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我有一个光栅图像（Tiff 格式）和一个转换为数组的 shapefile 格式的多边形区域。我希望找到一种优雅的方法来创建一个数组，其中多边形边界内的所有元素都有 1 值，多边形外的所有元素都有值 0。我的最终目标是用从 shapefile 派生的数组屏蔽从图像派生的数组.

我有以下问题，感谢您的帮助：

在使用np.zeros((ds.RasterYSize, ds.RasterXSize))和我的多边形边界的地理空间坐标的像素位置创建一个空数组后，用 1 填充数组内的多边形的最佳解决方案是什么？

from osgeo import gdal, gdalnumeric, ogr, osr
import osgeo.gdal
import math
import numpy
import numpy as np

def world2Pixel(geoMatrix, x, y):
    """
    Uses a gdal geomatrix (gdal.GetGeoTransform()) to calculate
    the pixel location of a geospatial coordinate
    (source http://www2.geog.ucl.ac.uk/~plewis/geogg122/vectorMask.html)
    geoMatrix
    [0] = top left x (x Origin)
    [1] = w-e pixel resolution (pixel Width)
    [2] = rotation, 0 if image is "north up"
    [3] = top left y (y Origin)
    [4] = rotation, 0 if image is "north up"
    [5] = n-s pixel resolution (pixel Height)

    """
    ulX = geoMatrix[0]
    ulY = geoMatrix[3]
    xDist = geoMatrix[1]
    yDist = geoMatrix[5]
    rtnX = geoMatrix[2]
    rtnY = geoMatrix[4]
    pixel = np.round((x - ulX) / xDist).astype(np.int)
    line = np.round((ulY - y) / xDist).astype(np.int)
    return (pixel, line)

# Open the image as a read only image
ds = osgeo.gdal.Open(inFile,gdal.GA_ReadOnly)
# Get image georeferencing information.
geoMatrix = ds.GetGeoTransform()
ulX = geoMatrix[0] # top left x (x Origin)
ulY = geoMatrix[3] # top left y (y Origin)
xDist = geoMatrix[1] # w-e pixel resolution (pixel Width)
yDist = geoMatrix[5] # n-s pixel resolution (pixel Height)
rtnX = geoMatrix[2] # rotation, 0 if image is "north up"
rtnY = geoMatrix[4] #rotation, 0 if image is "north up"

# open shapefile (= border of are of interest)
shp = osgeo.ogr.Open(poly)
source_shp = ogr.GetDriverByName("Memory").CopyDataSource(shp, "")
# get the coordinates of the points from the boundary of the shapefile
source_layer = source_shp.GetLayer(0)
feature = source_layer.GetNextFeature()
geometry = feature.GetGeometryRef()
pts = geometry.GetGeometryRef(0)
points = []
for p in range(pts.GetPointCount()):
   points.append((pts.GetX(p), pts.GetY(p)))
pnts = np.array(points).transpose()

print pnts
pnts
array([[  558470.28969598,   559495.31976318,   559548.50931402,
    559362.85560495,   559493.99688721,   558958.22572622,
    558529.58862305,   558575.0174293 ,   558470.28969598],
    [ 6362598.63707171,  6362629.15167236,  6362295.16466266,
    6362022.63453845,  6361763.96246338,  6361635.8559779 ,
    6361707.07684326,  6362279.69352024,  6362598.63707171]])

# calculate the pixel location of a geospatial coordinate (= define the border of my polygon)

pixels, line = world2Pixel(geoMatrix,pnts[0],pnts[1])
pixels
array([17963, 20013, 20119, 19748, 20010, 18939, 18081, 18172, 17963])
line
array([35796, 35734, 36402, 36948, 37465, 37721, 37579, 36433, 35796])

#create an empty array with value zero using 
data = np.zeros((ds.RasterYSize, ds.RasterXSize))
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2 回答 2

2

这本质上是一个多边形内的点问题。

这是一个小库来解决这个问题。它来自此页面，并进行了一些修改以使其更具可读性。

点子

#From http://www.ariel.com.au/a/python-point-int-poly.html
# Modified by Nick ODell
from collections import namedtuple

def point_in_polygon(target, poly):
    """x,y is the point to test. poly is a list of tuples comprising the polygon."""
    point = namedtuple("Point", ("x", "y"))
    line = namedtuple("Line", ("p1", "p2"))
    target = point(*target)

    inside = False
    # Build list of coordinate pairs
    # First, turn it into named tuples

    poly = map(lambda p: point(*p), poly)

    # Make two lists, with list2 shifted forward by one and wrapped around
    list1 = poly
    list2 = poly[1:] + [poly[0]]
    poly = map(line, list1, list2)

    for l in poly:
        p1 = l.p1
        p2 = l.p2

        if p1.y == p2.y:
            # This line is horizontal and thus not relevant.
            continue
        if max(p1.y, p2.y) < target.y <= min(p1.y, p2.y):
            # This line is too high or low
            continue
        if target.x < max(p1.x, p2.x):
            # Ignore this line because it's to the right of our point
            continue
        # Now, the line still might be to the right of our target point, but 
        # still to the right of one of the line endpoints.
        rise = p1.y - p2.y
        run =  p1.x - p2.x
        try:
            slope = rise/float(run)
        except ZeroDivisionError:
            slope = float('inf')

        # Find the x-intercept, that is, the place where the line we are
        # testing equals the y value of our target point.

        # Pick one of the line points, and figure out what the run between it
        # and the target point is.
        run_to_intercept = target.x - p1.x
        x_intercept = p1.x + run_to_intercept / slope
        if target.x < x_intercept:
            # We almost crossed the line.
            continue

        inside = not inside

    return inside

if __name__ == "__main__":
    poly = [(2,2), (1,-1), (-1,-1), (-1, 1)]
    print point_in_polygon((1.5, 0), poly)
于 2012-11-16T19:17:11.890 回答
2

接受的答案对我不起作用。

我最终使用了 shapely 库。

sudo pip install shapely

代码：

import shapely.geometry 

poly = shapely.geometry.Polygon([(2,2), (1,-1), (-1,-1), (-1, 1)])
point = shapely.geometry.Point(1.5, 0)

point.intersects(poly)
于 2015-07-09T18:41:39.607 回答