If you are curious to know all the available methods, you can run. > import faker > fake faker() > fake.name() 'Lucy Cechtelar' > fake.address() '0535 Lisa Flats South Michele, MI 38477'. method on the object to get the required data. I was trying to search for it all over but could not find an example of doing this with PySpark. In the below example, we have created a faker object called fake and then ran the name, address, etc. You can do this using either zipWithIndex() or rownumber() (depending on the amount and kind of your data) but in every case there is a catch regarding performance. Say I have a pandas DataFrame like so: df = pd.DataFrame()Īnd I want to add a column with uuids that are the same if the name is the same. In Python, we can convert UUID to string and vice versa using str class and we can obtain string format removing the hyphen that is used for separation of components in UUID using string method replace () by replacing - with say for example. Adding sequential unique IDs to a Spark Dataframe is not very straight-forward, especially considering the distributed nature of it. If everyone on this earth were to generate a UUID per second, thats 10,000,000,000365246060 315360000000000000 UUIDs per year, which is quite close to 258. So now I use this: from pyspark.sql import functions as F df.withColumn ('uuid', F.expr ('uuid ()')) This is nicer and should be faster since it uses native Spark SQL. Examples: > SELECT true false > SELECT false true > SELECT NULL NULL Since: 1.0.0 expr1 expr2 - Returns true if expr1 is not equal to expr2, or false otherwise. Assuming perfect randomness, you can expect the first collision at around 261 generated UUIDs (thats the square root of 2122). I understand that Pandas can do something like what i want very easily, but if i want to achieve giving a unique UUID to each row of my pyspark dataframe based on a specific column attribute, how do I do that? import uuid udf def createrandomid (): return str (uuid.uuid4 ()) But as of Spark 3.0.0 there is a Spark SQL for random uuids. Is there no way to currently generate a UUID in a PySpark dataframe based on unique value of a field? Use monotonicallyincreasingid () for unique, but not consecutive numbers.
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