由于此格式不是标准格式,您需要使用unix_timestamp函数来解析字符串并将其转换为时间戳类型:
import org.apache.spark.sql.functions._
val df = Seq(
Tuple1("04-NOV-16 03.36.13.000000000 PM"),
Tuple1("06-NOV-15 03.42.21.000000000 PM"),
Tuple1("05-NOV-15 03.32.05.000000000 PM"),
Tuple1("06-NOV-15 03.32.14.000000000 AM")
).toDF("stringCol")
val pattern = "dd-MMM-yy hh.mm.ss.S a"
val newDF = df
.withColumn("timestampCol", unix_timestamp(df("stringCol"), pattern).cast("timestamp"))
.orderBy("timestampCol")
newDF.show(false)
结果:
+-------------------------------+---------------------+
|stringCol |timestampCol |
+-------------------------------+---------------------+
|05-NOV-15 03.32.05.000000000 PM|2015-11-05 15:32:05.0|
|06-NOV-15 03.32.14.000000000 AM|2015-11-06 03:32:14.0|
|06-NOV-15 03.42.21.000000000 PM|2015-11-06 15:42:21.0|
|04-NOV-16 03.36.13.000000000 PM|2016-11-04 15:36:13.0|
+-------------------------------+---------------------+
有关unix_timestamp和其他实用函数的详细信息可以在此处找到。
要构建时间戳格式,可以参考SimpleDateFormatter文档
编辑1:正如pheeleeppoo所说,您可以直接按表达式排序,而不是创建一个新列,假设您只想在数据框中保留字符串类型的列:
val newDF = df.orderBy(unix_timestamp(df("stringCol"), pattern).cast("timestamp"))
编辑2:请注意,unix_timestamp函数的精度为秒,如果毫秒非常重要,则可以使用UDF:
def myUDF(p: String) = udf(
(value: String) => {
val dateFormat = new SimpleDateFormat(p)
val parsedDate = dateFormat.parse(value)
new java.sql.Timestamp(parsedDate.getTime())
}
)
val pattern = "dd-MMM-yy hh.mm.ss.S a"
val newDF = df.withColumn("timestampCol", myUDF(pattern)(df("stringCol"))).orderBy("timestampCol")