pyspark：转换RDD [DenseVector]到数据帧

Question

我有以下RDD：

rdd.take（5）给我：

[DenseVector([9.2463, 1.0, 0.392, 0.3381, 162.6437, 7.9432, 8.3397, 11.7699]), 
DenseVector([9.2463, 1.0, 0.392, 0.3381, 162.6437, 7.9432, 8.3397, 11.7699]), 
DenseVector([5.0, 20.0, 0.3444, 0.3295, 54.3122, 4.0, 4.0, 9.0]), 
DenseVector([9.2463, 1.0, 0.392, 0.3381, 162.6437, 7.9432, 8.3397, 11.7699]), 
DenseVector([9.2463, 2.0, 0.392, 0.3381, 162.6437, 7.9432, 8.3397, 11.7699])] 

我想使它成为一个数据帧应该看如：

------------------------------------------------------------------- 
| features              | 
------------------------------------------------------------------- 
| [9.2463, 1.0, 0.392, 0.3381, 162.6437, 7.9432, 8.3397, 11.7699] | 
|-----------------------------------------------------------------| 
| [9.2463, 1.0, 0.392, 0.3381, 162.6437, 7.9432, 8.3397, 11.7699] | 
|-----------------------------------------------------------------| 
| [5.0, 20.0, 0.3444, 0.3295, 54.3122, 4.0, 4.0, 9.0]    | 
|-----------------------------------------------------------------| 
| [9.2463, 1.0, 0.392, 0.3381, 162.6437, 7.9432, 8.3397, 11.7699] | 
|-----------------------------------------------------------------| 
| [9.2463, 2.0, 0.392, 0.3381, 162.6437, 7.9432, 8.3397, 11.7699] | 
|-----------------------------------------------------------------| 

这可能吗？我试图使用df_new = sqlContext.createDataFrame(rdd,['features'])，但它没有奏效。有没有人有任何建议？谢谢！

Answer 1

地图tuples第一：

rdd.map(lambda x: (x,)).toDF(["features"]) 

请记住，作为星火2.0的有两种不同的Vector实施的ml算法需要pyspark.ml.Vector。