在pyspark中,如何在将broadcast变量应用于值列表时,将具有JSON的输入RDD转换为以下指定的输出?
输入
std
广播变量
[{'id': 1, 'title': "Foo", 'items': ['a','b','c']}, {'id': 2, 'title': "Bar", 'items': ['a','b','d']}]
所需的输出
[('a': 5), ('b': 12), ('c': 42), ('d': 29)]
答案 0 :(得分:1)
编辑:最初,我的印象是传递给map
函数的函数会自动广播,但是在阅读了一些文档后,我不再确定。
在任何情况下,您都可以定义广播变量:
bv = [('a', 5), ('b', 12), ('c', 42), ('d', 29)]
# turn into a dictionary
bv = dict(bv)
broadcastVar = sc.broadcast(bv)
print(broadcastVar.value)
#{'a': 5, 'c': 42, 'b': 12, 'd': 29}
现在是available on all machines as a read-only variable。您可以使用broascastVar.value
访问字典:
例如:
import json
rdd = sc.parallelize(
[
'{"id": 1, "title": "Foo", "items": ["a","b","c"]}',
'{"id": 2, "title": "Bar", "items": ["a","b","d"]}'
]
)
def myMapper(row):
# define the order of the values for your output
key_order = ["id", "title", "items"]
# load the json string into a dict
d = json.loads(row)
# replace the items using the broadcast variable dict
d["items"] = [broadcastVar.value.get(item) for item in d["items"]]
# return the values in order
return tuple(d[k] for k in key_order)
print(rdd.map(myMapper).collect())
#[(1, u'Foo', [5, 12, 42]), (2, u'Bar', [5, 12, 29])]