这样怎么样？请注意 Result 列被视为字符串类型而不是dict。

import pyspark.sql.functions as f
from pyspark.sql.types import Row

row_list = [Row(age=24, payloadId=1, salary=2900)]
row_add = {'age_condition':True,'salary_condition':True}

spark.createDataFrame(row_list) \
  .withColumn('Result', f.lit(str(row_add))) \
  .collect()

[Row(age=24, payloadId=1, salary=2900, Result="{'age_condition': True, 'salary_condition': True}")]

我不知道你为什么要把字典添加到dataframe列中，从而使事情复杂化，你应该添加两个新列 age_condition 以及 salary_condition 而不是布尔类型。

这应该是你想要的。。。

from pyspark.sql.types import *

schema = StructType([StructField("dict", StructType([StructField("age_condition", BooleanType(), True), StructField("salary_condition", BooleanType(), True)]), True)])

newDf = spark.createDataFrame([{'age_condition':True,'salary_condition':True}], schema=schema)

df = spark.read.json("/whatever/json/path")

df.crossJoin(newDf) #no of records is same as in df as no of records in newDf is 1