Question

我正在尝试访问PipelineRDD中包含的值以下是我的开始：

1。 RDD =（密钥，代码，值）

data = [(11720, (u'I50800', 0.08229813664596274)), (11720, (u'I50801', 0.03076923076923077))]

*强调文字* 2。我需要它按第一个值分组并将其转换为（key，tuple），其中tuple =（code，value）

testFeatures = lab_FeatureTuples = labEvents.select（＆＃39; ITEMID＆＃39;，＆＃39; SUBJECT_ID＆＃39;，＆＃39; NORM_ITEM_CNT＆＃39;）\ .orderBy（＆＃39; SUBJECT_ID＆＃39;＆＃39; ITEMID＆＃39;）\ .rdd.map（lambda（ITEMID，SUBJECT_ID，NORM_ITEM_CNT）:( SUBJECT_ID，（ITEMID，NORM_ITEM_CNT））） .groupByKey（）

testFeatures =  [(11720, [(u'I50800', 0.08229813664596274)),  (u'I50801', 0.03076923076923077)])]

在元组=（代码，值）上，我希望得到以下内容：

从中创建一个sparseVector，以便我可以将它用于SVM模型

result.take（1）

Answer 1

这是一种方法：

import pyspark
import pyspark.sql.functions as sf
import pyspark.sql.types as sparktypes
sc = pyspark.SparkContext()
sqlc = pyspark.SQLContext(sc)

data = [(11720, (u'I50800', 0.08229813664596274)), 
        (11720, (u'I50801', 0.03076923076923077))]
rdd = sc.parallelize(data)

df = sqlc.createDataFrame(rdd,  ['idx', 'tuple'])
df.show()

给出，

+-----+--------------------+
|  idx|               tuple|
+-----+--------------------+
|11720|[I50800,0.0822981...|
|11720|[I50801,0.0307692...|
+-----+--------------------+

现在定义pyspark用户定义的函数：

extract_tuple_0 = sf.udf(lambda x: x[0], returnType=sparktypes.StringType())
extract_tuple_1 = sf.udf(lambda x: x[1], returnType=sparktypes.FloatType())
df = df.withColumn('tup0', extract_tuple_0(sf.col('tuple')))

df = df.withColumn('tup1', extract_tuple_1(sf.col('tuple')))
df.show()

给出：

+-----+--------------------+----------+------+
|  idx|               tuple|      tup1|  tup0|
+-----+--------------------+----------+------+
|11720|[I50800,0.0822981...|0.08229814|I50800|
|11720|[I50801,0.0307692...|0.03076923|I50801|
+-----+--------------------+----------+------+

pySpark如何访问（key，tuple）RDD（python）中元组中的值

1 个答案: