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NumPy | Jupyter Notebook & Data Associated

 by   KeithGalli Jupyter Notebook Version: Current License: No License

 by   KeithGalli Jupyter Notebook Version: Current License: No License

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kandi X-RAY | NumPy Summary

NumPy is a Jupyter Notebook library typically used in Big Data, Numpy, Jupyter, Pandas applications. NumPy has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.
Jupyter Notebook & Data Associated with my Tutorial video on the Python NumPy Library.
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  • NumPy has a low active ecosystem.
  • It has 330 star(s) with 488 fork(s). There are 19 watchers for this library.
  • It had no major release in the last 12 months.
  • NumPy has no issues reported. There are 1 open pull requests and 0 closed requests.
  • It has a neutral sentiment in the developer community.
  • The latest version of NumPy is current.
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  • NumPy has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
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  • NumPy does not have a standard license declared.
  • Check the repository for any license declaration and review the terms closely.
  • Without a license, all rights are reserved, and you cannot use the library in your applications.
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NumPy Key Features

Jupyter Notebook & Data Associated with my Tutorial video on the Python NumPy Library

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Community Discussions

Trending Discussions on Big Data
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Trending Discussions on Big Data

QUESTION

How to group unassociated content

Asked 2022-Apr-15 at 12:43

I have a hive table that records user behavior

like this

userid behavior timestamp url
1 view 1650022601 url1
1 click 1650022602 url2
1 click 1650022614 url3
1 view 1650022617 url4
1 click 1650022622 url5
1 view 1650022626 url7
2 view 1650022628 url8
2 view 1650022631 url9

About 400GB is added to the table every day.

I want to order by timestamp asc, then one 'view' is in a group between another 'view' like this table, the first 3 lines belong to a same group , then subtract the timestamps, like 1650022614 - 1650022601 as the view time.

How to do this?

i try lag and lead function, or scala like this

        val pairRDD: RDD[(Int, String)] = record.map(x => {
            if (StringUtil.isDateString(x.split("\\s+")(0))) {
                partition = partition + 1
                (partition, x)
            } else {
                (partition, x)
            }
        })

or java like this

        LongAccumulator part = spark.sparkContext().longAccumulator("part");

        JavaPairRDD<Long, Row> pairRDD = spark.sql(sql).coalesce(1).javaRDD().mapToPair((PairFunction<Row, Long, Row>) row -> {
            if (row.getAs("event") == "pageview") {
                part.add(1L);
            }
        return new Tuple2<>(part.value(), row);
        });

but when a dataset is very large, this code just stupid.

save me plz

ANSWER

Answered 2022-Apr-15 at 12:43

If you use dataframe, you can build partition by using window that sum a column whose value is 1 when you change partition and 0 if you don't change partition.

You can transform a RDD to a dataframe with sparkSession.createDataframe() method as explained in this answer

Back to your problem. In you case, you change partition every time column behavior is equal to "view". So we can start with this condition:

import org.apache.spark.sql.functions.col

val df1 = df.withColumn("is_view", (col("behavior") === "view").cast("integer"))

You get the following dataframe:

+------+--------+----------+----+-------+
|userid|behavior|timestamp |url |is_view|
+------+--------+----------+----+-------+
|1     |view    |1650022601|url1|1      |
|1     |click   |1650022602|url2|0      |
|1     |click   |1650022614|url3|0      |
|1     |view    |1650022617|url4|1      |
|1     |click   |1650022622|url5|0      |
|1     |view    |1650022626|url7|1      |
|2     |view    |1650022628|url8|1      |
|2     |view    |1650022631|url9|1      |
+------+--------+----------+----+-------+

Then you use a window ordered by timestamp to sum over the is_view column:

import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions.sum

val df2 = df1.withColumn("partition", sum("is_view").over(Window.partitionBy("userid").orderBy("timestamp")))

Which get you the following dataframe:

+------+--------+----------+----+-------+---------+
|userid|behavior|timestamp |url |is_view|partition|
+------+--------+----------+----+-------+---------+
|1     |view    |1650022601|url1|1      |1        |
|1     |click   |1650022602|url2|0      |1        |
|1     |click   |1650022614|url3|0      |1        |
|1     |view    |1650022617|url4|1      |2        |
|1     |click   |1650022622|url5|0      |2        |
|1     |view    |1650022626|url7|1      |3        |
|2     |view    |1650022628|url8|1      |1        |
|2     |view    |1650022631|url9|1      |2        |
+------+--------+----------+----+-------+---------+

Then, you just have to aggregate per userid and partition:

import org.apache.spark.sql.functions.{max, min}

val result = df2.groupBy("userid", "partition")
  .agg((max("timestamp") - min("timestamp")).as("duration"))

And you get the following results:

+------+---------+--------+
|userid|partition|duration|
+------+---------+--------+
|1     |1        |13      |
|1     |2        |5       |
|1     |3        |0       |
|2     |1        |0       |
|2     |2        |0       |
+------+---------+--------+

The complete scala code:

import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions.{col, max, min, sum}

val result = df
  .withColumn("is_view", (col("behavior") === "view").cast("integer"))
  .withColumn("partition", sum("is_view").over(Window.partitionBy("userid").orderBy("timestamp")))
  .groupBy("userid", "partition")
  .agg((max("timestamp") - min("timestamp")).as("duration"))

Source https://stackoverflow.com/questions/71883786

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install NumPy

You can download it from GitHub.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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