Pyspark udf function error in lambda function
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I have written a udf function below and it throws me an error. Please help.
Below is my dataset;
df1 = sqlContext.range(0, 1000)
.withColumn('normal1',func.abs(10*func.round(randn(seed=1),2)))
.withColumn('normal2',func.abs(100*func.round(randn(seed=2),2)))
.withColumn('normal3',func.abs(func.round(randn(seed=3),2)))
df1 = df1.withColumn('Y',when(df1.normal1*df1.normal2*df1.normal3>750, 1)
.otherwise(0))
udf function below:
from pyspark.sql import types as T
balancingRatio=0.8
calculateWeights = udf(lambda d:(1 * balancingRatio) if d==0 else (1 * (1.0 - balancingRatio)),T.IntegerType())
weightedDataset = df1.withColumn('classWeightCol', calculateWeights('Y'))
weightedDataset.show()
It takes some time and throw me an error;
Py4JJavaError: An error occurred while calling o670.showString.
: org.apache.spark.SparkException: Job aborted due to stage failure: Task 0
in stage 25.0 failed 1 times, most recent failure: Lost task 0.0 in stage
25.0 (TID 427, localhost, executor driver): org.apache.spark.SparkException:
Python worker failed to connect back.
What might be the problem?
Thank you.
A simple example on internet that I found is not working also
maturity_udf = udf(lambda age: "adult" if age >=18 else "child",
T.StringType())
df = sqlContext.createDataFrame([{'name': 'Alice', 'age': 1}])
df.withColumn("maturity", maturity_udf(df.age)).show()
Not: I got python 3.7.1 and spark 2.4
lambda error-handling pyspark user-defined-functions
asked Nov 24 '18 at 12:06
melikmelik
175213
|
show 2 more comments
0
I have written a udf function below and it throws me an error. Please help.
Below is my dataset;
df1 = sqlContext.range(0, 1000)
.withColumn('normal1',func.abs(10*func.round(randn(seed=1),2)))
.withColumn('normal2',func.abs(100*func.round(randn(seed=2),2)))
.withColumn('normal3',func.abs(func.round(randn(seed=3),2)))
df1 = df1.withColumn('Y',when(df1.normal1*df1.normal2*df1.normal3>750, 1)
.otherwise(0))
udf function below:
from pyspark.sql import types as T
balancingRatio=0.8
calculateWeights = udf(lambda d:(1 * balancingRatio) if d==0 else (1 * (1.0 - balancingRatio)),T.IntegerType())
weightedDataset = df1.withColumn('classWeightCol', calculateWeights('Y'))
weightedDataset.show()
It takes some time and throw me an error;
Py4JJavaError: An error occurred while calling o670.showString.
: org.apache.spark.SparkException: Job aborted due to stage failure: Task 0
in stage 25.0 failed 1 times, most recent failure: Lost task 0.0 in stage
25.0 (TID 427, localhost, executor driver): org.apache.spark.SparkException:
Python worker failed to connect back.
What might be the problem?
Thank you.
A simple example on internet that I found is not working also
maturity_udf = udf(lambda age: "adult" if age >=18 else "child",
T.StringType())
df = sqlContext.createDataFrame([{'name': 'Alice', 'age': 1}])
df.withColumn("maturity", maturity_udf(df.age)).show()
Not: I got python 3.7.1 and spark 2.4
lambda error-handling pyspark user-defined-functions
asked Nov 24 '18 at 12:06
melikmelik
175213
What isT.IntegerType()exactly? shouldn't be justIntegerType()?
– Ali AzG
Nov 24 '18 at 12:52
from pyspark.sql import types as T
– melik
Nov 24 '18 at 12:53
What is your pyspark version?
– Ali AzG
Nov 24 '18 at 12:56
my spark version is 2.4
– melik
Nov 24 '18 at 13:01
It seems some version issues. try installing version 2.3 and try again.
– Ali AzG
Nov 24 '18 at 13:03
|
show 2 more comments
0
0
0
I have written a udf function below and it throws me an error. Please help.
