PySpark - 轉換的RDD成一個鍵值對RDD，與值列表是

Question

我有一個RDD與元組的形式是：

[("a1","b1","c1","d1","e1"), ("a2","b2","c2","d2","e2"), ... 

我想是要變換成關鍵 - 值對RDD，其中，所述第一場將首先串（鍵）和第二場字符串（值）的列表，即，欲把它轉化爲以下形式：

[("a1",["b1","c1","d1","e1"]), ("a2",["b2","c2","d2","e2"]), ... 

Answer 1

>>> rdd = sc.parallelize([("a1","b1","c1","d1","e1"), ("a2","b2","c2","d2","e2")]) 

>>> result = rdd.map(lambda x: (x[0], list(x[1:]))) 

>>> print result.collect() 
[('a1', ['b1', 'c1', 'd1', 'e1']), ('a2', ['b2', 'c2', 'd2', 'e2'])] 

說明lambda x: (x[0], list(x[1:]))：

1. x[0]將使所述第一元件是 輸出的第一個元素
2. x[1:]將使第一個除外的所有元素是 在第二元件
3. list(x[1:])將迫使該要一個列表 ，因爲默認將是一個元組