I have a very big pyspark.sql.dataframe.DataFrame named df. I need some way of enumerating records- thus, being able to access record with certain index. (or select group of records with indexes range)
In pandas, I could make just
indexes=[2,3,6,7]
df[indexes]
Here I want something similar, (and without converting dataframe to pandas)
The closest I can get to is:
Enumerating all the objects in the original dataframe by:
indexes=np.arange(df.count())
df_indexed=df.withColumn('index', indexes)
QUESTIONS:
Would it work later to make something like:
indexes=[2,3,6,7]
df1.where("index in indexes").collect()
Any faster and simpler way to deal with it?
Create PySpark DataFrame from Pandas Due to parallel execution on all cores on multiple machines, PySpark runs operations faster than Pandas, hence we often required to covert Pandas DataFrame to PySpark (Spark with Python) for better performance.
It doesn't work because:
withColumn
should be a Column
not a collection. np.array
won't work here"index in indexes"
as a SQL expression to where
indexes
is out of scope and it is not resolved as a valid identifierPySpark >= 1.4.0
You can add row numbers using respective window function and query using Column.isin
method or properly formated query string:
from pyspark.sql.functions import col, rowNumber
from pyspark.sql.window import Window
w = Window.orderBy()
indexed = df.withColumn("index", rowNumber().over(w))
# Using DSL
indexed.where(col("index").isin(set(indexes)))
# Using SQL expression
indexed.where("index in ({0})".format(",".join(str(x) for x in indexes)))
It looks like window functions called without PARTITION BY
clause move all data to the single partition so above may be not the best solution after all.
Any faster and simpler way to deal with it?
Not really. Spark DataFrames don't support random row access.
PairedRDD
can be accessed using lookup
method which is relatively fast if data is partitioned using HashPartitioner
. There is also indexed-rdd project which supports efficient lookups.
Edit:
Independent of PySpark version you can try something like this:
from pyspark.sql import Row
from pyspark.sql.types import StructType, StructField, LongType
row = Row("char")
row_with_index = Row("char", "index")
df = sc.parallelize(row(chr(x)) for x in range(97, 112)).toDF()
df.show(5)
## +----+
## |char|
## +----+
## | a|
## | b|
## | c|
## | d|
## | e|
## +----+
## only showing top 5 rows
# This part is not tested but should work and save some work later
schema = StructType(
df.schema.fields[:] + [StructField("index", LongType(), False)])
indexed = (df.rdd # Extract rdd
.zipWithIndex() # Add index
.map(lambda ri: row_with_index(*list(ri[0]) + [ri[1]])) # Map to rows
.toDF(schema)) # It will work without schema but will be more expensive
# inSet in Spark < 1.3
indexed.where(col("index").isin(indexes))
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