I want to add a column in a DataFrame with some arbitrary value (that is the same for each row). I get an error when I use withColumn as follows:

dt.withColumn('new_column', 10).head(5)

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-50-a6d0257ca2be> in <module>()
      1 dt = (messages
      2     .select(messages.fromuserid, messages.messagetype, floor(messages.datetime/(1000*60*5)).alias("dt")))
----> 3 dt.withColumn('new_column', 10).head(5)

/Users/evanzamir/spark-1.4.1/python/pyspark/sql/dataframe.pyc in withColumn(self, colName, col)
   1166         [Row(age=2, name=u'Alice', age2=4), Row(age=5, name=u'Bob', age2=7)]
   1167         """
-> 1168         return self.select('*', col.alias(colName))
   1169 
   1170     @ignore_unicode_prefix

AttributeError: 'int' object has no attribute 'alias'

It seems that I can trick the function into working as I want by adding and subtracting one of the other columns (so they add to zero) and then adding the number I want (10 in this case):

dt.withColumn('new_column', dt.messagetype - dt.messagetype + 10).head(5)

[Row(fromuserid=425, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=47019141, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=49746356, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=93506471, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=80488242, messagetype=1, dt=4809600.0, new_column=10)]

This is supremely hacky, right? I assume there is a more legit way to do this?

Spark 2.2+

Spark 2.2 introduces typedLit to support Seq, Map, and Tuples (SPARK-19254) and following calls should be supported (Scala):

import org.apache.spark.sql.functions.typedLit

df.withColumn("some_array", typedLit(Seq(1, 2, 3)))
df.withColumn("some_struct", typedLit(("foo", 1, 0.3)))
df.withColumn("some_map", typedLit(Map("key1" -> 1, "key2" -> 2)))

Spark 1.3+ (lit), 1.4+ (array, struct), 2.0+ (map):

The second argument for DataFrame.withColumn should be a Column so you have to use a literal:

from pyspark.sql.functions import lit

df.withColumn('new_column', lit(10))

If you need complex columns you can build these using blocks like array:

from pyspark.sql.functions import array, create_map, struct

df.withColumn("some_array", array(lit(1), lit(2), lit(3)))
df.withColumn("some_struct", struct(lit("foo"), lit(1), lit(.3)))
df.withColumn("some_map", create_map(lit("key1"), lit(1), lit("key2"), lit(2)))

Exactly the same methods can be used in Scala.

import org.apache.spark.sql.functions.{array, lit, map, struct}

df.withColumn("new_column", lit(10))
df.withColumn("map", map(lit("key1"), lit(1), lit("key2"), lit(2)))

To provide names for structs use either alias on each field:

df.withColumn(
    "some_struct",
    struct(lit("foo").alias("x"), lit(1).alias("y"), lit(0.3).alias("z"))
 )

or cast on the whole object

df.withColumn(
    "some_struct", 
    struct(lit("foo"), lit(1), lit(0.3)).cast("struct<x: string, y: integer, z: double>")
 )

It is also possible, although slower, to use an UDF.

Note:

The same constructs can be used to pass constant arguments to UDFs or SQL functions.

In spark 2.2 there are two ways to add constant value in a column in DataFrame:

1) Using lit

2) Using typedLit.

The difference between the two is that typedLit can also handle parameterized scala types e.g. List, Seq, and Map

Sample DataFrame:

val df = spark.createDataFrame(Seq((0,"a"),(1,"b"),(2,"c"))).toDF("id", "col1")

+---+----+
| id|col1|
+---+----+
|  0|   a|
|  1|   b|
+---+----+

1) Using lit: Adding constant string value in new column named newcol:

import org.apache.spark.sql.functions.lit
val newdf = df.withColumn("newcol",lit("myval"))

Result:

+---+----+------+
| id|col1|newcol|
+---+----+------+
|  0|   a| myval|
|  1|   b| myval|
+---+----+------+

2) Using typedLit:

import org.apache.spark.sql.functions.typedLit
df.withColumn("newcol", typedLit(("sample", 10, .044)))

Result:

+---+----+-----------------+
| id|col1|           newcol|
+---+----+-----------------+
|  0|   a|[sample,10,0.044]|
|  1|   b|[sample,10,0.044]|
|  2|   c|[sample,10,0.044]|
+---+----+-----------------+

I have a Spark DataFrame (using PySpark 1.5.1) and would like to add a new column.

