rdd.map(lambda x: (x[0], (x[1],x[2]))).groupByKey()\
.map(lambda x:(x[0], list(x[1]))).toDF()\
.withColumnRenamed('_1', 'user_id')\
.withColumnRenamed('_2', 'purchase_info')
结果如下:
答案 0 :(得分:3)
Spark 2.0 +
使用collect_list
和struct
:
from pyspark.sql.functions import collect_list, struct, col
df = sc.parallelize([
(1, 100, "2012-01-01 00:00:00"),
(1, 200, "2016-04-04 00:00:01")
]).toDF(["user_id","product_id","timestamp_gmt"])
pi = (collect_list(struct(col("product_id"), col("timestamp_gmt")))
.alias("purchase_info"))
df.groupBy("user_id").agg(pi)
Spark< 2.0 强>
使用Rows
:
(df
.select("user_id", struct(col("product_id"), col("timestamp_gmt")))
.rdd.groupByKey()
.toDF(["user_id", "purchase_info"]))
这可以说是更优雅但是应该具有类似的效果来替换你传递给map
的函数:
lambda x: (x[0], Row(product_id=x[1], timestamp_gmt=x[2]))
另一方面,这些不是字典(MapType
),而是structs
(StructType
)。