如何在PySpark中更改数据帧列名?

我来自熊猫的背景，我习惯了从CSV文件读取数据到一个dataframe，然后简单地改变列名使用简单的命令有用的东西:

df.columns = new_column_name_list

然而，这在使用sqlContext创建的PySpark数据框架中是行不通的。我能想到的唯一解决办法是:

df = sqlContext.read.format("com.databricks.spark.csv").options(header='false', inferschema='true', delimiter='\t').load("data.txt")
oldSchema = df.schema
for i,k in enumerate(oldSchema.fields):
  k.name = new_column_name_list[i]
df = sqlContext.read.format("com.databricks.spark.csv").options(header='false', delimiter='\t').load("data.txt", schema=oldSchema)

这基本上是定义变量两次，首先推断模式，然后重命名列名，然后用更新的模式再次加载数据框架。

有没有更好更有效的方法来做到这一点，就像我们对熊猫做的那样?

我的Spark版本是1.5.0

当前回答

df。withColumnRenamed(“年龄”、“age2”)

2018-06-28 14:49:43

其他回答

另一种重命名一个列的方法(使用import pyspark.sql.functions as F):

df = df.select( '*', F.col('count').alias('new_count') ).drop('count')

2018-06-20 14:24:12


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()

2021-03-28 04:50:24

df = df.withColumnRenamed("colName", "newColName")\
       .withColumnRenamed("colName2", "newColName2")

使用这种方式的优点:对于一个很长的列列表，您只需要更改几个列名。这在这些场景中非常方便。在连接具有重复列名的表时非常有用。

2016-03-30 07:25:17

df。withColumnRenamed(“年龄”、“age2”)

2018-06-28 14:49:43

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。

2018-04-13 15:17:41

如何在PySpark中更改数据帧列名?

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