如何将mllib矩阵转换为spark数据框? [英] How to convert a mllib matrix to a spark dataframe?

问题描述

我想在齐柏林飞艇笔记本中漂亮地打印相关结果:

I want to pretty print the result of a correlation in a zeppelin notebook:

val Row(coeff: Matrix) = Correlation.corr(data, "features").head

实现此目的的一种方法是将结果转换为DataFrame，并将每个值放在单独的列中，然后调用z.show().

One of the ways to achieve this is to convert the result into a DataFrame with each value in a separate column and call z.show().

但是，请查看

However, looking into the Matrix api I don't see any way to do this.

还有另一种直接的方法可以实现这一目标吗?

Is there another straight forward way to achieve this?

数据框有50列.仅仅转换为字符串将无济于事，因为输出会被截断.

The dataframe has 50 columns. Just converting to a string would not help as the output get truncated.

推荐答案

使用

Using the toString method should be the easiest and fastest way if you simply want to print the matrix. You can change the output by inputting the maximum number of lines to print as well as max line width. You can change the formatting by splitting on new lines and ",". For example:

val matrix = Matrices.dense(2,3, Array(1.0, 2.0, 3.0, 4.0, 5.0, 6.0))
matrix.toString
  .split("\n")
  .map(_.trim.split(" ").filter(_ != "").mkString("[", ",", "]"))
  .mkString("\n")

这将给出以下内容:

[1.0,3.0,5.0]
[2.0,4.0,6.0]

---

但是，如果要将矩阵转换为DataFrame，最简单的方法是先创建RDD，然后使用toDF().

---

However, if you want to convert the matrix to an DataFrame, the easiest way would be to first create an RDD and then use toDF().

val matrixRows = matrix.rowIter.toSeq.map(_.toArray)
val df = spark.sparkContext.parallelize(matrixRows).toDF("Row")

然后将每个值放在单独的列中，您可以执行以下操作

Then to put each value in a separate column you can do the following

val numOfCols = matrixRows.head.length
val df2 = (0 until numOfCols).foldLeft(df)((df, num) => 
    df.withColumn("Col" + num, $"Row".getItem(num)))
  .drop("Row")
df2.show(false)

使用示例数据的结果:

+----+----+----+
|Col0|Col1|Col2|
+----+----+----+
|1.0 |3.0 |5.0 |
|2.0 |4.0 |6.0 |
+----+----+----+

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