Converting Pandas dataframe into Spark dataframe error

I have tried this with your data and it is working :

%pyspark
import pandas as pd
from pyspark.sql import SQLContext
print sc
df = pd.read_csv("test.csv")
print type(df)
print df
sqlCtx = SQLContext(sc)
sqlCtx.createDataFrame(df).show()

Type related errors can be avoided by imposing a schema as follows:

note: a text file was created (test.csv) with the original data (as above) and hypothetical column names were inserted (“col1″,”col2″,…,”col25”).

import pyspark
from pyspark.sql import SparkSession
import pandas as pd

spark = SparkSession.builder.appName('pandasToSparkDF').getOrCreate()

pdDF = pd.read_csv("test.csv")

contents of the pandas data frame:

       col1     col2    col3    col4    col5    col6    col7    col8   ... 
0      10000001 1       0       1       12:35   OK      10002   1      ...
1      10000001 2       0       1       12:36   OK      10002   1      ...
2      10000002 1       0       4       12:19   PA      10003   1      ...

Next, create the schema:

from pyspark.sql.types import *

mySchema = StructType([ StructField("col1", LongType(), True)
                       ,StructField("col2", IntegerType(), True)
                       ,StructField("col3", IntegerType(), True)
                       ,StructField("col4", IntegerType(), True)
                       ,StructField("col5", StringType(), True)
                       ,StructField("col6", StringType(), True)
                       ,StructField("col7", IntegerType(), True)
                       ,StructField("col8", IntegerType(), True)
                       ,StructField("col9", IntegerType(), True)
                       ,StructField("col10", IntegerType(), True)
                       ,StructField("col11", StringType(), True)
                       ,StructField("col12", StringType(), True)
                       ,StructField("col13", IntegerType(), True)
                       ,StructField("col14", IntegerType(), True)
                       ,StructField("col15", IntegerType(), True)
                       ,StructField("col16", IntegerType(), True)
                       ,StructField("col17", IntegerType(), True)
                       ,StructField("col18", IntegerType(), True)
                       ,StructField("col19", IntegerType(), True)
                       ,StructField("col20", IntegerType(), True)
                       ,StructField("col21", IntegerType(), True)
                       ,StructField("col22", IntegerType(), True)
                       ,StructField("col23", IntegerType(), True)
                       ,StructField("col24", IntegerType(), True)
                       ,StructField("col25", IntegerType(), True)])

Note: True (implies nullable allowed)

create the pyspark dataframe:

df = spark.createDataFrame(pdDF,schema=mySchema)

confirm the pandas data frame is now a pyspark data frame:

type(df)

output:

pyspark.sql.dataframe.DataFrame

Aside:

To address Kate’s comment below – to impose a general (String) schema you can do the following:

df=spark.createDataFrame(pdDF.astype(str)) 

I received a similar error message once, in my case it was because my pandas dataframe contained NULLs. I will recommend to try & handle this in pandas before converting to spark (this resolved the issue in my case).

I made this script, It worked for my 10 pandas Data frames

from pyspark.sql.types import *

# Auxiliar functions
def equivalent_type(f):
    if f == 'datetime64[ns]': return TimestampType()
    elif f == 'int64': return LongType()
    elif f == 'int32': return IntegerType()
    elif f == 'float64': return DoubleType()
    elif f == 'float32': return FloatType()
    else: return StringType()

def define_structure(string, format_type):
    try: typo = equivalent_type(format_type)
    except: typo = StringType()
    return StructField(string, typo)

# Given pandas dataframe, it will return a spark's dataframe.
def pandas_to_spark(pandas_df):
    columns = list(pandas_df.columns)
    types = list(pandas_df.dtypes)
    struct_list = []
    for column, typo in zip(columns, types): 
      struct_list.append(define_structure(column, typo))
    p_schema = StructType(struct_list)
    return sqlContext.createDataFrame(pandas_df, p_schema)

You can see it also in this gist

With this you just have to call spark_df = pandas_to_spark(pandas_df)

In spark version >= 3 you can convert pandas dataframes to pyspark dataframe in one line

use spark.createDataFrame(pandasDF)

dataset = pd.read_csv("data/AS/test_v2.csv")

sparkDf = spark.createDataFrame(dataset);

if you are confused about spark session variable,
spark session is as follows

sc = SparkContext.getOrCreate(SparkConf().setMaster("local[*]"))

spark = SparkSession 
    .builder 
    .getOrCreate()

I cleaned up / simplified the top answer a bit:

import pyspark.sql.types as ps_types


def get_equivalent_spark_type(pandas_type):
    """
        This method will retrieve the corresponding spark type given a pandas
        type.

        Args:
            pandas_type (str): pandas data type

        Returns:
            spark data type
    """
    type_map = {
        'datetime64[ns]': ps_types.TimestampType(),
        'int64': ps_types.LongType(),
        'int32': ps_types.IntegerType(),
        'float64': ps_types.DoubleType(),
        'float32': ps_types.FloatType()}
    if pandas_type not in type_map:
        return ps_types.StringType()
    else:
        return type_map[pandas_type]


def pandas_to_spark(spark, pandas_df):
    """
        This method will return a spark dataframe given a pandas dataframe.

        Args:
            spark (pyspark.sql.session.SparkSession): pyspark session
            pandas_df (pandas.core.frame.DataFrame): pandas DataFrame

        Returns:
            equivalent spark DataFrame
    """
    columns = list(pandas_df.columns)
    types = list(pandas_df.dtypes)
    p_schema = ps_types.StructType([
        ps_types.StructField(column, get_equivalent_spark_type(pandas_type))
        for column, pandas_type in zip(columns, types)])

    return spark.createDataFrame(pandas_df, p_schema)