列表理解+ f-string:

df = df.toDF(*[f'n_{c}' for c in df.columns])

简单的列表理解:

df = df.toDF(*[c.lower() for c in df.columns])

我们可以使用各种方法重命名列名。

首先，让我们创建一个简单的数据框架。

df = spark.createDataFrame([("x", 1), ("y", 2)], 
                                  ["col_1", "col_2"])

现在我们试着把col_1重命名为col_3。PFB的几个方法也一样。

# Approach - 1 : using withColumnRenamed function.
df.withColumnRenamed("col_1", "col_3").show()

# Approach - 2 : using alias function.
df.select(df["col_1"].alias("col3"), "col_2").show()

# Approach - 3 : using selectExpr function.
df.selectExpr("col_1 as col_3", "col_2").show()

# Rename all columns
# Approach - 4 : using toDF function. Here you need to pass the list of all columns present in DataFrame.
df.toDF("col_3", "col_2").show()

这是输出。

+-----+-----+
|col_3|col_2|
+-----+-----+
|    x|    1|
|    y|    2|
+-----+-----+

我希望这能有所帮助。

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()

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)

方法1:

df = df.withColumnRenamed("old_column_name", "new_column_name")

方法2: 如果你想做一些计算并重命名新值

df = df.withColumn("old_column_name", F.when(F.col("old_column_name") > 1, F.lit(1)).otherwise(F.col("old_column_name"))
df = df.drop("new_column_name", "old_column_name")