Below is my dataset;
df1 = sqlContext.range(0, 1000)
.withColumn('normal1',func.abs(10*func.round(randn(seed=1),2)))
.withColumn('normal2',func.abs(100*func.round(randn(seed=2),2)))
.withColumn('normal3',func.abs(func.round(randn(seed=3),2)))
df1 = df1.withColumn('Y',when(df1.normal1*df1.normal2*df1.normal3>750, 1)
.otherwise(0))
udf function below:
from pyspark.sql import types as T
balancingRatio=0.8
calculateWeights = udf(lambda d:(1 * balancingRatio) if d==0 else (1 * (1.0 - balancingRatio)),T.IntegerType())
weightedDataset = df1.withColumn('classWeightCol', calculateWeights('Y'))
weightedDataset.show()
It takes some time and throw me an error;
Py4JJavaError: An error occurred while calling o670.showString.
: org.apache.spark.SparkException: Job aborted due to stage failure: Task 0
in stage 25.0 failed 1 times, most recent failure: Lost task 0.0 in stage
25.0 (TID 427, localhost, executor driver): org.apache.spark.SparkException:
Python worker failed to connect back.
What might be the problem?
Thank you.
A simple example on internet that I found is not working also
maturity_udf = udf(lambda age: "adult" if age >=18 else "child",
T.StringType())
df = sqlContext.createDataFrame([{'name': 'Alice', 'age': 1}])
df.withColumn("maturity", maturity_udf(df.age)).show()
Not: I got python 3.7.1 and spark 2.4
lambda error-handling pyspark user-defined-functions
asked Nov 24 '18 at 12:06
melikmelik
175213
I have written a udf function below and it throws me an error. Please help.
Below is my dataset;
df1 = sqlContext.range(0, 1000)
.withColumn('normal1',func.abs(10*func.round(randn(seed=1),2)))
.withColumn('normal2',func.abs(100*func.round(randn(seed=2),2)))
.withColumn('normal3',func.abs(func.round(randn(seed=3),2)))
df1 = df1.withColumn('Y',when(df1.normal1*df1.normal2*df1.normal3>750, 1)
.otherwise(0))
udf function below:
from pyspark.sql import types as T
balancingRatio=0.8
calculateWeights = udf(lambda d:(1 * balancingRatio) if d==0 else (1 * (1.0 - balancingRatio)),T.IntegerType())
weightedDataset = df1.withColumn('classWeightCol', calculateWeights('Y'))
weightedDataset.show()
It takes some time and throw me an error;
Py4JJavaError: An error occurred while calling o670.showString.
: org.apache.spark.SparkException: Job aborted due to stage failure: Task 0
in stage 25.0 failed 1 times, most recent failure: Lost task 0.0 in stage
25.0 (TID 427, localhost, executor driver): org.apache.spark.SparkException:
Python worker failed to connect back.
What might be the problem?
Thank you.
A simple example on internet that I found is not working also
maturity_udf = udf(lambda age: "adult" if age >=18 else "child",
T.StringType())
df = sqlContext.createDataFrame([{'name': 'Alice', 'age': 1}])
df.withColumn("maturity", maturity_udf(df.age)).show()
Not: I got python 3.7.1 and spark 2.4
lambda error-handling pyspark user-defined-functions
lambda error-handling pyspark user-defined-functions
asked Nov 24 '18 at 12:06
melikmelik
175213
asked Nov 24 '18 at 12:06
melikmelik
175213
edited Nov 24 '18 at 14:53
melik
asked Nov 24 '18 at 12:06
melikmelik
175213
asked Nov 24 '18 at 12:06
melikmelik
175213
asked Nov 24 '18 at 12:06
melikmelik
175213
175213
What isT.IntegerType()exactly? shouldn't be justIntegerType()?
– Ali AzG
Nov 24 '18 at 12:52
from pyspark.sql import types as T
– melik
Nov 24 '18 at 12:53
What is your pyspark version?
– Ali AzG
Nov 24 '18 at 12:56
my spark version is 2.4
– melik
Nov 24 '18 at 13:01
It seems some version issues. try installing version 2.3 and try again.
