247
votes

I come from pandas background and am used to reading data from CSV files into a dataframe and then simply changing the column names to something useful using the simple command:

df.columns = new_column_name_list

However, the same doesn't work in pyspark dataframes created using sqlContext. The only solution I could figure out to do this easily is the following:

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)

This is basically defining the variable twice and inferring the schema first then renaming the column names and then loading the dataframe again with the updated schema.

Is there a better and more efficient way to do this like we do in pandas ?

My spark version is 1.5.0

18

18 Answers

410
votes

There are many ways to do that:

  • Option 1. Using selectExpr.

     data = sqlContext.createDataFrame([("Alberto", 2), ("Dakota", 2)], 
                                       ["Name", "askdaosdka"])
     data.show()
     data.printSchema()
    
     # Output
     #+-------+----------+
     #|   Name|askdaosdka|
     #+-------+----------+
     #|Alberto|         2|
     #| Dakota|         2|
     #+-------+----------+
    
     #root
     # |-- Name: string (nullable = true)
     # |-- askdaosdka: long (nullable = true)
    
     df = data.selectExpr("Name as name", "askdaosdka as age")
     df.show()
     df.printSchema()
    
     # Output
     #+-------+---+
     #|   name|age|
     #+-------+---+
     #|Alberto|  2|
     #| Dakota|  2|
     #+-------+---+
    
     #root
     # |-- name: string (nullable = true)
     # |-- age: long (nullable = true)
    
  • Option 2. Using withColumnRenamed, notice that this method allows you to "overwrite" the same column. For Python3, replace xrange with range.

     from functools import reduce
    
     oldColumns = data.schema.names
     newColumns = ["name", "age"]
    
     df = reduce(lambda data, idx: data.withColumnRenamed(oldColumns[idx], newColumns[idx]), xrange(len(oldColumns)), data)
     df.printSchema()
     df.show()
    
  • Option 3. using alias, in Scala you can also use as.

     from pyspark.sql.functions import col
    
     data = data.select(col("Name").alias("name"), col("askdaosdka").alias("age"))
     data.show()
    
     # Output
     #+-------+---+
     #|   name|age|
     #+-------+---+
     #|Alberto|  2|
     #| Dakota|  2|
     #+-------+---+
    
  • Option 4. Using sqlContext.sql, which lets you use SQL queries on DataFrames registered as tables.

     sqlContext.registerDataFrameAsTable(data, "myTable")
     df2 = sqlContext.sql("SELECT Name AS name, askdaosdka as age from myTable")
    
     df2.show()
    
     # Output
     #+-------+---+
     #|   name|age|
     #+-------+---+
     #|Alberto|  2|
     #| Dakota|  2|
     #+-------+---+
    
226
votes
df = df.withColumnRenamed("colName", "newColName")\
       .withColumnRenamed("colName2", "newColName2")

Advantage of using this way: With long list of columns you would like to change only few column names. This can be very convenient in these scenarios. Very useful when joining tables with duplicate column names.

83
votes

If you want to change all columns names, try df.toDF(*cols)

61
votes

In case you would like to apply a simple transformation on all column names, this code does the trick: (I am replacing all spaces with underscore)

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

df = df.toDF(*new_column_name_list)

Thanks to @user8117731 for toDf trick.

18
votes

If you want to rename a single column and keep the rest as it is:

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])
18
votes

df.withColumnRenamed('age', 'age2')

16
votes

this is the approach that I used:

create pyspark session:

import pyspark
from pyspark.sql import SparkSession
spark = SparkSession.builder.appName('changeColNames').getOrCreate()

create dataframe:

df = spark.createDataFrame(data = [('Bob', 5.62,'juice'),  ('Sue',0.85,'milk')], schema = ["Name", "Amount","Item"])

view df with column names:

df.show()
+----+------+-----+
|Name|Amount| Item|
+----+------+-----+
| Bob|  5.62|juice|
| Sue|  0.85| milk|
+----+------+-----+

create a list with new column names:

newcolnames = ['NameNew','AmountNew','ItemNew']

change the column names of the df:

for c,n in zip(df.columns,newcolnames):
    df=df.withColumnRenamed(c,n)

view df with new column names:

df.show()
+-------+---------+-------+
|NameNew|AmountNew|ItemNew|
+-------+---------+-------+
|    Bob|     5.62|  juice|
|    Sue|     0.85|   milk|
+-------+---------+-------+
12
votes

I made an easy to use function to rename multiple columns for a pyspark dataframe, in case anyone wants to use it:

def renameCols(df, old_columns, new_columns):
    for old_col,new_col in zip(old_columns,new_columns):
        df = df.withColumnRenamed(old_col,new_col)
    return df

old_columns = ['old_name1','old_name2']
new_columns = ['new_name1', 'new_name2']
df_renamed = renameCols(df, old_columns, new_columns)

Be careful, both lists must be the same length.

8
votes

Another way to rename just one column (using import pyspark.sql.functions as F):

df = df.select( '*', F.col('count').alias('new_count') ).drop('count')
5
votes

You can use the following function to rename all the columns of your dataframe.

def df_col_rename(X, to_rename, replace_with):
    """
    :param X: spark dataframe
    :param to_rename: list of original names
    :param replace_with: list of new names
    :return: dataframe with updated names
    """
    import pyspark.sql.functions as F
    mapping = dict(zip(to_rename, replace_with))
    X = X.select([F.col(c).alias(mapping.get(c, c)) for c in to_rename])
    return X

In case you need to update only a few columns' names, you can use the same column name in the replace_with list

To rename all columns

df_col_rename(X,['a', 'b', 'c'], ['x', 'y', 'z'])

To rename a some columns

df_col_rename(X,['a', 'b', 'c'], ['a', 'y', 'z'])
3
votes

I use this one:

from pyspark.sql.functions import col
df.select(['vin',col('timeStamp').alias('Date')]).show()
3
votes

We can use various approaches to rename the column name.

First, let create a simple DataFrame.

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

Now let's try to rename col_1 to col_3. PFB a few approaches to do the same.

# 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()

Here is the output.

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

I hope this helps.

3
votes

You can put into for loop, and use zip to pairs each column name in two array.

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)
3
votes

Method 1:

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

Method 2: If you want to do some computation and rename the new values

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")
2
votes

I like to use a dict to rename the df.

rename = {'old1': 'new1', 'old2': 'new2'}
for col in df.schema.names:
    df = df.withColumnRenamed(col, rename[col])
1
votes

For a single column rename, you can still use toDF(). For example,

df1.selectExpr("SALARY*2").toDF("REVISED_SALARY").show()
1
votes

There are multiple approaches you can use:

  1. df1=df.withColumn("new_column","old_column").drop(col("old_column"))

  2. df1=df.withColumn("new_column","old_column")

  3. df1=df.select("old_column".alias("new_column"))

0
votes

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()