Spark：如何通过mapInPandas正确转换数据帧 [英] Spark: How to correctly transform dataframe by mapInPandas
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问题描述
我正在尝试使用最新的Spark 3.0.1函数转换10k行的Spark数据帧mapInPandas。
预期输出：映射的PANDAS_Function()将一行转换为三行，因此输出Transform_df应具有30k行
当前输出：我得到3行1核和24行8核。
输入：Response_sdf有10k行
+-----+-------------------------------------------------------------------+
|url |content |
+-----+-------------------------------------------------------------------+
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
|api_1|{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } |
|api_2|{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }|
+-----+-------------------------------------------------------------------+
only showing top 20 rows
Input respond_sdf has 10000 rows
输出A)3行-1核-.master(‘local[1]’)
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } (0 + 1) / 1]
+-----+---+---+
| api| A| B|
+-----+---+---+
|api_1| 1| 4|
|api_1| 2| 5|
|api_1| 3| 6|
+-----+---+---+
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
Output transformed_df has 3 rows
输出B)24行-8核-.master(‘local[8]’)
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } (0 + 1) / 1]
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
+-----+---+---+
| api| A| B|
+-----+---+---+
|api_1| 1| 4|
|api_1| 2| 5|
|api_1| 3| 6|
|api_1| 1| 4|
|api_1| 2| 5|
|api_1| 3| 6|
|api_1| 1| 4|
|api_1| 2| 5|
|api_1| 3| 6|
|api_1| 1| 4|
|api_1| 2| 5|
|api_1| 3| 6|
|api_1| 1| 4|
|api_1| 2| 5|
|api_1| 3| 6|
|api_1| 1| 4|
|api_1| 2| 5|
|api_1| 3| 6|
|api_1| 1| 4|
|api_1| 2| 5|
+-----+---+---+
only showing top 20 rows
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] } (3 + 5) / 8]
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }
Output transformed_df has 24 rows
示例代码：
#### IMPORT PYSPARK ###
import pandas as pd
import pyspark
from pyspark.sql import Row
from pyspark.sql.types import StructType, StructField, IntegerType,StringType
spark = pyspark.sql.SparkSession.builder.appName("test")
.master('local[1]')
.getOrCreate()
sc = spark.sparkContext
####### INPUT DATAFRAME WITH LIST OF JSONS ########################
# Create list with 10k nested tuples(url,content)
rdd_list = [('api_1',"{'api': ['api_1', 'api_1', 'api_1'],'A': [1,2,3], 'B': [4,5,6] }"),
(' api_2', "{'api': ['api_2', 'api_2', 'api_2'],'A': [7,8,9], 'B': [10,11,12] }")]*5000
schema = StructType([
StructField('url', StringType(), True),
StructField('content', StringType(), True)
])
#Create input dataframe with 10k rows
jsons = sc.parallelize(rdd_list)
respond_sdf = spark.createDataFrame(jsons, schema)
respond_sdf.show(truncate=False)
print(f'Input respond_sdf has {respond_sdf.count()} rows')
####### TRANSFORMATION DATAFRAME ########################
# Pandas transformation function returning pandas dataframe
def pandas_function(iter):
for df in iter:
print(df['content'][0])
yield pd.DataFrame(eval(df['content'][0]))
transformed_df = respond_sdf.mapInPandas(pandas_function, "api string, A int, B int")
transformed_df.show()
print(f' Output transformed_df has {transformed_df.count()} rows')
print(f'Expected output dataframe should has 30k rows')
相关讨论链接： How to yield pandas dataframe rows to spark dataframe
推荐答案
很抱歉，在我对您上一个问题的回答中，使用mapInPandas的部分不正确。我认为下面这个函数是编写 pandas 函数的正确方式。上次我犯了一个错误，因为我之前认为iter是行的迭代，但实际上它是数据帧的迭代。
def pandas_function(iter):
for df in iter:
yield pd.concat(pd.DataFrame(x) for x in df['content'].map(eval))
(PS感谢here的回答。)
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