在Pyspark df中将字典键添加为列名,并将字典值添加为该列的常量值

时间:2018-12-04 07:23:24

标签: pyspark apache-spark-sql pyspark-sql

我有一个字典x = {'colA': 20, 'colB': 30}和一个pyspark df。

ID Value
1  ABC
1  BCD
1  AKB
2  CAB
2  AIK
3  KIB 

我想使用x创建df1,如下所示:

ID Value colA colB
1  ABC    20.0  30.0
1  BCD    20.0  30.0
1  AKB    20.0  30.0
2  CAB    20.0  30.0
...

任何想法如何做到Pyspark。 我知道我可以像这样创建一个常量列,

df1 = df.withColumn('colA', lit(20.0))
df1 = df1.withColumn('colB', lit(30.0))

但不确定从字典执行此操作的动态过程

2 个答案:

答案 0 :(得分:1)

如下浏览字典

df1 = df  
for key in x:
    df1 = df1.withColumn(key, lit(x[key]))

答案 1 :(得分:1)

有很多方法可以隐藏循环,但是执行过程是相同的。例如,您可以使用select

from pyspark.sql.functions import lit

df2 = df.select("*", *[lit(val).alias(key) for key, val in x.items()])
df2.show()
#+---+-----+----+----+
#| ID|Value|colB|colA|
#+---+-----+----+----+
#|  1|  ABC|  30|  20|
#|  1|  BCD|  30|  20|
#|  1|  AKB|  30|  20|
#|  2|  CAB|  30|  20|
#|  2|  AIK|  30|  20|
#|  3|  KIB|  30|  20|
#+---+-----+----+----+

或者functools.reducewithColumn

from functools import reduce
df3 = reduce(lambda df, key: df.withColumn(key, lit(x[key])), x, df)
df3.show()
# Same as above

或将pyspark.sql.functions.structselect() and the "*" syntax

from pyspark.sql.functions import struct
df4 = df.withColumn('x', struct([lit(val).alias(key) for key, val in x.items()]))\
    .select("ID", "Value", "x.*")
df4.show()
#Same as above

但是,如果您查看这些方法的执行计划,您会发现它们是完全相同的:

df2.explain()
#== Physical Plan ==
#*Project [ID#44L, Value#45, 30 AS colB#151, 20 AS colA#152]
#+- Scan ExistingRDD[ID#44L,Value#45]

df3.explain()
#== Physical Plan ==
#*Project [ID#44L, Value#45, 30 AS colB#102, 20 AS colA#107]
#+- Scan ExistingRDD[ID#44L,Value#45]

df4.explain()
#== Physical Plan ==
#*Project [ID#44L, Value#45, 30 AS colB#120, 20 AS colA#121]
#+- Scan ExistingRDD[ID#44L,Value#45]

如果您比较@anil的answer中的循环方法,则:

df1 = df  
for key in x:
    df1 = df1.withColumn(key, lit(x[key]))
df1.explain()
#== Physical Plan ==
#*Project [ID#44L, Value#45, 30 AS colB#127, 20 AS colA#132]
#+- Scan ExistingRDD[ID#44L,Value#45]

您会发现它也是一样。