在实践中,性能影响几乎与您省略了partitionBy子句相同.所有记录将被洗牌到一个分区,在本地排序并逐个顺序迭代.
差异仅在于总共创建的分区数.让我们举例说明使用包含10个分区和1000个记录的简单数据集的示例：
df = spark.range(0, 1000, 1, 10).toDF("index").withColumn("col1", f.randn(42))
如果您定义没有partition by子句的框架
w_unpart = Window.orderBy(f.col("index").asc())
并使用滞后
df_lag_unpart = df.withColumn(
"diffs_col1", f.lag("col1", 1).over(w_unpart) - f.col("col1")
)
总共只有一个分区：
df_lag_unpart.rdd.glom().map(len).collect()
[1000]
与具有虚拟索引的帧定义相比(与您的代码相比简化了一点：
w_part = Window.partitionBy(f.lit(0)).orderBy(f.col("index").asc())
将使用等于spark.sql.shuffle.partitions的分区数：
spark.conf.set("spark.sql.shuffle.partitions", 11)
df_lag_part = df.withColumn(
"diffs_col1", f.lag("col1", 1).over(w_part) - f.col("col1")
)
df_lag_part.rdd.glom().count()
11
只有一个非空分区：
df_lag_part.rdd.glom().filter(lambda x: x).count()
1
遗憾的是,没有通用的解决方案可以用来解决PySpark中的这个问题.这只是实现的固有机制与分布式处理模型相结合.
由于索引列是顺序的,因此您可以生成每个块具有固定数量记录的人工分区键：
rec_per_block = df.count() // int(spark.conf.get("spark.sql.shuffle.partitions"))
df_with_block = df.withColumn(
"block", (f.col("index") / rec_per_block).cast("int")
)
并用它来定义框架规范：
w_with_block = Window.partitionBy("block").orderBy("index")
df_lag_with_block = df_with_block.withColumn(
"diffs_col1", f.lag("col1", 1).over(w_with_block) - f.col("col1")
)
这将使用预期的分区数：
df_lag_with_block.rdd.glom().count()
11
大致统一的数据分布(我们无法避免哈希冲突)：
df_lag_with_block.rdd.glom().map(len).collect()
[0, 180, 0, 90, 90, 0, 90, 90, 100, 90, 270]
但是在块边界上有许多空白：
df_lag_with_block.where(f.col("diffs_col1").isNull()).count()
12
由于边界易于计算：
from itertools import chain
boundary_idxs = sorted(chain.from_iterable(
# Here we depend on sequential identifiers
# This could be generalized to any monotonically increasing
# id by taking min and max per block
(idx - 1, idx) for idx in
df_lag_with_block.groupBy("block").min("index")
.drop("block").rdd.flatMap(lambda x: x)
.collect()))[2:] # The first boundary doesn't carry useful inf.
你总是可以选择：
missing = df_with_block.where(f.col("index").isin(boundary_idxs))
并分别填写：
# We use window without partitions here. Since number of records
# will be small this won't be a performance issue
# but will generate "Moving all data to a single partition" warning
missing_with_lag = missing.withColumn(
"diffs_col1", f.lag("col1", 1).over(w_unpart) - f.col("col1")
).select("index", f.col("diffs_col1").alias("diffs_fill"))
并加入：
combined = (df_lag_with_block
.join(missing_with_lag, ["index"], "leftouter")
.withColumn("diffs_col1", f.coalesce("diffs_col1", "diffs_fill")))
获得理想的结果：
mismatched = combined.join(df_lag_unpart, ["index"], "outer").where(
combined["diffs_col1"] != df_lag_unpart["diffs_col1"]
)
assert mismatched.count() == 0
)