I’ve tried the following without any success:

type(randomed_hours) # => list

# Create in Python and transform to RDD

new_col = pd.DataFrame(randomed_hours, columns=['new_col'])

spark_new_col = sqlContext.createDataFrame(new_col)

my_df_spark.withColumn("hours", spark_new_col["new_col"])

Also got an error using this:

my_df_spark.withColumn("hours",  sc.parallelize(randomed_hours))

So how do I add a new column (based on Python vector) to an existing DataFrame with PySpark?

You cannot add an arbitrary column to a DataFrame in Spark. New columns can be created only by using literals (other literal types are described in How to add a constant column in a Spark DataFrame?)

from pyspark.sql.functions import lit

df = sqlContext.createDataFrame(
    [(1, "a", 23.0), (3, "B", -23.0)], ("x1", "x2", "x3"))

df_with_x4 = df.withColumn("x4", lit(0))
df_with_x4.show()

## +---+---+-----+---+
## | x1| x2|   x3| x4|
## +---+---+-----+---+
## |  1|  a| 23.0|  0|
## |  3|  B|-23.0|  0|
## +---+---+-----+---+

transforming an existing column:

from pyspark.sql.functions import exp

df_with_x5 = df_with_x4.withColumn("x5", exp("x3"))
df_with_x5.show()

## +---+---+-----+---+--------------------+
## | x1| x2|   x3| x4|                  x5|
## +---+---+-----+---+--------------------+
## |  1|  a| 23.0|  0| 9.744803446248903E9|
## |  3|  B|-23.0|  0|1.026187963170189...|
## +---+---+-----+---+--------------------+

included using join:

from pyspark.sql.functions import exp

lookup = sqlContext.createDataFrame([(1, "foo"), (2, "bar")], ("k", "v"))
df_with_x6 = (df_with_x5
    .join(lookup, col("x1") == col("k"), "leftouter")
    .drop("k")
    .withColumnRenamed("v", "x6"))

## +---+---+-----+---+--------------------+----+
## | x1| x2|   x3| x4|                  x5|  x6|
## +---+---+-----+---+--------------------+----+
## |  1|  a| 23.0|  0| 9.744803446248903E9| foo|
## |  3|  B|-23.0|  0|1.026187963170189...|null|
## +---+---+-----+---+--------------------+----+

or generated with function / udf:

from pyspark.sql.functions import rand

df_with_x7 = df_with_x6.withColumn("x7", rand())
df_with_x7.show()

## +---+---+-----+---+--------------------+----+-------------------+
## | x1| x2|   x3| x4|                  x5|  x6|                 x7|
## +---+---+-----+---+--------------------+----+-------------------+
## |  1|  a| 23.0|  0| 9.744803446248903E9| foo|0.41930610446846617|
## |  3|  B|-23.0|  0|1.026187963170189...|null|0.37801881545497873|
## +---+---+-----+---+--------------------+----+-------------------+

Performance-wise, built-in functions (pyspark.sql.functions), which map to Catalyst expression, are usually preferred over Python user defined functions.

If you want to add content of an arbitrary RDD as a column you can

• add row numbers to existing data frame
• call zipWithIndex on RDD and convert it to data frame
• join both using index as a join key

To add a column using a UDF:

df = sqlContext.createDataFrame(
    [(1, "a", 23.0), (3, "B", -23.0)], ("x1", "x2", "x3"))

from pyspark.sql.functions import udf
from pyspark.sql.types import *

def valueToCategory(value):
   if   value == 1: return 'cat1'
   elif value == 2: return 'cat2'
   ...
   else: return 'n/a'

# NOTE: it seems that calls to udf() must be after SparkContext() is called
udfValueToCategory = udf(valueToCategory, StringType())
df_with_cat = df.withColumn("category", udfValueToCategory("x1"))
df_with_cat.show()

## +---+---+-----+---------+
## | x1| x2|   x3| category|
## +---+---+-----+---------+
## |  1|  a| 23.0|     cat1|
## |  3|  B|-23.0|      n/a|
## +---+---+-----+---------+

For Spark 2.0

# assumes schema has 'age' column 
df.select('*', (df.age + 10).alias('agePlusTen'))

There are multiple ways we can add a new column in pySpark.

Let’s first create a simple DataFrame.

date = [27, 28, 29, None, 30, 31]
df = spark.createDataFrame(date, IntegerType())

Now let’s try to double the column value and store it in a new column. PFB few different approaches to achieve the same.

# Approach - 1 : using withColumn function
df.withColumn("double", df.value * 2).show()

# Approach - 2 : using select with alias function.
df.select("*", (df.value * 2).alias("double")).show()

# Approach - 3 : using selectExpr function with as clause.
df.selectExpr("*", "value * 2 as double").show()

# Approach - 4 : Using as clause in SQL statement.
df.createTempView("temp")
spark.sql("select *, value * 2 as double from temp").show()

For more examples and explanation on spark DataFrame functions, you can visit my blog.

