environment variables PYSPARK_PYTHON and PYSPARK_DRIVER_PYTHON
By the way, if you use PyCharm, you could add PYSPARK_PYTHON and PYSPARK_DRIVER_PYTHON to run/debug configurations per image below
By the way, if you use PyCharm, you could add PYSPARK_PYTHON and PYSPARK_DRIVER_PYTHON to run/debug configurations per image below
You should use & / | operators and be careful about operator precedence (== has lower precedence than bitwise AND and OR):
You should use & / | operators and be careful about operator precedence (== has lower precedence than bitwise AND and OR):
One possible reason is JAVA_HOME is not set because java is not installed. I encountered the same issue. It says at sc = pyspark.SparkConf(). I solved it by running which is from https://www.digitalocean.com/community/tutorials/how-to-install-java-with-apt-get-on-ubuntu-16-04
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”) … Read more
Reading the Spark documentation I found an easier solution. Since version 1.4 of spark there is a function drop(col) which can be used in pyspark on a dataframe. You can use it in two ways df.drop(‘age’) df.drop(df.age) Pyspark Documentation – Drop
Simple dataframe creation: According to official doc: when schema is a list of column names, the type of each column will be inferred from data. (example above ↑) When schema is pyspark.sql.types.DataType or a datatype string, it must match the real data. (examples below ↓) Additionally, you can create your dataframe from Pandas dataframe, schema will be inferred from Pandas … Read more
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| … Read more
See, There are two ways to convert an RDD to DF in Spark. toDF() and createDataFrame(rdd, schema) I will show you how you can do that dynamically. toDF() The toDF() command gives you the way to convert an RDD[Row] to a Dataframe. The point is, the object Row() can receive a **kwargs argument. So, there is an easy way to do that. This way you … Read more
copy paste the application Id from the spark scheduler, for instance application_1428487296152_25597 connect to the server that have launch the job yarn application -kill application_1428487296152_25597