Using Spark I'm reading a csv and want to apply a function to a column on the csv. I have some code that works but it's very hacky. What is the proper way to do this?
My code
SparkContext().addPyFile("myfile.py")
spark = SparkSession\
.builder\
.appName("myApp")\
.getOrCreate()
from myfile import myFunction
df = spark.read.csv(sys.argv[1], header=True,
mode="DROPMALFORMED",)
a = df.rdd.map(lambda line: Row(id=line[0], user_id=line[1], message_id=line[2], message=myFunction(line[3]))).toDF()
I would like to be able to just call the function on the column name instead of mapping each row to line and then calling the function on line[index].
I'm using Spark version 2.0.1
You can simply use User Defined Functions (udf) combined with a withColumn :
from pyspark.sql.types import IntegerType
from pyspark.sql.functions import udf
udf_myFunction = udf(myFunction, IntegerType()) # if the function returns an int
df = df.withColumn("message", udf_myFunction("_3")) #"_3" being the column name of the column you want to consider
This will add a new column to the dataframe df containing the result of myFunction(line[3]).
Related
I have an use case to map the elements of a pyspark column based on a condition.
Going through this documentation pyspark column, i could not find a function for pyspark column to execute map function.
So tried to use the pyspark dataFrame map function, but not being able to convert the pyspark column to a dataframe
Note: The reason i am using the pyspark column is because i get that as an input from a library(Great expectations) which i use.
#column_condition_partial(engine=SparkDFExecutionEngine)
def _spark(cls, column, ts_formats, **kwargs):
return column.isin([3])
# need to replace the above logic with a map function
# like column.map(lambda x: __valid_date(x))
_spark function arguments are passed from the library
What i have,
A pyspark column with timestamp strings
What i require,
A Pyspark column with boolean(True/False) for each element based on validating the timestamp format
example for dataframe,
df.rdd.map(lambda x: __valid_date(x)).toDF()
__valid_date function returns True/False
So, i either need to convert the pyspark column into dataframe to use the above map function or is there any map function available for the pyspark column?
Looks like you need to return a column object that the framework will use for validation.
I have not used Great expectations, but maybe you can define an UDF for transforming your column. Something like this:
import pyspark.sql.functions as F
import pyspark.sql.types as T
valid_date_udf = udf(lambda x: __valid_date(x), T.BooleanType())
#column_condition_partial(engine=SparkDFExecutionEngine)
def _spark(cls, column, ts_formats, **kwargs):
return valid_date_udf(column)
i have a dataset like below
and i would like to create a dataframe using above dataset like below
First you need stack your dataframe, group by var_name and apply collect_list
import pyspark.sql.functions as f
expr_columns = ', '.join(map(lambda col: '"{col}", {col}'.format(col=col), df.columns))
expr = "stack(2, {columns}) as (var_name, values)".format(columns=expr_columns)
df_stack = df.selectExpr(expr)
df_final = df_stack.groupBy("var_name").agg(f.collect_list(f.col("values")))
I have a data frame in PySpark called df. I have registered this df as a temptable like below.
df.registerTempTable('mytempTable')
date=datetime.now().strftime('%Y-%m-%d %H:%M:%S')
Now from this temp table I will get certain values, like max_id of a column id
min_id = sqlContext.sql("select nvl(min(id),0) as minval from mytempTable").collect()[0].asDict()['minval']
max_id = sqlContext.sql("select nvl(max(id),0) as maxval from mytempTable").collect()[0].asDict()['maxval']
Now I will collect all these values like below.
test = ("{},{},{}".format(date,min_id,max_id))
I found that test is not a data frame but it is a str string
>>> type(test)
<type 'str'>
Now I want save this test as a file in HDFS. I would also like to append data to the same file in hdfs.
How can I do that using PySpark?
FYI I am using Spark 1.6 and don't have access to Databricks spark-csv package.
