为QGIS 2空间连接创建空间索引(PyQGIS)

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1 回答
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提问于 2025-04-18 06:34

我写了一段代码,用于在QGIS 2和2.2中进行简单的空间连接(就是找出那些在某个缓冲区内的点,并获取这个缓冲区的属性)。不过,我想使用QgsSpatialIndex来加快这个过程。接下来我该怎么做呢:

pointProvider = self.pointLayer.dataProvider()
rotateProvider = self.rotateBUFF.dataProvider()

all_point = pointProvider.getFeatures()
point_spIndex = QgsSpatialIndex()
for feat in all_point:
    point_spIndex.insertFeature(feat)

all_line = rotateProvider.getFeatures()
line_spIndex = QgsSpatialIndex()
for feat in all_line:
    line_spIndex.insertFeature(feat)

rotate_IDX = self.rotateBUFF.fieldNameIndex('bearing')
point_IDX = self.pointLayer.fieldNameIndex('bearing')

self.pointLayer.startEditing()
for rotatefeat in self.rotateBUFF.getFeatures():
    for pointfeat in self.pointLayer.getFeatures():
        if pointfeat.geometry().intersects(rotatefeat.geometry()) == True:
            pointID = pointfeat.id()
bearing = rotatefeat.attributes()[rotate_IDX]
self.pointLayer.changeAttributeValue(pointID, point_IDX, bearing)
self.pointLayer.commitChanges()

1 个回答

1

要进行这种空间连接,你可以使用QgsSpatialIndex中的intersects(QgsRectangle)函数,这个函数可以帮你找到符合条件的特征ID列表,或者使用nearestNeighbor(QgsPoint,n)函数来获取离某个点最近的n个邻居的特征ID列表。

因为你只想要那些在缓冲区内的点,所以intersects函数看起来最合适。我还没有测试过如果用一个退化的边界框(就是一个点)是否可以。如果不行的话,你可以在你的点周围创建一个非常小的边界框。

intersects函数会返回所有与给定矩形相交的特征,所以你需要对这些候选特征进行进一步的测试,以确认它们是否真的相交。

你的外层循环应该是针对这些点的(你想要从它们所在的缓冲区中添加属性值到每个点上)。

# If degenerate rectangles are allowed, delta could be 0,
# if not, choose a suitable, small value
delta = 0.1
# Loop through the points
for point in all_point:
    # Create a search rectangle
    # Assuming that all_point consist of QgsPoint
    searchRectangle = QgsRectangle(point.x() - delta, point.y()  - delta, point.x() + delta, point.y() + delta)
    # Use the search rectangle to get candidate buffers from the buffer index
    candidateIDs = line_index.intesects(searchRectangle)
    # Loop through the candidate buffers to find the first one that contains the point
    for candidateID in candidateIDs:
        candFeature == rotateProvider.getFeatures(QgsFeatureRequest(candidateID)).next()
        if candFeature.geometry().contains(point):
            # Do something useful with the point - buffer pair

            # No need to look further, so break
            break

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