from pyspark.sql.types import StructType,StructField, StringType, IntegerType

CreatingDataFrame = [("James","Sales","NY",90000,34,10000),
    ("Michael","Sales","NY",86000,56,20000),
    ("Robert","Sales","CA",81000,30,23000),
    ("Maria","Finance","CA",90000,24,23000),
    ("Raman","Finance","CA",99000,40,24000),
    ("Scott","Finance","NY",83000,36,19000),
    ("Jen","Finance","NY",79000,53,15000),
    ("Jeff","Marketing","CA",80000,25,18000),
    ("Kumar","Marketing","NY",91000,50,21000)
  ]

schema = StructType([ \
    StructField("employee_name",StringType(),True), \
    StructField("department",StringType(),True), \
    StructField("state",StringType(),True), \
    StructField("salary", IntegerType(), True), \
    StructField("age", StringType(), True), \
    StructField("bonus", IntegerType(), True) \
  ])

 
OurData = spark.createDataFrame(data=CreatingDataFrame,schema=schema)

OurData.show()

# COMMAND ----------

GrouppedBonusData=OurData.groupBy("department").sum("bonus")


# COMMAND ----------

GrouppedBonusData.show()


# COMMAND ----------

GrouppedBonusData.printSchema()

# COMMAND ----------

from pyspark.sql.functions import col

BonusColumnRenamed = GrouppedBonusData.select(col("department").alias("department"), col("sum(bonus)").alias("Total_Bonus"))
BonusColumnRenamed.show()

# COMMAND ----------

GrouppedBonusData.groupBy("department").count().show()

# COMMAND ----------

GrouppedSalaryData=OurData.groupBy("department").sum("salary")

# COMMAND ----------

GrouppedSalaryData.show()

# COMMAND ----------

from pyspark.sql.functions import col

SalaryColumnRenamed = GrouppedSalaryData.select(col("department").alias("Department"), col("sum(salary)").alias("Total_Salary"))
SalaryColumnRenamed.show()

如果你想重命名一个列，并保持其他列不变:

from pyspark.sql.functions import col
new_df = old_df.select(*[col(s).alias(new_name) if s == column_to_change else s for s in old_df.columns])

如果你想对所有列名应用一个简单的转换，这段代码可以做到:(我用下划线替换所有空格)

new_column_name_list= list(map(lambda x: x.replace(" ", "_"), df.columns))

df = df.toDF(*new_column_name_list)

感谢@user8117731的toDf技巧。

试试下面的方法。下面的方法允许您重命名多个文件的列

参考:https://www.linkedin.com/pulse/pyspark-methods-rename-columns-kyle-gibson/

df_initial = spark.read.load('com.databricks.spark.csv')
    
    rename_dict = {
      'Alberto':'Name',
      'Dakota':'askdaosdka'
    }
    
    df_renamed = df_initial \
    .select([col(c).alias(rename_dict.get(c, c)) for c in df_initial.columns])

    
     rename_dict = {
       'FName':'FirstName',
       'LName':'LastName',
       'DOB':'BirthDate'
        }

     return df.select([col(c).alias(rename_dict.get(c, c)) for c in df.columns])


df_renamed = spark.read.load('/mnt/datalake/bronze/testData') \
.transform(renameColumns)

我喜欢使用字典重命名df。

rename = {'old1': 'new1', 'old2': 'new2'}
for col in df.schema.names:
    df = df.withColumnRenamed(col, rename[col])

您可以放入for循环，并使用zip将两个数组中的每个列名配对。

new_name = ["id", "sepal_length_cm", "sepal_width_cm", "petal_length_cm", "petal_width_cm", "species"]

new_df = df
for old, new in zip(df.columns, new_name):
    new_df = new_df.withColumnRenamed(old, new)