I am stuck in a problem where I wanna do PCA on a Pyspark Dataframe column. The name of the column is ‘features’ where each row is a SparseVector.
This is the flow:
Df - name of the pyspark df
Features - name of column
Snippet of the rdd
[Row(features=SparseVector(2,{1:50.0})),
Row(features=SparseVector(2,{0:654.0, 1:20.0}))],
from pyspark.mllib.linalg.distributed import RowMatrix
i = RowMatrix(df.select(‘features’).rdd)
ipc = i.computePrincipalComponents(2)
Error Message
You are getting an RDD[Row] object where your Row is Row(features=SparseVector(2,{1:50.0})).
You need an RDD[SparseVector], so you should change your line:
i = RowMatrix(df.select(‘features’).rdd)
to
i = RowMatrix(df.select(‘features’).rdd.map(lambda x: x[0]))
which will return RDD[SparseVector]
Related
How to create a column with json structure based on other columns of a pyspark dataframe.
For example, I want to achieve the below in pyspark dataframe. I am able to do this on pandas dataframe as below, but how do I do the same on pyspark dataframe
df = {'Address': ['abc', 'dvf', 'bgh'], 'zip': [34567, 12345, 78905], 'state':['VA', 'TN', 'MA']}
df = pd.DataFrame(df, columns = ['Address', 'zip', 'state'])
lst = ['Address', 'zip']
df['new_col'] = df[lst].apply(lambda x: x.to_json(), axis = 1)
Expected output
Assuming your pyspark dataframe is named df, use the struct function to construct a struct, and then use the to_json function to convert it to a json string.
import pyspark.sql.functions as F
....
lst = ['Address', 'zip']
df = df.withColumn('new_col', F.to_json(F.struct(*[F.col(c) for c in lst])))
df.show(truncate=False)
I have a dataframe with headers for example outputDF. I now want to take outputDF.columns and create a new dataframe with just one row which contains column names.
I then want to union both these dataframes with option("head=false") which spark can then write to a HDFS.
How do i do that?
below is an example
Val df = spark.read.csv("path")
val newDf = df.columns.toSeq.toDF
val unoindf= df.union(newDf);
I know this is probably to be a stupid question. I have the following code:
from pyspark.sql import SparkSession
rows = [1,2,3]
df = SparkSession.createDataFrame(rows)
df.printSchema()
df.show()
But I got an error:
createDataFrame() missing 1 required positional argument: 'data'
I don't understand why this happens because I already supplied 'data', which is the variable rows.
Thanks
You have to create SparkSession instance using the build pattern and use it for creating dataframe, check
https://spark.apache.org/docs/2.2.1/api/python/pyspark.sql.html#pyspark.sql.SparkSession
spark= SparkSession.builder.getOrCreate()
Below are the steps to create pyspark dataframe using createDataFrame
Create sparksession
spark = SparkSession.builder.appName('SparkByExamples.com').getOrCreate()
Create data and columns
columns = ["language","users_count"]
data = [("Java", "20000"), ("Python", "100000"), ("Scala", "3000")]
Creating DataFrame from RDD
rdd = spark.sparkContext.parallelize(data)
df= spark.createDataFrame(rdd).toDF(*columns)
the second approach, Directly creating dataframe
df2 = spark.createDataFrame(data).toDF(*columns)
Try
row = [(1,), (2,), (3,)]
?
If I am not wrong createDataFrame() takes 2 lists as input: first list is the data and second list is the column names. The data must be a lists of list of tuples, where each tuple is a row of the dataframe.
I have an empty dataframe with schema already created.
I'm trying to add the columns to this dataframe from a new dataframe to the existing columns in a for loop.
k schema - |ID|DATE|REPORTID|SUBMITTEDDATE|
for(data <- 0 to range-1){
val c = df2.select(substring(col("value"), str(data)._2, str(data)._3).alias(str(data)._1)).toDF()
//c.show()
k = c.withColumn(str(data)._1, c(str(data)._1))
}
k.show()
But the k dataframe has just one column, but it should have all the 4 columns populated with values.
I think the last line in for loop is replacing exisitng columns in the dataframe.
Can somebody help me with this?
Thanks!!
Add your logic and conditions and create new dataframe
val dataframe2 = dataframe1.select("A","B",C)
Copying few columns of a dataframe to another one is not possible in spark.
Although there are few alternatives to attain the same
1. You need to join both the dataframe based on some join condition.
2. Convert bot the data frame to json and do RDD Union
val rdd = df1.toJSON.union(df2.toJSON)
val dfFinal = spark.read.json(rdd)
I'm new to Pyspark and I'm trying to add a new column to my existing dataframe. The new column should contain only 4 fixed values (e.g. 1,2,3,4) and I'd like to randomly pick one of the values for each row.
How can I do that?
Pyspark dataframes are immutable, so you have to return a new one (e.g. you can't just assign to it the way you can with Pandas dataframes). To do what you want use a udf:
from pyspark.sql.functions import udf
import numpy as np
df = <original df>
udf_randint = udf(np.random.randint(1, 4))
df_new = df.withColumn("random_num": udf_randint)