问题描述
我正在使用pyspark分析数据集，即使我使用的是未广播的变量，对于以下代码为何仍能正常工作，我感到有些惊讶。
I'm using pyspark to analyse a dataset and I'm a little bit surprised as to why the following code works correctly even though I'm using a variable that was not broadcast.
有问题的变量是 video ，该变量用于函数 filter
The variable in question is video, that's used in function filter, after the join.
seed = random.randint(0,999)
# df is a dataframe
# video is just one randomly sampled element
video = df.sample(False,0.001,seed).head()
# just a python list
otherVideos = [ (22,0.32),(213,0.43) ]
# transform the python list into an rdd
resultsRdd = sc.parallelize(similarVideos)
rdd = df.rdd.map(lambda row: (row.video_id,row.title))
# perform a join between resultsRdd and rdd
# note that video.title was NOT broadcast
(resultsRdd
.join(rdd)
.filter(lambda pair: pair[1][1] != video.title) # HERE!!!
.takeOrdered(10, key= lambda pair: -pair[1][0]))
我在独立模式下使用pyspark，以下是pyspark的参数-submit：
I'm using pyspark in standalone mode, with the following arguments to pyspark-submit:
-num-executors 12 --executor-cores 4 --executor-memory 1g --master local [*]
此外，我正在 jupyter （新的ipython笔记本）上运行以前的代码。
Also, I'm running the previous code on jupyter (new ipython-notebooks).
推荐答案
[重新发布评论作为答案。]
[Reposting comment as an answer.]
对于这个概念，我认为这个是一个很不错的阅读。基本上，您不需要在RDD范围之外广播所有变量，因为闭包（在您的情况下 video ）将被序列化并发送给每个执行者和任务以进行访问在任务执行期间。当正在广播的数据集很大时，广播变量很有用，因为它将作为只读缓存存在，该缓存位于执行程序上，而不会在执行程序上运行的每个任务都进行序列化/发送/反序列化。
For this concept, I think this link on understanding closures is a pretty good read. Essentially, you do not need to broadcast all variables outside the scope of an RDD since the closure (in your case video) will be serialized and sent to each executor and task for access during task execution. Broadcast variables are useful when the dataset being broadcast is large because it will exist as a read-only cache that will sit on the executor and not be serialized/sent/deserialized with each task run on that executor.
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