I hope this helps.

You can define a new udf when adding a column_name:

u_f = F.udf(lambda :yourstring,StringType())
a.select(u_f().alias('column_name')

from pyspark.sql.functions import udf
from pyspark.sql.types import *
func_name = udf(
    lambda val: val, # do sth to val
    StringType()
)
df.withColumn('new_col', func_name(df.old_col))

I would like to offer a generalized example for a very similar use case:

Use Case: I have a csv consisting of:

First|Third|Fifth
data|data|data
data|data|data
...billion more lines

I need to perform some transformations and the final csv needs to look like

First|Second|Third|Fourth|Fifth
data|null|data|null|data
data|null|data|null|data
...billion more lines

I need to do this because this is the schema defined by some model and I need for my final data to be interoperable with SQL Bulk Inserts and such things.

so:

1) I read the original csv using spark.read and call it “df”.

2) I do something to the data.

3) I add the null columns using this script:

outcols = []
for column in MY_COLUMN_LIST:
    if column in df.columns:
        outcols.append(column)
    else:
        outcols.append(lit(None).cast(StringType()).alias('{0}'.format(column)))

df = df.select(outcols)

In this way, you can structure your schema after loading a csv (would also work for reordering columns if you have to do this for many tables).

The simplest way to add a column is to use “withColumn”. Since the dataframe is created using sqlContext, you have to specify the schema or by default can be available in the dataset. If the schema is specified, the workload becomes tedious when changing every time.

Below is an example that you can consider:

from pyspark.sql import SQLContext
from pyspark.sql.types import *
sqlContext = SQLContext(sc) # SparkContext will be sc by default 

# Read the dataset of your choice (Already loaded with schema)
Data = sqlContext.read.csv("/path", header = True/False, schema = "infer", sep = "delimiter")

# For instance the data has 30 columns from col1, col2, ... col30. If you want to add a 31st column, you can do so by the following:
Data = Data.withColumn("col31", "Code goes here")

# Check the change 
Data.printSchema()

We can add additional columns to DataFrame directly with below steps:

from pyspark.sql.functions import when
df = spark.createDataFrame([["amit", 30], ["rohit", 45], ["sameer", 50]], ["name", "age"])
df = df.withColumn("profile", when(df.age >= 40, "Senior").otherwise("Executive"))
df.show()

I’m trying to filter a PySpark dataframe that has None as a row value:

df.select('dt_mvmt').distinct().collect()

[Row(dt_mvmt=u'2016-03-27'),
 Row(dt_mvmt=u'2016-03-28'),
 Row(dt_mvmt=u'2016-03-29'),
 Row(dt_mvmt=None),
 Row(dt_mvmt=u'2016-03-30'),
 Row(dt_mvmt=u'2016-03-31')]

and I can filter correctly with an string value:

df[df.dt_mvmt == '2016-03-31']
# some results here

but this fails:

df[df.dt_mvmt == None].count()
0
df[df.dt_mvmt != None].count()
0

But there are definitely values on each category. What’s going on?

You can use Column.isNull / Column.isNotNull:

df.where(col("dt_mvmt").isNull())

df.where(col("dt_mvmt").isNotNull())

If you want to simply drop NULL values you can use na.drop with subset argument:

df.na.drop(subset=["dt_mvmt"])

Equality based comparisons with NULL won’t work because in SQL NULL is undefined so any attempt to compare it with another value returns NULL:

sqlContext.sql("SELECT NULL = NULL").show()
## +-------------+
## |(NULL = NULL)|
## +-------------+
## |         null|
## +-------------+


sqlContext.sql("SELECT NULL != NULL").show()
## +-------------------+
## |(NOT (NULL = NULL))|
## +-------------------+
## |               null|
## +-------------------+

The only valid method to compare value with NULL is IS / IS NOT which are equivalent to the isNull / isNotNull method calls.

Try to just use isNotNull function.

df.filter(df.dt_mvmt.isNotNull()).count()

To obtain entries whose values in the dt_mvmt column are not null we have

df.filter("dt_mvmt is not NULL")

and for entries which are null we have

df.filter("dt_mvmt is NULL")

If you want to keep with the Pandas syntex this worked for me.

df = df[df.dt_mvmt.isNotNull()]

if column = None

COLUMN_OLD_VALUE
----------------
None
1
None
100
20
------------------

Use create a temptable on data frame:

sqlContext.sql("select * from tempTable where column_old_value='None' ").show()

So use : column_old_value='None'

There are multiple ways you can remove/filter the null values from a column in DataFrame.

