pyspark：汇总列中最频繁的值

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问题描述

  aggregrated_table = df_input.groupBy('city', 'income_bracket') \
        .agg(
       count('suburb').alias('suburb'),
       sum('population').alias('population'),
       sum('gross_income').alias('gross_income'),
       sum('no_households').alias('no_households'))

想按城市和收入等级分组，但要在每个城市内郊区有不同的收入等级。如何按每个城市最常见的收入等级分组？

Would like to group by city and income bracket but within each city certain suburbs have different income brackets. How do I group by the most frequently occurring income bracket per city?

例如：

city1 suburb1 income_bracket_10 
city1 suburb1 income_bracket_10 
city1 suburb2 income_bracket_10 
city1 suburb3 income_bracket_11 
city1 suburb4 income_bracket_10

将按收入_支架_10分组

Would be grouped by income_bracket_10

推荐答案

使用聚合前使用窗口函数可能会达到目的：

Using a window function before aggregating might do the trick:

from pyspark.sql import Window
import pyspark.sql.functions as psf

w = Window.partitionBy('city')
aggregrated_table = df_input.withColumn(
    "count", 
    psf.count("*").over(w)
).withColumn(
    "rn", 
    psf.row_number().over(w.orderBy(psf.desc("count")))
).filter("rn = 1").groupBy('city', 'income_bracket').agg(
   psf.count('suburb').alias('suburb'),
   psf.sum('population').alias('population'),
   psf.sum('gross_income').alias('gross_income'),
   psf.sum('no_households').alias('no_households'))

您还可以在汇总后使用窗口函数，因为您要保留（城市，收入居中）发生次数。

you can also use a window function after aggregating since you're keeping a count of (city, income_bracket) occurrences.

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Income