Pyspark:使用Pyspark.sql.dataframe.dataframe需要很长时间

Question

我有一个pyspark.sql.dataframe.DataFrame如下所示

df.show()
+--------------------+----+----+---------+----------+---------+----------+---------+
|                  ID|Code|bool|      lat|       lon|       v1|        v2|       v3|
+--------------------+----+----+---------+----------+---------+----------+---------+
|5ac52674ffff34c98...|IDFA|   1|42.377167| -71.06994|17.422535|1525319638|36.853622|
|5ac52674ffff34c98...|IDFA|   1| 42.37747|-71.069824|17.683573|1525319639|36.853622|
|5ac52674ffff34c98...|IDFA|   1| 42.37757| -71.06942|22.287935|1525319640|36.853622|
|5ac52674ffff34c98...|IDFA|   1| 42.37761| -71.06943|19.110023|1525319641|36.853622|
|5ac52674ffff34c98...|IDFA|   1|42.377243| -71.06952|18.904774|1525319642|36.853622|
|5ac52674ffff34c98...|IDFA|   1|42.378254| -71.06948|20.772903|1525319643|36.853622|
|5ac52674ffff34c98...|IDFA|   1| 42.37801| -71.06983|18.084948|1525319644|36.853622|
|5ac52674ffff34c98...|IDFA|   1|42.378693| -71.07033| 15.64326|1525319645|36.853622|
|5ac52674ffff34c98...|IDFA|   1|42.378723|-71.070335|21.093477|1525319646|36.853622|
|5ac52674ffff34c98...|IDFA|   1| 42.37868| -71.07034|21.851894|1525319647|36.853622|
|5ac52674ffff34c98...|IDFA|   1|42.378716| -71.07029|20.583202|1525319648|36.853622|
|5ac52674ffff34c98...|IDFA|   1| 42.37872| -71.07067|19.738768|1525319649|36.853622|
|5ac52674ffff34c98...|IDFA|   1|42.379112| -71.07097|20.480911|1525319650|36.853622|
|5ac52674ffff34c98...|IDFA|   1| 42.37952|  -71.0708|20.526752|1525319651| 44.93808|
|5ac52674ffff34c98...|IDFA|   1| 42.37902| -71.07056|20.534052|1525319652| 44.93808|
|5ac52674ffff34c98...|IDFA|   1|42.380203|  -71.0709|19.921381|1525319653| 44.93808|
|5ac52674ffff34c98...|IDFA|   1| 42.37968|-71.071144| 20.12599|1525319654| 44.93808|
|5ac52674ffff34c98...|IDFA|   1|42.379696| -71.07114|18.760069|1525319655| 36.77853|
|5ac52674ffff34c98...|IDFA|   1| 42.38011| -71.07123|19.155525|1525319656| 36.77853|
|5ac52674ffff34c98...|IDFA|   1| 42.38022|  -71.0712|16.978994|1525319657| 36.77853|
+--------------------+----+----+---------+----------+---------+----------+---------+
only showing top 20 rows

如果尝试count

%%time
df.count()

CPU times: user 4 ms, sys: 0 ns, total: 4 ms
Wall time: 28.1 s

30241272

现在如果我取df的一个子集，计数的时间甚至更长

id0 = df.first().ID  ## First ID
tmp = df.filter( (df['ID'] == id0) )

%%time
tmp.count()

CPU times: user 12 ms, sys: 0 ns, total: 12 ms
Wall time: 1min 33s
Out[6]:
3299

Answer 1

你的问题很难回答，也很棘手

为了重现你的行为，我用一个大数据集进行了测试

问题描述

我在一个大型数据集中测试了以下两个案例：

# Case 1
df.count() # Execution time: 37secs

# Case 2
df.filter((df['ID'] == id0)).count() #Execution time: 1.39 min

解释

让我们看看只有.count()的物理计划：

== Physical Plan ==
*(2) HashAggregate(keys=[], functions=[count(1)], output=[count#38L])
+- Exchange SinglePartition
   +- *(1) HashAggregate(keys=[], functions=[partial_count(1)], output=[count#41L])
      +- *(1) FileScan csv [] Batched: false, Format: CSV, Location: InMemoryFileIndex[file:...], PartitionFilters: [], PushedFilters: [], ReadSchema: struct<>

让我们先查看物理计划，然后再查看.filter()：

== Physical Plan ==
*(2) HashAggregate(keys=[], functions=[count(1)], output=[count#61L])
+- Exchange SinglePartition
   +- *(1) HashAggregate(keys=[], functions=[partial_count(1)], output=[count#64L])
      +- *(1) Project
         +- *(1) Filter (isnotnull(ID#11) && (ID#11 = Muhammed MacIntyre))
            +- *(1) FileScan csv [ID#11] Batched: false, Format: CSV, Location: InMemoryFileIndex[file:...], PartitionFilters: [], PushedFilters: [IsNotNull(ID), EqualTo(ID,Muhammed MacIntyre)], ReadSchema: struct<_c1:string>

通常，Spark when counts of rows映射count=1的行，并减少所有映射器以创建最终行数

在案例2中Spark必须首先过滤，然后为每个分区创建部分计数，然后再进行另一个阶段将这些计数相加。因此，对于相同的行，在第二种情况下，Spark也会进行过滤，这会影响大型数据集中的计算时间。Spark是一个用于分布式处理的框架，没有像Pandas这样的索引，可以在不传递所有行的情况下极快地进行过滤

总结

在这种简单的情况下，您不能做很多事情来提高执行时间。 您可以使用不同的配置设置（例如#spark.sql.shuffle.partitions、# spark.default.parallelism、# of executors、# executor memory等）尝试应用程序

Answer 2

这是因为spark是lazily evaluated。调用tmp.count（）时，这是一个操作步骤。换句话说，tmp.count的计时也包括过滤时间。如果要真正比较这两个计数，请尝试以下方法：

%%time
df.count()

id0 = df.first().ID  ## First ID
tmp = df.filter( (df['ID'] == id0) )
tmp.persist().show()

%%time
tmp.count()

这里的重要组件是执行计数之前的tmp.persist（）.show（）。这将执行筛选并缓存结果。这样，tmp.count（）只包含实际的计数时间