Lets create a simple DataFrame with below code:

date = ['2016-03-27','2016-03-28','2016-03-29', None, '2016-03-30','2016-03-31']
df = spark.createDataFrame(date, StringType())

Now you can try one of the below approach to filter out the null values.

# Approach - 1
df.filter("value is not null").show()

# Approach - 2
df.filter(col("value").isNotNull()).show()

# Approach - 3
df.filter(df["value"].isNotNull()).show()

# Approach - 4
df.filter(df.value.isNotNull()).show()

# Approach - 5
df.na.drop(subset=["value"]).show()

# Approach - 6
df.dropna(subset=["value"]).show()

# Note: You can also use where function instead of a filter.

You can also check the section “Working with NULL Values” on my blog for more information.

I hope it helps.

PySpark provides various filtering options based on arithmetic, logical and other conditions. Presence of NULL values can hamper further processes. Removing them or statistically imputing them could be a choice.

Below set of code can be considered:

# Dataset is df
# Column name is dt_mvmt
# Before filtering make sure you have the right count of the dataset
df.count() # Some number

# Filter here
df = df.filter(df.dt_mvmt.isNotNull())

# Check the count to ensure there are NULL values present (This is important when dealing with large dataset)
df.count() # Count should be reduced if NULL values are present

I would also try:

df = df.dropna(subset=["dt_mvmt"])

If you want to filter out records having None value in column then see below example:

df=spark.createDataFrame([[123,"abc"],[234,"fre"],[345,None]],["a","b"])

Now filter out null value records:

df=df.filter(df.b.isNotNull())

df.show()

If you want to remove those records from DF then see below:

df1=df.na.drop(subset=['b'])

df1.show()

None/Null is a data type of the class NoneType in pyspark/python so, Below will not work as you are trying to compare NoneType object with string object

df[df.dt_mvmt == None].count() 0 df[df.dt_mvmt != None].count() 0

correct

df=df.where(col(“dt_mvmt”).isNotNull()) returns all records with dt_mvmt as None/Null

I have a dataframe with column as String. I wanted to change the column type to Double type in PySpark.

Following is the way, I did:

toDoublefunc = UserDefinedFunction(lambda x: x,DoubleType())
changedTypedf = joindf.withColumn("label",toDoublefunc(joindf['show']))

Just wanted to know, is this the right way to do it as while running through Logistic Regression, I am getting some error, so I wonder, is this the reason for the trouble.

There is no need for an UDF here. Column already provides cast method with DataType instance :

from pyspark.sql.types import DoubleType

changedTypedf = joindf.withColumn("label", joindf["show"].cast(DoubleType()))

or short string:

changedTypedf = joindf.withColumn("label", joindf["show"].cast("double"))

where canonical string names (other variations can be supported as well) correspond to simpleString value. So for atomic types:

from pyspark.sql import types 

for t in ['BinaryType', 'BooleanType', 'ByteType', 'DateType', 
          'DecimalType', 'DoubleType', 'FloatType', 'IntegerType', 
           'LongType', 'ShortType', 'StringType', 'TimestampType']:
    print(f"{t}: {getattr(types, t)().simpleString()}")

BinaryType: binary
BooleanType: boolean
ByteType: tinyint
DateType: date
DecimalType: decimal(10,0)
DoubleType: double
FloatType: float
IntegerType: int
LongType: bigint
ShortType: smallint
StringType: string
TimestampType: timestamp

and for example complex types

types.ArrayType(types.IntegerType()).simpleString()   

'array<int>'

types.MapType(types.StringType(), types.IntegerType()).simpleString()

'map<string,int>'

Preserve the name of the column and avoid extra column addition by using the same name as input column:

changedTypedf = joindf.withColumn("show", joindf["show"].cast(DoubleType()))

Given answers are enough to deal with the problem but I want to share another way which may be introduced the new version of Spark (I am not sure about it) so given answer didn’t catch it.

We can reach the column in spark statement with col("colum_name") keyword:

from pyspark.sql.functions import col , column
changedTypedf = joindf.withColumn("show", col("show").cast("double"))

pyspark version:

  df = <source data>
  df.printSchema()

  from pyspark.sql.types import *

  # Change column type
  df_new = df.withColumn("myColumn", df["myColumn"].cast(IntegerType()))
  df_new.printSchema()
  df_new.select("myColumn").show()

the solution was simple –

toDoublefunc = UserDefinedFunction(lambda x: float(x),DoubleType())
changedTypedf = joindf.withColumn("label",toDoublefunc(joindf['show']))