Pandas API on Spark is supported from Spark 3.2.0 (https://issues.apache.org/jira/browse/SPARK-34849), but the method you try to use on Dataframe was implemented later and introduced in the version 3.3.0.
It was introduced with this ticket https://issues.apache.org/jira/browse/SPARK-37337 in commit: https://github.com/apache/spark/commit/bc7d55fc1046a55df61fdb380629699e9959fcc6
Which basically makes changes to naming and depreciation/undepreciation of the methods.
### What changes were proposed in this pull request?
The PR is proposed to:
- Undeprecate (Spark)DataFrame.to_koalas
- Deprecate (Spark)DataFrame.to_pandas_like and introduce (Spark)DataFrame.pandas_api instead.
### Why are the changes needed?
Currently, (Spark)DataFrame.to_pandas_on_spark is too long to memorize and inconvenient to call.
With the proposal of the PR, we may improve the user experience and make APIs more developer-friendly.
### Does this PR introduce _any_ user-facing change?
Yes.
(Spark)DataFrame.pandas_api is introduced.
(Spark)DataFrame.to_pandas_on_spark is deprecated.
(Spark)DataFrame.to_koalas is undeprecated.
For the Spark 3.2.1 you can check those:
type(df.to_koalas())
/databricks/spark/python/pyspark/sql/dataframe.py:2964: FutureWarning: DataFrame.to_koalas is deprecated. Use DataFrame.to_pandas_on_spark instead.
warnings.warn(
Out[5]: pyspark.pandas.frame.DataFrame
or this one:
type(df.to_pandas_on_spark())
Out[6]: pyspark.pandas.frame.DataFrame
Answer from Paweł Tajs on Stack Overflowwhen you put .show() at the end, it is not a pyspark data frame anymore.
Remove it and it should work.
tx_ecommerce =tx_df.filter(tx_df["POS_Cardholder_Presence"]=="ECommerce")
tx_ecommerce.toPandas()
you can do this to read a parquet file:
import pandas as pd
txt = pd.read_parquet("/data/file.parquet")
txt_ecommerce = txt.loc[txt.POS_Cardholder_Presence =="ECommerce"]
Check your DataFrame with data.columns
It should print something like this
Index([u'regiment', u'company', u'name',u'postTestScore'], dtype='object')
Check for hidden white spaces..Then you can rename with
data = data.rename(columns={'Number ': 'Number'})
I think the column name that contains "Number" is something like " Number" or "Number ". I'm assuming you might have a residual space in the column name. Please run print "<{}>".format(data.columns[1]) and see what you get. If it's something like < Number>, it can be fixed with:
data.columns = data.columns.str.strip()
See pandas.Series.str.strip
In general, AttributeError: 'DataFrame' object has no attribute '...', where ... is some column name, is caused because . notation has been used to reference a nonexistent column name or pandas method.
pandas methods are accessed with a .. pandas columns can also be accessed with a . (e.g. data.col) or with brackets (e.g. ['col'] or [['col1', 'col2']]).
data.columns = data.columns.str.strip() is a fast way to quickly remove leading and trailing spaces from all column names. Otherwise verify the column or attribute is correctly spelled.
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 have a spark dataframe, not a pandas dataframe. To add new column to the spark dataframe:
import pyspark.sql.functions as F
from pyspark.sql.types import IntegerType
df = df.withColumn('new_column', F.udf(some_map.get, IntegerType())(some_column_name))
df.show()
"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)
The Iris Dataset from Sklearn is in Sklearn's Bunch format:
print(type(iris))
print(iris.keys())
output:
<class 'sklearn.utils.Bunch'>
dict_keys(['data', 'target', 'target_names', 'DESCR', 'feature_names', 'filename'])
So, that's why you can access it as:
x=iris.data
y=iris.target
But when you read the CSV file as DataFrame as mentioned by you:
iris = pd.read_csv('iris.csv',header=None).iloc[:,2:4]
iris.head()
output is:
2 3
0 petal_length petal_width
1 1.4 0.2
2 1.4 0.2
3 1.3 0.2
4 1.5 0.2
Here the column names are '1' and '2'.
First of all you should read the CSV file as:
df = pd.read_csv('iris.csv')
you should not include header=None as your csv file includes the column names i.e. the headers.
So, now what you can do is something like this:
X = df.iloc[:, [2, 3]] # Will give you columns 2 and 3 i.e 'petal_length' and 'petal_width'
y = df.iloc[:, 4] # Label column i.e 'species'
or if you want to use the column names then:
X = df[['petal_length', 'petal_width']]
y = df.iloc['species']
Also, if you want to convert labels from string to numerical format use sklearn LabelEncoder
from sklearn import preprocessing
le = preprocessing.LabelEncoder()
y = le.fit_transform(y)