在实际性能的影响将是几乎一样的,如果你省略partitionBy
条款都没有。所有记录都会被混洗到一个单独的分区,在本地排序并逐一依次迭代。
区别只在于创建的分区总数。让我们用简单的数据集与10个分区和1000个记录表明,与一个例子:
df = spark.range(0, 1000, 1, 10).toDF("index").withColumn("col1", f.randn(42))
如果您by子句定义框架不分区
w_unpart = Window.orderBy(f.col("index").asc())
与lag
df_lag_unpart = df.withColumn(
"diffs_col1", f.lag("col1", 1).over(w_unpart) - f.col("col1")
)
使用总共只有一个分区:
df_lag_unpart.rdd.glom().map(len).collect()
[1000]
与用哑指标该帧定义(简化的比特相比,您的代码:
w_part = Window.partitionBy(f.lit(0)).orderBy(f.col("index").asc())
将使用等于分区数到spark.sql.shuffle.partitions
:
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
与只有一个非空分区:
df_lag_part.rdd.glom().filter(lambda x: x).count()
1
不幸的是,它可以用来在PySpark解决这个问题没有通用的解决方案。这只是实现的一种内在机制,与分布式处理模型相结合。
由于index
列是连续的,你可以产生人工分区键与固定数量的每块的记录:
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")
)
,并用它来定义帧规定:
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")
)
这将使用预期数量分区:
df_lag_with_block.rdd.glom().count()
11
与大致均匀数据分布(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]
但对块边界的多项空白:
df_lag_with_block.where(f.col("diffs_col1").isNull()).count()
12
由于边界很容易计算:
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.
你可以随时选择:
missing = df_with_block.where(f.col("index").isin(boundary_idxs))
并分别填补这些:
# 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"))
和join
:
combined = (df_lag_with_block
.join(missing_with_lag, ["index"], "leftouter")
.withColumn("diffs_col1", f.coalesce("diffs_col1", "diffs_fill")))
得到期望的结果:
mismatched = combined.join(df_lag_unpart, ["index"], "outer").where(
combined["diffs_col1"] != df_lag_unpart["diffs_col1"]
)
assert mismatched.count() == 0