I have a dataframe from which I need to create a new dataframe with a small change in the schema by doing the following operation.
>>> X = spark.createDataFrame([[1,2], [3,4]], ['a', 'b'])
>>> schema_new = X.schema.add('id_col', LongType(), False)
>>> _X = X.rdd.zipWithIndex().map(lambda l: list(l[0]) + [l[1]]).toDF(schema_new)
The problem is that in the above operation, the schema of X
gets changed inplace. So when I print X.columns
I get
>>> X.columns
['a', 'b', 'id_col']
but the values in X
are still the same
>>> X.show()
+---+---+
| a| b|
+---+---+
| 1| 2|
| 3| 4|
+---+---+
To avoid changing the schema of X
, I tried creating a copy of X
using three ways
- using copy
and deepcopy
methods from the copy
module
- simply using _X = X
The copy
methods failed and returned a
RecursionError: maximum recursion depth exceeded
The assignment method also doesn't work
>>> _X = X
>>> id(_X) == id(X)
True
Since their id
are the same, creating a duplicate dataframe doesn't really help here and the operations done on _X
reflect in X
.
So my question really is two fold
how to change the schema outplace (that is without making any changes to X
)?
and more importantly, how to create a duplicate of a pyspark dataframe?
Note:
This question is a followup to this post
As explained in the answer to the other question, you could make a deepcopy of your initial schema. We can then modify that copy and use it to initialize the new DataFrame
_X
:
import pyspark.sql.functions as F
from pyspark.sql.types import LongType
import copy
X = spark.createDataFrame([[1,2], [3,4]], ['a', 'b'])
_schema = copy.deepcopy(X.schema)
_schema.add('id_col', LongType(), False) # modified inplace
_X = X.rdd.zipWithIndex().map(lambda l: list(l[0]) + [l[1]]).toDF(_schema)
Now let's check:
print('Schema of X: ' + str(X.schema))
print('Schema of _X: ' + str(_X.schema))
Output:
Schema of X: StructType(List(StructField(a,LongType,true),StructField(b,LongType,true)))
Schema of _X: StructType(List(StructField(a,LongType,true),
StructField(b,LongType,true),StructField(id_col,LongType,false)))
Note that to copy a DataFrame
you can just use _X = X
. Whenever you add a new column with e.g. withColumn
, the object is not altered in place, but a new copy is returned.
Hope this helps!