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

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")

其他回答

你可以使用多种方法:

df1 = df . withColumn(“new_column”、“old_column)。放下坳(old_column”) df1 = df . withColumn(“new_column”、“old_column”) 选择(“old_column df1 = df。别名(new_column”)

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()

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")

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)