如何在字典中使用 pyspark.sql.functions.when() 的多个条件？

Question

我想根据字典中的值生成一个 when 子句。它与正在做的事情非常相似How do I use multiple conditions with pyspark.sql.funtions.when()?

只有我想传递一个列和值的字典

假设我有一个字典：

{
  'employed': 'Y',
  'athlete': 'N'
}

我想用那个 dict 来生成等价的：

df.withColumn("call_person",when((col("employed") == "Y") & (col("athlete") == "N"), "Y")

所以最终结果是：

+---+-----------+--------+-------+
| id|call_person|employed|athlete|
+---+-----------+--------+-------+
|  1|     Y     |    Y   |   N   |
|  2|     N     |    Y   |   Y   |
|  3|     N     |    N   |   N   |
+---+-----------+--------+-------+

请注意我想以编程方式执行此操作的部分原因是我有不同长度的字典（条件数）

Answer 1

使用reduce()函数：

from functools import reduce
from pyspark.sql.functions import when, col

# dictionary
d = {
  'employed': 'Y',
  'athlete': 'N'
}

# set up the conditions, multiple conditions merged with `&`
cond = reduce(lambda x,y: x&y, [ col(c) == v for c,v in d.items() if c in df.columns ])

# set up the new column
df.withColumn("call_person", when(cond, "Y").otherwise("N")).show()
+---+--------+-------+-----------+
| id|employed|athlete|call_person|
+---+--------+-------+-----------+
|  1|       Y|      N|          Y|
|  2|       Y|      Y|          N|
|  3|       N|      N|          N|
+---+--------+-------+-----------+

Answer 2

您也可以直接访问字典项：

dict ={
  'code': 'b',
  'amt': '4'
  }
list = [(1, 'code'),(1,'amt')]
df=spark.createDataFrame(list, ['id', 'dict_key'])

from pyspark.sql.functions import udf
from pyspark.sql.types import StringType

user_func =  udf (lambda x: dict.get(x), StringType())
newdf = df.withColumn('new_column',user_func(df.dict_key))

>>> newdf.show();
+---+--------+----------+
| id|dict_key|new_column|
+---+--------+----------+
|  1|    code|         b|
|  1|     amt|         4|
+---+--------+----------+

或广播字典

broadcast_dict = sc.broadcast(dict)

def my_func(key):
    return broadcast_dict.value.get(key)

new_my_func = udf(my_func, StringType())

newdf = df.withColumn('new_column',new_my_func(df.dict_key))
>>> newdf.show();
+---+--------+----------+
| id|dict_key|new_column|
+---+--------+----------+
|  1|    code|         b|
|  1|     amt|         4|
+---+--------+----------+