是的,你可以使用类似SpatialSpark。它仅适用于Spark 1.6.1,但您可以使用BroadcastSpatialJoin创建效率极高的RTree
。
下面是使用SpatialSpark与PySpark我的一个例子,以检查是否不同多边形在彼此或相交的:
from ast import literal_eval as make_tuple
print "Java Spark context version:", sc._jsc.version()
spatialspark = sc._jvm.spatialspark
rectangleA = Polygon([(0, 0), (0, 10), (10, 10), (10, 0)])
rectangleB = Polygon([(-4, -4), (-4, 4), (4, 4), (4, -4)])
rectangleC = Polygon([(7, 7), (7, 8), (8, 8), (8, 7)])
pointD = Point((-1, -1))
def geomABWithId():
return sc.parallelize([
(0L, rectangleA.wkt),
(1L, rectangleB.wkt)
])
def geomCWithId():
return sc.parallelize([
(0L, rectangleC.wkt)
])
def geomABCWithId():
return sc.parallelize([
(0L, rectangleA.wkt),
(1L, rectangleB.wkt),
(2L, rectangleC.wkt)])
def geomDWithId():
return sc.parallelize([
(0L, pointD.wkt)
])
dfAB = sqlContext.createDataFrame(geomABWithId(), ['id', 'wkt'])
dfABC = sqlContext.createDataFrame(geomABCWithId(), ['id', 'wkt'])
dfC = sqlContext.createDataFrame(geomCWithId(), ['id', 'wkt'])
dfD = sqlContext.createDataFrame(geomDWithId(), ['id', 'wkt'])
# Supported Operators: Within, WithinD, Contains, Intersects, Overlaps, NearestD
SpatialOperator = spatialspark.operator.SpatialOperator
BroadcastSpatialJoin = spatialspark.join.BroadcastSpatialJoin
joinRDD = BroadcastSpatialJoin.apply(sc._jsc, dfABC._jdf, dfAB._jdf, SpatialOperator.Within(), 0.0)
joinRDD.count()
results = joinRDD.collect()
map(lambda result: make_tuple(result.toString()), results)
# [(0, 0), (1, 1), (2, 0)] read as:
# ID 0 is within 0
# ID 1 is within 1
# ID 2 is within 0
注线
joinRDD = BroadcastSpatialJoin.apply(sc._jsc, dfABC._jdf, dfAB._jdf, SpatialOperator.Within(), 0.0)
的最后一个参数是一个缓冲器值,在你的情况下,这将是你想要使用的宽容。如果您使用纬度/经度,它可能会是一个非常小的数字,因为它是一个径向系统,并且取决于您想要的公差,您需要输入calculate based on lat/lon for your area of interest。
使用k-d树可以开始。 –
我工作过类似的用例。我们使用'GeoHashes'的属性来定义单元格,并定义每个单元格的过程。这仍然是一个广泛的问题。也许你可以展示你目前的方法的代码来推动讨论? – maasg
@LostInOverflow - 当然,通过它。 – OrangeRind