Converting Pandas DataFrame to Spark DataFrame
I am not sure if this question is still relevant to the current version of pySpark, but here is the solution I worked out a couple weeks after posting this question. The code is rather ugly and possibly inefficient, but I am posting it here due to the continued interest in this question.:
from pyspark import SparkContextfrom pyspark.sql import HiveContextfrom pyspark import SparkConffrom py4j.protocol import Py4JJavaErrormyConf = SparkConf(loadDefaults=True)sc = SparkContext(conf=myConf)hc = HiveContext(sc)def chunks(lst, k): """Yield k chunks of close to equal size""" n = len(lst) / k for i in range(0, len(lst), n): yield lst[i: i + n]def reconstruct_rdd(lst, num_parts): partitions = chunks(lst, num_parts) for part in range(0, num_parts - 1): print "Partition ", part, " started..." partition = next(partitions) # partition is a list of lists if part == 0: prime_rdd = sc.parallelize(partition) else: second_rdd = sc.parallelize(partition) prime_rdd = prime_rdd.union(second_rdd) print "Partition ", part, " complete!" return prime_rdddef build_col_name_list(len_cols): name_lst = [] for i in range(1, len_cols): idx = "_" + str(i) name_lst.append(idx) return name_lstdef set_spark_df_header(header, sdf): oldColumns = build_col_name_lst(len(sdf.columns)) newColumns = header sdf = reduce(lambda sdf, idx: sdf.withColumnRenamed(oldColumns[idx], newColumns[idx]), xrange(len(oldColumns)), sdf) return sdfdef convert_pdf_matrix_to_sdf(pdf, sdf_header, num_of_parts): try: sdf = hc.createDataFrame(pdf) except ValueError: lst = pdf.values.tolist() #Need to convert to list of list to parallelize try: rdd = sc.parallelize(lst) except Py4JJavaError: rdd = reconstruct_rdd(lst, num_of_parts) sdf = hc.createDataFrame(rdd) sdf = set_spark_df_header(sdf_header, sdf) return sdf
to_sparse(fill_value=0)
is basically obsolete. Just use standard variant
sqlContext.createDataFrame(pd.DataFrame(csc_mat.todense()))
and as long as types are compatible you'd be fine.