– Ali AzG
Nov 24 '18 at 13:03
|
show 2 more comments
What isT.IntegerType()exactly? shouldn't be justIntegerType()?
– Ali AzG
Nov 24 '18 at 12:52
from pyspark.sql import types as T
– melik
Nov 24 '18 at 12:53
What is your pyspark version?
– Ali AzG
Nov 24 '18 at 12:56
my spark version is 2.4
– melik
Nov 24 '18 at 13:01
It seems some version issues. try installing version 2.3 and try again.
– Ali AzG
Nov 24 '18 at 13:03
What is
T.IntegerType() exactly? shouldn't be just IntegerType()?– Ali AzG
Nov 24 '18 at 12:52
What is
T.IntegerType() exactly? shouldn't be just IntegerType()?– Ali AzG
Nov 24 '18 at 12:52
from pyspark.sql import types as T
– melik
Nov 24 '18 at 12:53
from pyspark.sql import types as T
– melik
Nov 24 '18 at 12:53
What is your pyspark version?
– Ali AzG
Nov 24 '18 at 12:56
What is your pyspark version?
– Ali AzG
Nov 24 '18 at 12:56
my spark version is 2.4
– melik
Nov 24 '18 at 13:01
my spark version is 2.4
– melik
Nov 24 '18 at 13:01
It seems some version issues. try installing version 2.3 and try again.
– Ali AzG
Nov 24 '18 at 13:03
It seems some version issues. try installing version 2.3 and try again.
– Ali AzG
Nov 24 '18 at 13:03
|
show 2 more comments
1 Answer
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active
oldest
votes
1
You need to disable fork safety by setting the OBJC_DISABLE_INITIALIZE_FORK_SAFETY variable to YES This solved the issue for me.
import os
os.environ['OBJC_DISABLE_INITIALIZE_FORK_SAFETY'] = 'YES'
answered Jan 11 at 20:54
Smit ShahSmit Shah
262
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You need to disable fork safety by setting the OBJC_DISABLE_INITIALIZE_FORK_SAFETY variable to YES This solved the issue for me.
import os
os.environ['OBJC_DISABLE_INITIALIZE_FORK_SAFETY'] = 'YES'
answered Jan 11 at 20:54
Smit ShahSmit Shah
262
add a comment |
1
You need to disable fork safety by setting the OBJC_DISABLE_INITIALIZE_FORK_SAFETY variable to YES This solved the issue for me.
import os
os.environ['OBJC_DISABLE_INITIALIZE_FORK_SAFETY'] = 'YES'
answered Jan 11 at 20:54
Smit ShahSmit Shah
262
add a comment |
1
1
1
You need to disable fork safety by setting the OBJC_DISABLE_INITIALIZE_FORK_SAFETY variable to YES This solved the issue for me.
import os
os.environ['OBJC_DISABLE_INITIALIZE_FORK_SAFETY'] = 'YES'
answered Jan 11 at 20:54
Smit ShahSmit Shah
262
You need to disable fork safety by setting the OBJC_DISABLE_INITIALIZE_FORK_SAFETY variable to YES This solved the issue for me.
import os
os.environ['OBJC_DISABLE_INITIALIZE_FORK_SAFETY'] = 'YES'
answered Jan 11 at 20:54
Smit ShahSmit Shah
262
answered Jan 11 at 20:54
Smit ShahSmit Shah
262
answered Jan 11 at 20:54
Smit ShahSmit Shah
262
answered Jan 11 at 20:54
Smit ShahSmit Shah
262
262
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What is
T.IntegerType()exactly? shouldn't be justIntegerType()?– Ali AzG
Nov 24 '18 at 12:52
from pyspark.sql import types as T
– melik
Nov 24 '18 at 12:53
What is your pyspark version?
– Ali AzG
Nov 24 '18 at 12:56
my spark version is 2.4
– melik
Nov 24 '18 at 13:01
It seems some version issues. try installing version 2.3 and try again.
– Ali AzG
Nov 24 '18 at 13:03