我如何在 SparkR 中按小时分组？

Question

我正在尝试使用 SparkR 和 Spark 2.1.0 按时间汇总一些日期。我的数据如下：

                       created_at
1  Sun Jul 31 22:25:01 +0000 2016
2  Sun Jul 31 22:25:01 +0000 2016
3  Fri Jun 03 10:16:57 +0000 2016
4  Mon May 30 19:23:55 +0000 2016
5  Sat Jun 11 21:00:07 +0000 2016
6  Tue Jul 12 16:31:46 +0000 2016
7  Sun May 29 19:12:26 +0000 2016
8  Sat Aug 06 11:04:29 +0000 2016
9  Sat Aug 06 11:04:29 +0000 2016
10 Sat Aug 06 11:04:29 +0000 2016

我希望输出为：

Hour      Count
22         2
10         1
19         1
11         3
....

我试过了：

sumdf <- summarize(groupBy(df, df$created_at), count = n(df$created_at))
head(select(sumdf, "created_at", "count"),10)

但是分组到最接近的秒数：

                       created_at count
1  Sun Jun 12 10:24:54 +0000 2016     1
2  Tue Aug 09 14:12:35 +0000 2016     2
3  Fri Jul 29 19:22:03 +0000 2016     2
4  Mon Jul 25 21:05:05 +0000 2016     2

我试过了：

sumdf <- summarize(groupBy(df, hr=hour(df$created_at)), count = n(hour(df$created_at)))
head(select(sumdf, "hour(created_at)", "count"),20)

但这给出了：

  hour(created_at) count
1               NA     0

我试过了：

sumdf <- summarize(groupBy(df, df$created_at), count = n(hour(df$created_at)))
head(select(sumdf, "created_at", "count"),10)

但这给出了：

                       created_at count
1  Sun Jun 12 10:24:54 +0000 2016     0
2  Tue Aug 09 14:12:35 +0000 2016     0
3  Fri Jul 29 19:22:03 +0000 2016     0
4  Mon Jul 25 21:05:05 +0000 2016     0
...

如何使用小时函数来实现，或者有更好的方法吗？

Answer 1

这是SCALA的代码，我想你可以参考一下。

    var index = ss.sparkContext.parallelize( Seq(
  (1,"Sun Jul 31 22:25:01 +0000 2016"),
  (2,"Sun Jul 31 22:25:01 +0000 2016"),
  (3,"Fri Jun 03 10:16:57 +0000 2016"),
  (4,"Mon May 30 19:23:55 +0000 2016"),
  (5,"Sat Jun 11 21:00:07 +0000 2016"),
  (6,"Tue Jul 12 16:31:46 +0000 2016"),
  (7,"Sun May 29 19:12:26 +0000 2016"),
  (8,"Sat Aug 06 11:04:29 +0000 2016"),
  (9,"Sat Aug 06 11:04:29 +0000 2016"),
  (10,"Sat Aug 06 11:04:29 +0000 2016"))
).toDF("ID", "time")

val getHour = udf( (s : String) => {
  s.substring( 11, 13)
})
index.withColumn("hour", getHour($"time")).groupBy( "hour").agg( count("*").as("count")).show

Answer 2

假设您的本地 table 是 df，这里真正的问题是从您的 created_at 列中提取小时，然后使用您的分组代码。为此，您可以使用 dapply:

library(SparkR)
sc1 <- sparkR.session()
df2 <- createDataFrame(df)

#with dapply you need to specify the schema i.e. the data.frame that will come out
#of the applied function - i.e. substringDF in our case
schema <- structType(structField('created_at', 'string'), structField('time', 'string'))

#a function that will be applied to each partition of the spark data frame.
#remember that each partition is a data.frame itself.
substringDF <- function(DF) {

 DF$time <- substr(DF$created_at, 15, 16)

 DF

}

#and then we use the above in dapply
df3 <- dapply(df2, substringDF, schema)
head(df3)
#                        created_at time
#1 1  Sun Jul 31 22:25:01 +0000 2016   22
#2 2  Sun Jul 31 22:25:01 +0000 2016   22
#3 3  Fri Jun 03 10:16:57 +0000 2016   10
#4 4  Mon May 30 19:23:55 +0000 2016   19
#5 5  Sat Jun 11 21:00:07 +0000 2016   21
#6 6  Tue Jul 12 16:31:46 +0000 2016   16

然后只需应用您的正常分组代码：

sumdf <- summarize(groupBy(df3, df3$time), count = n(df3$time))
head(select(sumdf, "time", "count"))
#  time count
#1   11     3
#2   22     2
#3   16     1
#4   19     2
#5   10     1
#6   21     1

Answer 3

我会用 to_timestamp (Spark 2.2) 或 unix_timestamp %>% cast("timestamp") （早期版本）解析日期并访问 hour:

df <- createDataFrame(data.frame(created_at="Sat Aug 19 12:33:26 +0000 2017"))
head(count(group_by(df, 
  alias(hour(to_timestamp(column("created_at"), "EEE MMM d HH:mm:ss Z yyyy")), "hour")
)))
##  hour count
## 1   14     1

我如何在 SparkR 中按小时分组？

How do I group by hour in SparkR?

r

apache-spark

sparkr