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[Docs] Fix typo for Python section on unifying Kafka streams #9545

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[Docs] Fix typo for Python section on unifying Kafka streams #9545

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4 changes: 2 additions & 2 deletions docs/streaming-programming-guide.md
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Original file line number Diff line number Diff line change
Expand Up @@ -1948,8 +1948,8 @@ unifiedStream.print();
{% highlight python %}
numStreams = 5
kafkaStreams = [KafkaUtils.createStream(...) for _ in range (numStreams)]
unifiedStream = streamingContext.union(kafkaStreams)
unifiedStream.print()
unifiedStream = streamingContext.union(*kafkaStreams)
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@srowen srowen Nov 9, 2015

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@C-kang actually I don't know Python syntax well enough to be sure about this -- this is varargs syntax?

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@theckang theckang Nov 9, 2015

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@srowen right. The method is: def union(self, *dstreams), so it expects a variable number of dstreams passed in as arguments. If you pass in a list of dstreams without unpacking it, then vargs= [kafkaStreams, ] as opposed to vargs = [stream1, stream2, stream3, ...]. In the method then, len(vargs) == 1, so it just returns [kafkaStreams] as the unifiedStream without actually performing the union. And then unifiedStream.pprint() will fail because unifiedStream is not a dstream but a list.

unifiedStream.pprint()
{% endhighlight %}
</div>
</div>
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