Question

我正在使用Spark 1.5.1

当我这样做时

df <- createDataFrame(sqlContext, iris)

#creating a new column for category "Setosa"

df$Species1<-ifelse((df)[[5]]=="setosa",1,0)

head(df)

输出：新列创建

      Sepal.Length  Sepal.Width  Petal.Length  Petal.Width    Species
1     5.1           3.5           1.4          0.2            setosa
2     4.9           3.0           1.4          0.2            setosa
3     4.7           3.2           1.3          0.2            setosa
4     4.6           3.1           1.5          0.2            setosa
5     5.0           3.6           1.4          0.2            setosa
6     5.4           3.9           1.7          0.4            setosa

但是当我将虹膜数据集保存为CSV文件并尝试读取并将其转换为sparkR数据帧时

df <- read.df(sqlContext,"/Users/devesh/Github/deveshgit2/bdaml/data/iris/",
              source = "com.databricks.spark.csv",header = "true",inferSchema = "true")

现在当我尝试创建新列

时

df$Species1<-ifelse((df)[[5]]=="setosa",1,0)

我收到以下错误：

16/02/05 12:11:01 ERROR RBackendHandler: col on 922 failed Error in select(x, x$"*", alias(col, colName)) :    
error in evaluating the argument 'col' in selecting a method for function 'select': Error in invokeJava(isStatic = FALSE, objId$id, methodName, ...) :    
org.apache.spark.sql.AnalysisException: Cannot resolve column name "Sepal.Length" among (Sepal.Length, Sepal.Width, Petal.Length, Petal.Width, Species);    
at org.apache.spark.s

Answer 1

SparkSQL不支持嵌入点的名称。当您使用createDataFrame名称时会自动为您调整，对于其他方法，您必须明确提供架构：

schema <- structType(
  structField("Sepal_Length", "double"),
  structField("Sepal_Width", "double"),
  structField("Petal_Length", "double"),      
  structField("Petal_Width", "double"),
  structField("Species", "string"))

df <- read.df(sqlContext, path, source = "com.databricks.spark.csv",
  header="true", schema=schema)

在SparkR中使用createDataFrame和read.df时的不同行为

1 个答案: