CODEWITHSUNDEEP
pandasapache-sparkdataframeapache-zeppelin
Bala
Bala

Reputation: 685

converting pandas dataframes to spark dataframe in zeppelin

I am new to zeppelin. I have a usecase wherein i have a pandas dataframe.I need to visualize the collections using in-built chart of zeppelin I do not have a clear approach here. MY understanding is with zeppelin we can visualize the data if it is a RDD format. So, i wanted to convert to pandas dataframe into spark dataframe, and then do some querying (using sql), I will visualize. To start with, I tried to convert pandas dataframe to spark's but i failed

%pyspark
import pandas as pd
from pyspark.sql import SQLContext
print sc
df = pd.DataFrame([("foo", 1), ("bar", 2)], columns=("k", "v"))
print type(df)
print df
sqlCtx = SQLContext(sc)
sqlCtx.createDataFrame(df).show()

And I got the below error

Traceback (most recent call last): File "/tmp/zeppelin_pyspark.py", 
line 162, in <module> eval(compiledCode) File "<string>", 
line 8, in <module> File "/home/bala/Software/spark-1.5.0-bin-hadoop2.6/python/pyspark/sql/context.py", 
line 406, in createDataFrame rdd, schema = self._createFromLocal(data, schema) File "/home/bala/Software/spark-1.5.0-bin-hadoop2.6/python/pyspark/sql/context.py", 
line 322, in _createFromLocal struct = self._inferSchemaFromList(data) File "/home/bala/Software/spark-1.5.0-bin-hadoop2.6/python/pyspark/sql/context.py", 
line 211, in _inferSchemaFromList schema = _infer_schema(first) File "/home/bala/Software/spark-1.5.0-bin-hadoop2.6/python/pyspark/sql/types.py", 
line 829, in _infer_schema raise TypeError("Can not infer schema for type: %s" % type(row)) 
TypeError: Can not infer schema for type: <type 'str'>

Can someone please help me out here? Also, correct me if I am wrong anywhere.

Upvotes: 17

Views: 39284

Answers (3)

eddies
eddies

Reputation: 7473

The following works for me with Zeppelin 0.6.0, Spark 1.6.2 and Python 3.5.2:

%pyspark
import pandas as pd
df = pd.DataFrame([("foo", 1), ("bar", 2)], columns=("k", "v"))
z.show(sqlContext.createDataFrame(df))

which renders as:

enter image description here

Upvotes: 14

Jay Feng
Jay Feng

Reputation: 39

Try setting the SPARK_HOME and PYTHONPATH Variables in bash and then rerunning it

    export SPARK_HOME=path to spark
    export PYTHONPATH=$SPARK_HOME/python:$SPARK_HOME/python/build:$PYTHONPATH
    export PYTHONPATH=$SPARK_HOME/python/lib/py4j-0.8.2.1-src.zip:$PYTHONPATH

Upvotes: 0

leleplx
leleplx

Reputation: 71

I've just copied and pasted your code in a notebook and it works.

%pyspark
import pandas as pd
from pyspark.sql import SQLContext
print sc
df = pd.DataFrame([("foo", 1), ("bar", 2)], columns=("k", "v"))
print type(df)
print df
sqlCtx = SQLContext(sc)
sqlCtx.createDataFrame(df).show()

<pyspark.context.SparkContext object at 0x10b0a2b10>
<class 'pandas.core.frame.DataFrame'>
     k  v
0  foo  1
1  bar  2
+---+-+
|  k|v|
+---+-+
|foo|1|
|bar|2|
+---+-+

I am using this version: zeppelin-0.5.0-incubating-bin-spark-1.4.0_hadoop-2.3.tgz

Upvotes: 7

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