flatMap (word, (song, another_song)) 到 rdd
我有一个具有以下结构的 rdd:
[('song1', 'artist1', ['word1', 'word2', 'word3', 'word1']),
('song2', 'artist2', ['word1', 'word3', 'word1'])]
我想要映射与不同歌曲一起出现的每个单词,例如:
('word1', ('song1', 'song2')), ('word2', ('song1')), ('word3', ('song1', 'song2'))
我尝试使用 flatMap 和 reduceByKey 执行以下操作:
rdd_songs = rdd.map(lambda x: [(word, x[0]) for word in x[2]]).flatMap(lambda x: x).reduceByKey(lambda a, b: a+b)
输出是我想要的,但不完全这样做:
[('word1', 'song1song1song2'), ('word2', 'song1'), ('word3', 'song1song2')
我是告诉 flatMapValues 函数会很有用,但似乎也无法理解该方法,很好,任何帮助都会非常有帮助
I have a rdd with this structure:
[('song1', 'artist1', ['word1', 'word2', 'word3', 'word1']),
('song2', 'artist2', ['word1', 'word3', 'word1'])]
Where I want map every word that appears with the different songs, for example:
('word1', ('song1', 'song2')), ('word2', ('song1')), ('word3', ('song1', 'song2'))
What I have tried is the following using flatMap and reduceByKey:
rdd_songs = rdd.map(lambda x: [(word, x[0]) for word in x[2]]).flatMap(lambda x: x).reduceByKey(lambda a, b: a+b)
The output is what I want but not exactly doing this:
[('word1', 'song1song1song2'), ('word2', 'song1'), ('word3', 'song1song2')
I was told to that the flatMapValues function would be useful but also can't seem to understand the method, very well, any help would be very helpful
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