在 Spark 中将字符串字段转换为时间戳的更好方法

问题描述：

我有一个 CSV，其中一个字段是特定格式的日期时间.我不能直接在我的 Dataframe 中导入它，因为它需要是一个时间戳.所以我将它作为字符串导入并将其转换为 Timestamp 像这样

I have a CSV in which a field is datetime in a specific format. I cannot import it directly in my Dataframe because it needs to be a timestamp. So I import it as string and convert it into a Timestamp like this

import java.sql.Timestamp
import java.text.SimpleDateFormat
import java.util.Date
import org.apache.spark.sql.Row

def getTimestamp(x:Any) : Timestamp = {
    val format = new SimpleDateFormat("MM/dd/yyyy' 'HH:mm:ss")
    if (x.toString() == "") 
    return null
    else {
        val d = format.parse(x.toString());
        val t = new Timestamp(d.getTime());
        return t
    }
}

def convert(row : Row) : Row = {
    val d1 = getTimestamp(row(3))
    return Row(row(0),row(1),row(2),d1)
}

有没有更好、更简洁的方法来做到这一点，使用 Dataframe API 或 spark-sql?上面的方法需要创建一个RDD并再次为Dataframe提供schema.

Is there a better, more concise way to do this, with the Dataframe API or spark-sql? The above method requires the creation of an RDD and to give the schema for the Dataframe again.

答

Spark >= 2.2

从 2.2 开始，您可以直接提供格式字符串:

Since you 2.2 you can provide format string directly:

import org.apache.spark.sql.functions.to_timestamp

val ts = to_timestamp($"dts", "MM/dd/yyyy HH:mm:ss")

df.withColumn("ts", ts).show(2, false)

// +---+-------------------+-------------------+
// |id |dts                |ts                 |
// +---+-------------------+-------------------+
// |1  |05/26/2016 01:01:01|2016-05-26 01:01:01|
// |2  |#$@#@#             |null               |
// +---+-------------------+-------------------+

火花 >= 1.6，

您可以使用 Spark 1.5 中引入的日期处理函数.假设您有以下数据:

You can use date processing functions which have been introduced in Spark 1.5. Assuming you have following data:

val df = Seq((1L, "05/26/2016 01:01:01"), (2L, "#$@#@#")).toDF("id", "dts")

您可以使用 unix_timestamp 来解析字符串并将其转换为时间戳

You can use unix_timestamp to parse strings and cast it to timestamp

import org.apache.spark.sql.functions.unix_timestamp

val ts = unix_timestamp($"dts", "MM/dd/yyyy HH:mm:ss").cast("timestamp")

df.withColumn("ts", ts).show(2, false)

// +---+-------------------+---------------------+
// |id |dts                |ts                   |
// +---+-------------------+---------------------+
// |1  |05/26/2016 01:01:01|2016-05-26 01:01:01.0|
// |2  |#$@#@#             |null                 |
// +---+-------------------+---------------------+

如您所见，它涵盖了解析和错误处理.格式字符串应与 Java 兼容 SimpleDateFormat.

As you can see it covers both parsing and error handling. The format string should be compatible with Java SimpleDateFormat.

火花 >= 1.5，

你必须使用这样的东西:

You'll have to use use something like this:

unix_timestamp($"dts", "MM/dd/yyyy HH:mm:ss").cast("double").cast("timestamp")

或

(unix_timestamp($"dts", "MM/dd/yyyy HH:mm:ss") * 1000).cast("timestamp")

由于 SPARK-11724.

火花

您应该能够将这些与 expr 和 HiveContext 一起使用.

you should be able to use these with expr and HiveContext.

在 Spark 中将字符串字段转换为时间戳的更好方法

相关推荐