重命名 PySpark Dataframe 中的旋转和聚合列答案

【问题标题】：Rename pivoted and aggregated column in PySpark Dataframe重命名 PySpark Dataframe 中的旋转和聚合列
【发布时间】：2016-06-15 13:28:07
【问题描述】：

使用如下数据框：

from pyspark.sql.functions import avg, first

rdd = sc.parallelize(
    [
        (0, "A", 223,"201603", "PORT"), 
        (0, "A", 22,"201602", "PORT"), 
        (0, "A", 422,"201601", "DOCK"), 
        (1,"B", 3213,"201602", "DOCK"), 
        (1,"B", 3213,"201601", "PORT"), 
        (2,"C", 2321,"201601", "DOCK")
    ]
)
df_data = sqlContext.createDataFrame(rdd, ["id","type", "cost", "date", "ship"])

df_data.show()

我在它上面做了一个支点，

df_data.groupby(df_data.id, df_data.type).pivot("date").agg(avg("cost"), first("ship")).show()

+---+----+----------------+--------------------+----------------+--------------------+----------------+--------------------+
| id|type|201601_avg(cost)|201601_first(ship)()|201602_avg(cost)|201602_first(ship)()|201603_avg(cost)|201603_first(ship)()|
+---+----+----------------+--------------------+----------------+--------------------+----------------+--------------------+
|  2|   C|          2321.0|                DOCK|            null|                null|            null|                null|
|  0|   A|           422.0|                DOCK|            22.0|                PORT|           223.0|                PORT|
|  1|   B|          3213.0|                PORT|          3213.0|                DOCK|            null|                null|
+---+----+----------------+--------------------+----------------+--------------------+----------------+--------------------+

但是我得到了这些非常复杂的列名称。在聚合上应用alias 通常可以工作，但由于在这种情况下pivot 名称更糟：

+---+----+--------------------------------------------------------------+------------------------------------------------------------------+--------------------------------------------------------------+------------------------------------------------------------------+--------------------------------------------------------------+------------------------------------------------------------------+
| id|type|201601_(avg(cost),mode=Complete,isDistinct=false) AS cost#1619|201601_(first(ship)(),mode=Complete,isDistinct=false) AS ship#1620|201602_(avg(cost),mode=Complete,isDistinct=false) AS cost#1619|201602_(first(ship)(),mode=Complete,isDistinct=false) AS ship#1620|201603_(avg(cost),mode=Complete,isDistinct=false) AS cost#1619|201603_(first(ship)(),mode=Complete,isDistinct=false) AS ship#1620|
+---+----+--------------------------------------------------------------+------------------------------------------------------------------+--------------------------------------------------------------+------------------------------------------------------------------+--------------------------------------------------------------+------------------------------------------------------------------+
|  2|   C|                                                        2321.0|                                                              DOCK|                                                          null|                                                              null|                                                          null|                                                              null|
|  0|   A|                                                         422.0|                                                              DOCK|                                                          22.0|                                                              PORT|                                                         223.0|                                                              PORT|
|  1|   B|                                                        3213.0|                                                              PORT|                                                        3213.0|                                                              DOCK|                                                          null|                                                              null|
+---+----+--------------------------------------------------------------+------------------------------------------------------------------+--------------------------------------------------------------+------------------------------------------------------------------+--------------------------------------------------------------+------------------------------------------------------------------+

有没有办法在数据透视和聚合时动态重命名列名？

【问题讨论】：

标签： python apache-spark pyspark apache-spark-sql

【解决方案1】：

一个简单的正则表达式就可以解决问题：

import re

def clean_names(df):
    p = re.compile("^(\w+?)_([a-z]+)\((\w+)\)(?:\(\))?")
    return df.toDF(*[p.sub(r"\1_\3", c) for c in df.columns])

pivoted = df_data.groupby(...).pivot(...).agg(...)

