我的问题是由用例计算出spark数据帧中连续行之间的差异引起的。
例如,我有:
>>> df.show()
+-----+----------+
|index| col1|
+-----+----------+
| 0.0|0.58734024|
| 1.0|0.67304325|
| 2.0|0.85154736|
| 3.0| 0.5449719|
+-----+----------+
如果我选择使用“窗口”函数来计算这些,那么我可以这样做:
>>> winSpec = Window.partitionBy(df.index >= 0).orderBy(df.index.asc())
>>> import pyspark.sql.functions as f
>>> df.withColumn('diffs_col1', f.lag(df.col1, -1).over(winSpec) - df.col1).show()
+-----+----------+-----------+
|index| col1| diffs_col1|
+-----+----------+-----------+
| 0.0|0.58734024|0.085703015|
| 1.0|0.67304325| 0.17850411|
| 2.0|0.85154736|-0.30657548|
| 3.0| 0.5449719| null|
+-----+----------+-----------+
问题:我将数据框明确地划分为一个分区。这会对性能产生什么影响,如果有的话,为什么会这样,我又如何避免呢?因为当我不指定分区时,会收到以下警告:
16/12/24 13:52:27 WARN WindowExec: No Partition Defined for Window operation! Moving all data to a single partition, this can cause serious performance degradation.
最佳答案
实际上,对性能的影响几乎与完全省略partitionBy
子句的影响相同。所有记录都将被改组到一个分区,在本地进行排序,并一个接一个地依次迭代。
差异仅在于总共创建的分区数。让我们用一个带有10个分区和1000条记录的简单数据集的示例进行说明:
df = spark.range(0, 1000, 1, 10).toDF("index").withColumn("col1", f.randn(42))
如果您定义不带子句的框架
w_unpart = Window.orderBy(f.col("index").asc())
并与
lag
一起使用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]
与带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
仅具有一个非空分区:
df_lag_part.rdd.glom().filter(lambda x: x).count()
1
不幸的是,在PySpark中没有通用的解决方案可以用来解决此问题。这只是实现与分布式处理模型相结合的固有机制。
由于
index
列是连续的,因此您可以生成每个分区具有固定记录数的人工分区键: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")
)
并使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
具有大致均匀的数据分布(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.
您可以随时选择:
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"))
和
join
: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
关于apache-spark - 避免Spark窗口功能中单个分区模式的性能影响,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41313488/