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dl5txlt9 于 2021-05-27 发布在 Spark

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我想替换 "BMU 1 Cell 1 Temp. (C)" 与 "BMU_1_Cell_1_Temp_C" 也可以把柱子铸成双柱。
尝试了以下选项：
以单引号提供列名

val df= df1.withColumn("`BMU 1 Cell 1 Temp. (C)`",col("`BMU 1 Cell 1 Temp. (C)`").cast("Double")).withColumnRenamed("`BMU 1 Cell 1 Temp. (C)`","BMU_1_Cell_1_Temp_C")

val df= df1.withColumn("BMU 1 Cell 1 Temp. (C)",col("BMU 1 Cell 1 Temp. (C)").cast("Double")).withColumnRenamed("BMU 1 Cell 1 Temp. (C)","BMU_1_Cell_1_Temp_C").replaceAll("\\.","_"))

获取以下错误：

org.apache.spark.sql.AnalysisException: cannot resolve '`BMU 1 PCB Temp. (C)`'

你能帮我做这个吗。

scala DataFrame apache-spark databricks

来源：https://stackoverflow.com/questions/63598665/replacing-special-character-with-in-dataframe-column-name-in-scala

1条答案

按热度按时间

ne5o7dgx1#

这是我的spark 3.0.0。测试。

val df = spark.createDataFrame(Seq(("1", "123.456"))).toDF("id", "BMU 1 Cell 1 Temp. (C)")

df.withColumnRenamed("BMU 1 Cell 1 Temp. (C)", "BMU_1_Cell_1_Temp_C")
  .withColumn("BMU_1_Cell_1_Temp_C", $"BMU_1_Cell_1_Temp_C".cast("double")).show

+---+-------------------+
| id|BMU_1_Cell_1_Temp_C|
+---+-------------------+
|  1|            123.456|
+---+-------------------+

这两个都管用。

val df = spark.createDataFrame(Seq(("1", "123.456"))).toDF("id", "BMU 1 Cell 1 Temp. (C)")

val cols = df.columns.map(c => c.replaceAll("([.] )|[ ]", "_").replaceAll("[()]", ""))

df.toDF(cols: _*).withColumn("BMU_1_Cell_1_Temp_C", $"BMU_1_Cell_1_Temp_C".cast("double")).show()

+---+-------------------+
| id|BMU_1_Cell_1_Temp_C|
+---+-------------------+
|  1|            123.456|
+---+-------------------+

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