df.repartition没有列参数分区？

在PySpark中，重分区模块有一个可选的列参数，该参数会通过该键重分区数据框架。

我的问题是 - 当没有密钥时，Spark如何重新分配？我无法深入研究源代码，以找到Spark本身的位置。

def repartition(self, numPartitions, *cols):

"""
Returns a new :class:`DataFrame` partitioned by the given partitioning expressions. The
resulting DataFrame is hash partitioned.

:param numPartitions:
    can be an int to specify the target number of partitions or a Column.
    If it is a Column, it will be used as the first partitioning column. If not specified,
    the default number of partitions is used.

.. versionchanged:: 1.6
   Added optional arguments to specify the partitioning columns. Also made numPartitions
   optional if partitioning columns are specified.

>>> df.repartition(10).rdd.getNumPartitions()
10
>>> data = df.union(df).repartition("age")
>>> data.show()
+---+-----+
|age| name|
+---+-----+
|  5|  Bob|
|  5|  Bob|
|  2|Alice|
|  2|Alice|
+---+-----+
>>> data = data.repartition(7, "age")
>>> data.show()
+---+-----+
|age| name|
+---+-----+
|  2|Alice|
|  5|  Bob|
|  2|Alice|
|  5|  Bob|
+---+-----+
>>> data.rdd.getNumPartitions()
7
"""
if isinstance(numPartitions, int):
    if len(cols) == 0:
        return DataFrame(self._jdf.repartition(numPartitions), self.sql_ctx)
    else:
        return DataFrame(
            self._jdf.repartition(numPartitions, self._jcols(*cols)), self.sql_ctx)
elif isinstance(numPartitions, (basestring, Column)):
    cols = (numPartitions, ) + cols
    return DataFrame(self._jdf.repartition(self._jcols(*cols)), self.sql_ctx)
else:
    raise TypeError("numPartitions should be an int or Column")

例如：调用这些行完全没问题，但我不知道它的作用。它是整条线的哈希值吗？或许数据框中的第一栏？

df_2 = df_1\

   .where(sf.col('some_column') == 1)\
   .repartition(32)\
   .alias('df_2')