clean_names(pivoted).printSchema()
## root
##  |-- id: long (nullable = true)
##  |-- type: string (nullable = true)
##  |-- 201601_cost: double (nullable = true)
##  |-- 201601_ship: string (nullable = true)
##  |-- 201602_cost: double (nullable = true)
##  |-- 201602_ship: string (nullable = true)
##  |-- 201603_cost: double (nullable = true)
##  |-- 201603_ship: string (nullable = true)

如果您想保留函数名称，请将替换模式更改为例如\1_\2_\3。

【讨论】：

当你有很多列时，更重要的是，当你事先不知道列名时，例如由于某些 ETL 操作，您必须使用@zero323 方法

【解决方案2】：

一种简单的方法是在聚合函数之后使用别名。我从您创建的 df_data spark dataFrame 开始。

df_data.groupby(df_data.id, df_data.type).pivot("date").agg(avg("cost").alias("avg_cost"), first("ship").alias("first_ship")).show()
+---+----+---------------+-----------------+---------------+-----------------+---------------+-----------------+
| id|type|201601_avg_cost|201601_first_ship|201602_avg_cost|201602_first_ship|201603_avg_cost|201603_first_ship|
+---+----+---------------+-----------------+---------------+-----------------+---------------+-----------------+
|  1|   B|         3213.0|             PORT|         3213.0|             DOCK|           null|             null|
|  2|   C|         2321.0|             DOCK|           null|             null|           null|             null|
|  0|   A|          422.0|             DOCK|           22.0|             PORT|          223.0|             PORT|
+---+----+---------------+-----------------+---------------+-----------------+---------------+-----------------+

列名将采用“original_column_name_aliased_column_name”的形式。对于您的情况，original_column_name 将为 201601，aliased_column_name 将为 avg_cost，列名称为 201601_avg_cost（由下划线“_”链接）。

【讨论】：

这仅在您有超过 1 个聚合时才有效。就我而言，我选择了该列并在枢轴之前对其进行了转换。 IE。 df_data.select(($"date" + "_avg_cost").as("date"), $"cost")).pivot("date").agg(avg("cost"))
如何在agg 和alias 时将此下划线更改为空格？

【解决方案3】：

您可以直接为聚合设置别名：

pivoted = df_data \
    .groupby(df_data.id, df_data.type) \
    .pivot("date") \
    .agg(
       avg('cost').alias('cost'),
       first("ship").alias('ship')
    )

pivoted.printSchema()
##root
##|-- id: long (nullable = true)
##|-- type: string (nullable = true)
##|-- 201601_cost: double (nullable = true)
##|-- 201601_ship: string (nullable = true)
##|-- 201602_cost: double (nullable = true)
##|-- 201602_ship: string (nullable = true)
##|-- 201603_cost: double (nullable = true)
##|-- 201603_ship: string (nullable = true)

【讨论】：

我认为这应该是公认的答案。

【解决方案4】：

为此编写了一个简单快速的函数。享受！ :)

# This function efficiently rename pivot tables' urgly names
def rename_pivot_cols(rename_df, remove_agg):
    """change spark pivot table's default ugly column names at ease.
        Option 1: remove_agg = True: `2_sum(sum_amt)` --> `sum_amt_2`.
        Option 2: remove_agg = False: `2_sum(sum_amt)` --> `sum_sum_amt_2`
    """
    for column in rename_df.columns:
        if remove_agg == True:
            start_index = column.find('(')
            end_index = column.find(')')
            if (start_index > 0 and end_index > 0):
                rename_df = rename_df.withColumnRenamed(column, column[start_index+1:end_index]+'_'+column[:1])
        else:
            new_column = column.replace('(','_').replace(')','')
            rename_df = rename_df.withColumnRenamed(column, new_column[2:]+'_'+new_column[:1])   
    return rename_df

【讨论】：

【解决方案5】：

从 zero323 的修改版本，适用于 spark 2.4

import re

def clean_names(df):
    p = re.compile("^(\w+?)_([a-z]+)\((\w+)(,\s\w+)\)(:\s\w+)?")
    return df.toDF(*[p.sub(r"\1_\3", c) for c in df.columns])

当前列名类似于0_first(is_flashsale, false): int

【讨论】：