正在使用pyspark计算groupBy的总计数百分比

dm7nw8vv  于 2023-03-11  发布在  Spark
关注(0)|答案(4)|浏览(299)

我在pyspark中有下面的代码,结果是一个表,显示了一列的不同值和它们的计数。我希望有另一列显示每一行占总计数的百分比。我该怎么做呢?

difrgns = (df1
           .groupBy("column_name")
           .count()
           .sort(desc("count"))
           .show())

先谢了!

vi4fp9gy

vi4fp9gy1#

如果对注解中提到的窗口化不满意,则作为替代方案的示例是更好的方法:

# Running in Databricks, not all stuff required
from pyspark.sql import Row
from pyspark.sql import SparkSession
import pyspark.sql.functions as F
from pyspark.sql.types import *
#from pyspark.sql.functions import col

data = [("A", "X", 2, 100), ("A", "X", 7, 100), ("B", "X", 10, 100),
        ("C", "X", 1, 100), ("D", "X", 50, 100), ("E", "X", 30, 100)]
rdd = sc.parallelize(data)

someschema = rdd.map(lambda x: Row(c1=x[0], c2=x[1], val1=int(x[2]), val2=int(x[3])))

df = sqlContext.createDataFrame(someschema)

tot = df.count()

df.groupBy("c1") \
  .count() \
  .withColumnRenamed('count', 'cnt_per_group') \
  .withColumn('perc_of_count_total', (F.col('cnt_per_group') / tot) * 100 ) \
  .show()

退货:

+---+-------------+-------------------+
| c1|cnt_per_group|perc_of_count_total|
+---+-------------+-------------------+
|  E|            1| 16.666666666666664|
|  B|            1| 16.666666666666664|
|  D|            1| 16.666666666666664|
|  C|            1| 16.666666666666664|
|  A|            2|  33.33333333333333|
+---+-------------+-------------------+

我专注于Scala,这似乎更容易。也就是说,通过注解建议的解决方案使用Window,这是我在Scala中使用over()所做的。

ymdaylpp

ymdaylpp2#

df本身是一个更复杂的转换链,并且运行两次--首先计算总数,然后分组并计算百分比--开销太大时,可以利用窗口函数来实现类似的结果。下面是一个更通用的代码(扩展了bluephantomanswer),它可以与许多group-by维一起使用:

from pyspark.sql import Row
from pyspark.sql import SparkSession
from pyspark.sql.functions import *
from pyspark.sql.types import *
from pyspark.sql.window import Window

data = [("A", "X", 2, 100), ("A", "X", 7, 100), ("B", "X", 10, 100),
        ("C", "X", 1, 100), ("D", "X", 50, 100), ("E", "X", 30, 100)]
rdd = sc.parallelize(data)

someschema = rdd.map(lambda x: Row(c1=x[0], c2=x[1], val1=int(x[2]), val2=int(x[3])))

df = (sqlContext.createDataFrame(someschema)
      .withColumn('total_count', count('*').over(Window.partitionBy(<your N-1 dimensions here>)))
     .groupBy(<your N dimensions here>)
       .agg((count('*')/first(col('total_count'))).alias('percent_total'))
)

df.show()
pexxcrt2

pexxcrt23#

您可以使用groupby并使用agg进行聚合。例如,对于以下DataFrame:

+--------+-----+
|category|value|
+--------+-----+
|       a|    1|
|       b|    2|
|       a|    3|
+--------+-----+

您可以用途:

import pyspark.sql.functions as F

df.groupby('category').agg(
    (F.count('value')).alias('count'),
    (F.count('value') / df.count()).alias('percentage')
).show()

输出:

+--------+-----+------------------+
|category|count|        percentage|
+--------+-----+------------------+
|       b|    1|0.3333333333333333|
|       a|    2|0.6666666666666666|
+--------+-----+------------------+

或者,您可以使用SQL:

df.createOrReplaceTempView('df')

spark.sql(
    """
    SELECT category,
           COUNT(*) AS count,
           COUNT(*) / (SELECT COUNT(*) FROM df) AS ratio
    FROM df
    GROUP BY category
    """
).show()
yx2lnoni

yx2lnoni4#

更多的“美化”输出,消除多余的小数和排序

import pyspark.sql.functions as func
count_cl = data_fr.count()

data_fr \
.groupBy('col_name') \
.count() \
.withColumn('%', func.round((func.col('count')/count_cl)*100,2)) \
.orderBy('count', ascending=False) \
.show(4, False)
+--------------+-----+----+
    | col_name     |count|   %|
    +--------------------+----+
    |      C.LQQQQ |30957|8.91|
    |      C.LQQQQ |29688|8.54|
    |      C-LQQQQ |29625|8.52|
    |       CLQQQQ |29342|8.44|    
    +--------------------+----+

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