scala 从单个字符串创建 Spark DataFrame
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Creating a Spark DataFrame from a single string
提问by smeeb
I'm trying to take a hardcoded String and turn it into a 1-row Spark DataFrame (with a single column of type StringType) such that:
我正在尝试采用硬编码字符串并将其转换为 1 行 Spark DataFrame（具有单个类型的列StringType），以便：
String fizz = "buzz"
Would result with a DataFrame whose .show()method looks like:
将导致 DataFrame 的.show()方法如下所示：
+-----+
| fizz|
+-----+
| buzz|
+-----+
My best attempt thus far has been:
到目前为止，我最好的尝试是：
val rawData = List("fizz")
val df = sqlContext.sparkContext.parallelize(Seq(rawData)).toDF()
df.show()
But I get the following compiler error:
但我收到以下编译器错误：
java.lang.ClassCastException: org.apache.spark.sql.types.ArrayType cannot be cast to org.apache.spark.sql.types.StructType
at org.apache.spark.sql.SQLContext.createDataFrame(SQLContext.scala:413)
at org.apache.spark.sql.SQLImplicits.rddToDataFrameHolder(SQLImplicits.scala:155)
Any ideas as to where I'm going awry? Also, how do I set "buzz"as the row value for the fizzcolumn?
关于我要去哪里的任何想法？另外，如何设置"buzz"为列的行值fizz？
Update:
更新：
Trying:
试：
sqlContext.sparkContext.parallelize(rawData).toDF()
I get a DF that looks like:
我得到一个看起来像的 DF：
+----+
| _1|
+----+
|buzz|
+----+
回答by
Try:
尝试：
sqlContext.sparkContext.parallelize(rawData).toDF()
In 2.0 you can:
在 2.0 中，您可以：
import spark.implicits._
rawData.toDF
Optionally provide a sequence of names for toDF:
（可选）为 提供一系列名称toDF：
sqlContext.sparkContext.parallelize(rawData).toDF("fizz")