Here you go, you'll just need to concat your data with concat_ws and right it as a text:
query = """select concat_ws(',', date, nvl(min(id), 0), nvl(max(id), 0))
from mytempTable"""
sqlContext.sql(query).write("text").mode("append").save("/tmp/fooo")
Or even a better alternative :
from pyspark.sql import functions as f
(sqlContext
.table("myTempTable")
.select(f.concat_ws(",", f.first(f.lit(date)), f.min("id"), f.max("id")))
.coalesce(1)
.write.format("text").mode("append").save("/tmp/fooo"))
I'm using python on Spark and would like to get a csv into a dataframe.
The documentation for Spark SQL strangely does not provide explanations for CSV as a source.
I have found Spark-CSV, however I have issues with two parts of the documentation:
"This package can be added to Spark using the --jars command line option. For example, to include it when starting the spark shell: $ bin/spark-shell --packages com.databricks:spark-csv_2.10:1.0.3"
Do I really need to add this argument everytime I launch pyspark or spark-submit? It seems very inelegant. Isn't there a way to import it in python rather than redownloading it each time?
df = sqlContext.load(source="com.databricks.spark.csv", header="true", path = "cars.csv") Even if I do the above, this won't work. What does the "source" argument stand for in this line of code? How do I simply load a local file on linux, say "/Spark_Hadoop/spark-1.3.1-bin-cdh4/cars.csv"?
With more recent versions of Spark (as of, I believe, 1.4) this has become a lot easier. The expression sqlContext.read gives you a DataFrameReader instance, with a .csv() method:
df = sqlContext.read.csv("/path/to/your.csv")
Note that you can also indicate that the csv file has a header by adding the keyword argument header=True to the .csv() call. A handful of other options are available, and described in the link above.
from pyspark.sql.types import StringType
from pyspark import SQLContext
sqlContext = SQLContext(sc)
Employee_rdd = sc.textFile("\..\Employee.csv")
.map(lambda line: line.split(","))
Employee_df = Employee_rdd.toDF(['Employee_ID','Employee_name'])
Employee_df.show()
for Pyspark, assuming that the first row of the csv file contains a header
spark = SparkSession.builder.appName('chosenName').getOrCreate()
df=spark.read.csv('fileNameWithPath', mode="DROPMALFORMED",inferSchema=True, header = True)
Read the csv file in to a RDD and then generate a RowRDD from the original RDD.
Create the schema represented by a StructType matching the structure of Rows in the RDD created in Step 1.
Apply the schema to the RDD of Rows via createDataFrame method provided by SQLContext.
lines = sc.textFile("examples/src/main/resources/people.txt")
parts = lines.map(lambda l: l.split(","))
# Each line is converted to a tuple.
people = parts.map(lambda p: (p[0], p[1].strip()))
# The schema is encoded in a string.
schemaString = "name age"
fields = [StructField(field_name, StringType(), True) for field_name in schemaString.split()]
schema = StructType(fields)
# Apply the schema to the RDD.
schemaPeople = spark.createDataFrame(people, schema)
source: SPARK PROGRAMMING GUIDE
If you do not mind the extra package dependency, you could use Pandas to parse the CSV file. It handles internal commas just fine.
Dependencies:
from pyspark import SparkContext
from pyspark.sql import SQLContext
import pandas as pd
Read the whole file at once into a Spark DataFrame:
sc = SparkContext('local','example') # if using locally
sql_sc = SQLContext(sc)
pandas_df = pd.read_csv('file.csv') # assuming the file contains a header
# If no header:
# pandas_df = pd.read_csv('file.csv', names = ['column 1','column 2'])
s_df = sql_sc.createDataFrame(pandas_df)
Or, even more data-consciously, you can chunk the data into a Spark RDD then DF:
chunk_100k = pd.read_csv('file.csv', chunksize=100000)
for chunky in chunk_100k:
Spark_temp_rdd = sc.parallelize(chunky.values.tolist())
try:
Spark_full_rdd += Spark_temp_rdd
except NameError:
Spark_full_rdd = Spark_temp_rdd
del Spark_temp_rdd
Spark_DF = Spark_full_rdd.toDF(['column 1','column 2'])
Following Spark 2.0, it is recommended to use a Spark Session:
from pyspark.sql import SparkSession
from pyspark.sql import Row
# Create a SparkSession
spark = SparkSession \
.builder \
.appName("basic example") \
.config("spark.some.config.option", "some-value") \
.getOrCreate()
def mapper(line):
fields = line.split(',')
return Row(ID=int(fields[0]), field1=str(fields[1].encode("utf-8")), field2=int(fields[2]), field3=int(fields[3]))
lines = spark.sparkContext.textFile("file.csv")
df = lines.map(mapper)
