"sklearn.datasets" is a scikit package, where it contains a method load_iris().

load_iris(), by default return an object which holds data, target and other members in it. In order to get actual values you have to read the data and target content itself.

Whereas 'iris.csv', holds feature and target together.

FYI: If you set return_X_y as True in load_iris(), then you will directly get features and target.

from sklearn import datasets
data,target = datasets.load_iris(return_X_y=True)

Seaborn expects pandas dataframes and not spark dataframes. Change your code to:

g1= sns.scatterplot( x= "events", y= "ways" , data = x2.toPandas())

and it should work.

The syntax is valid with Pandas DataFrames but that attribute doesn't exist for the PySpark created DataFrames. You can check out this link for the documentation.

Usually, the collect() method or the .rdd attribute would help you with these tasks.

You can use the following snippet to produce the desired result:

http_path = sdf.rdd.map(lambda row: row['http_path'].split('?'))
api_param_df = pd.DataFrame([[row[0], np.nan] if len(row) == 1 else row for row in http_path.collect()], columns=["api", "param"])
sdf = pd.concat([sdf.toPandas()['raw'], api_param_df], axis=1)

Note that I removed the comments to make it more readable and I've also substituted the regex with a simple split.

I think your pandas version is older as 0.17.0.

See DataFrame.plot.scatter:

New in version 0.17.0.

In older version you can use:

df.plot(kind='scatter')

I think you're looking for dataset.plot not plt.

It means that you have an object of type Dataframe, and you are trying to call the attribute dataframe which does not exist.

The syntax you are using is for a pandas DataFrame. To achieve this for a spark DataFrame, you should use the withColumn() method. This works great for a wide range of well defined DataFrame functions, but it's a little more complicated for user defined mapping functions.

General Case

In order to define a udf, you need to specify the output data type. For instance, if you wanted to apply a function my_func that returned a string, you could create a udf as follows:

import pyspark.sql.functions as f
my_udf = f.udf(my_func, StringType())

Then you can use my_udf to create a new column like:

df = df.withColumn('new_column', my_udf(f.col("some_column_name")))

Another option is to use select:

df = df.select("*", my_udf(f.col("some_column_name")).alias("new_column"))

Specific Problem

Using a udf

In your specific case, you want to use a dictionary to translate the values of your DataFrame.

Here is a way to define a udf for this purpose:

some_map_udf = f.udf(lambda x: some_map.get(x, None), IntegerType())

Notice that I used dict.get() because you want your udf to be robust to bad inputs.

df = df.withColumn('new_column', some_map_udf(f.col("some_column_name")))

Using DataFrame functions

Sometimes using a udf is unavoidable, but whenever possible, using DataFrame functions is usually preferred.

Here is one option to do the same thing without using the udf.

The trick is to iterate over the items in some_map to create a list of pyspark.sql.functions.when() functions.

some_map_func = [f.when(f.col("some_column_name") == k, v) for k, v in some_map.items()]
print(some_map_func)
#[Column<CASE WHEN (some_column_name = a) THEN 0 END>,
# Column<CASE WHEN (some_column_name = c) THEN 1 END>,
# Column<CASE WHEN (some_column_name = b) THEN 1 END>]

Now you can use pyspark.sql.functions.coalesce() inside of a select:

df = df.select("*", f.coalesce(*some_map_func).alias("some_column_name"))

This works because when() returns null by default if the condition is not met, and coalesce() will pick the first non-null value it encounters. Since the keys of the map are unique, at most one column will be non-null.

You tried to use .Values with a capital v instead of .values. Changing the capital v to a lowercase v should solve fix the error you're getting.