Python 是否可以在leftOuterJoin上初始化空的默认值?
我有以下两个RDD:Python 是否可以在leftOuterJoin上初始化空的默认值?,python,apache-spark,pyspark,Python,Apache Spark,Pyspark,我有以下两个RDD: name_to_hour = sc.parallelize([("Amy", [7,8,7,18,19]), ("Dan", [6,7]), ("Emily", [1,2,3,7,7,7,2])]) name_biz = sc.parallelize(["Amy", "Brian", "Chris", "Dan", "Emily"]) 我想加入他们,所以我得到的rdd如下所示: [('Amy', [7, 8, 7, 18, 19]), ('Chris', []), ('
name_to_hour = sc.parallelize([("Amy", [7,8,7,18,19]), ("Dan", [6,7]), ("Emily", [1,2,3,7,7,7,2])])
name_biz = sc.parallelize(["Amy", "Brian", "Chris", "Dan", "Emily"])
我想加入他们,所以我得到的rdd如下所示:
[('Amy', [7, 8, 7, 18, 19]), ('Chris', []), ('Brian', []), ('Dan', [6, 7]), ('Emily', [1, 2, 3, 7, 7, 7, 2])]
我可以用我认为笨拙的解决方案实现这一点:
from pyspark import SparkContext
sc = SparkContext()
name_to_hour = sc.parallelize([("Amy", [7,8,7,18,19]), ("Dan", [6,7]), ("Emily", [1,2,3,7,7,7,2])])
name_biz = sc.parallelize(["Amy", "Brian", "Chris", "Dan", "Emily"])
temp = name_biz.map(lambda x: (x, []))
joined_rdd = temp.leftOuterJoin(name_to_hour)
def concat(my_tup):
if my_tup[1] is None:
return []
else:
return my_tup[1]
result_rdd = joined_rdd.map(lambda x: (x[0], concat(x[1])))
print "\033[0;34m{}\033[0m".format(result_rdd.collect())
有更好的方法吗
我在想,如果可以在
leftOuterJoin
上以某种方式指定,非空字段保留它们在name\u to\u hour
中的内容,空字段获得默认值[]
,我的问题就可以更容易地解决,但是我不认为有这种方法。解决这个问题的一种方法是利用Python列表的字典顺序。由于空列表总是“小于”非空列表,我们可以简单地创建一个联合
,并使用max
减少:
temp.union(name_to_hour).reduceByKey(max)
当然,这是假设键是唯一的