# Infer the schema, and register the DataFrame as a table.
schemaDf = spark.createDataFrame(df).cache()
schemaDf.createOrReplaceTempView("tablename")
I ran into similar problem. The solution is to add an environment variable named as "PYSPARK_SUBMIT_ARGS" and set its value to "--packages com.databricks:spark-csv_2.10:1.4.0 pyspark-shell". This works with Spark's Python interactive shell.
Make sure you match the version of spark-csv with the version of Scala installed. With Scala 2.11, it is spark-csv_2.11 and with Scala 2.10 or 2.10.5 it is spark-csv_2.10.
Hope it works.
Based on the answer by Aravind, but much shorter, e.g. :
lines = sc.textFile("/path/to/file").map(lambda x: x.split(","))
df = lines.toDF(["year", "month", "day", "count"])
With the current implementation(spark 2.X) you dont need to add the packages argument, You can use the inbuilt csv implementation
Additionally as the accepted answer you dont need to create an rdd then enforce schema that has 1 potential problem
When you read the csv as then it will mark all the fields as string and when you enforce the schema with an integer column you will get exception.
A better way to do the above would be
spark.read.format("csv").schema(schema).option("header", "true").load(input_path).show()
I have an rdd (we can call it myrdd) where each record in the rdd is of the form:
[('column 1',value), ('column 2',value), ('column 3',value), ... , ('column 100',value)]
I would like to convert this into a DataFrame in pyspark - what is the easiest way to do this?
How about use the toDF method? You only need add the field names.
df = rdd.toDF(['column', 'value'])
The answer by #dapangmao got me to this solution:
my_df = my_rdd.map(lambda l: Row(**dict(l))).toDF()
Take a look at the DataFrame documentation to make this example work for you, but this should work. I'm assuming your RDD is called my_rdd
from pyspark.sql import SQLContext, Row
sqlContext = SQLContext(sc)
# You have a ton of columns and each one should be an argument to Row
# Use a dictionary comprehension to make this easier
def record_to_row(record):
schema = {'column{i:d}'.format(i = col_idx):record[col_idx] for col_idx in range(1,100+1)}
return Row(**schema)
row_rdd = my_rdd.map(lambda x: record_to_row(x))
# Now infer the schema and you have a DataFrame
schema_my_rdd = sqlContext.inferSchema(row_rdd)
# Now you have a DataFrame you can register as a table
schema_my_rdd.registerTempTable("my_table")
I haven't worked much with DataFrames in Spark but this should do the trick
In pyspark, let's say you have a dataframe named as userDF.
>>> type(userDF)
<class 'pyspark.sql.dataframe.DataFrame'>
Lets just convert it to RDD (
userRDD = userDF.rdd
>>> type(userRDD)
<class 'pyspark.rdd.RDD'>
and now you can do some manipulations and call for example map function :
newRDD = userRDD.map(lambda x:{"food":x['favorite_food'], "name":x['name']})
Finally, lets create a DataFrame from resilient distributed dataset (RDD).
newDF = sqlContext.createDataFrame(newRDD, ["food", "name"])
>>> type(ffDF)
<class 'pyspark.sql.dataframe.DataFrame'>
That's all.
I was hitting this warning message before when I tried to call :
newDF = sc.parallelize(newRDD, ["food","name"] :
.../spark-2.0.0-bin-hadoop2.7/python/pyspark/sql/session.py:336: UserWarning: Using RDD of dict to inferSchema is deprecated. Use pyspark.sql.Row inst warnings.warn("Using RDD of dict to inferSchema is deprecated. "
So no need to do this anymore...