访问geoseries中的几何图形中的行

Question

我已经创建了两个geodataframe，我试图根据答案here找到一个geodataframe中每个点到另一个geodataframe中任意点之间的最短距离。在

然而，即使我认为我已经创建了一个几何列，我也不能访问这些行。我得到这个错误：

AttributeError: ("'Series' object has no attribute 'geometry'", u'occurred at index 0')

在GeoDataFrame.head的名称（）退货：

^{pr2}$

这是我非常不雅的剧本。在

f = h5py.File("temp_bin2x_outer_tagged.hdf", "r")
data = f["MDF/images/0/image"]
my_test = data[171, :, :]
val = filter.threshold_otsu(my_test)
binary = np.where(my_test > val, 1, 0)
outskel = skeletonize(binary)
x, y = np.where(outskel>0)
y_coord = y.tolist()
x_coord = x.tolist()
index = list(range(0,(len(x_coord))))
df = pd.DataFrame({"y_coord": y_coord, "x_coord": x_coord, "node": index})
df['Coordinates1'] = list(zip(df.x_coord, df.y_coord))
df['Coordinates1'] = df['Coordinates1'].apply(Point)
outer = geopandas.GeoDataFrame(df, geometry='Coordinates1')

f2 = h5py.File("temp_bin2x_inner_tagged.hdf", "r")
data2 = f2["MDF/images/0/image"]
my_test2 = data2[211, :, :]
val2 = filter.threshold_otsu(my_test2)
binary2 = np.where(my_test2 > val2, 1, 0)
binary2 = np.where(my_test2 > val2, 1, 0)
inskel = skeletonize(binary2)
x2, y2 = np.where(inskel>0)
y_coord2 = y2.tolist()
x_coord2 = x2.tolist()
index2 = list(range(0,(len(x_coord2))))
df2 = pd.DataFrame({"y_coord2": y_coord2, "x_coord2": x_coord2, "node": index2})
df2['Coordinates'] = list(zip(df2.x_coord2, df2.y_coord2))
df2['Coordinates'] = df2['Coordinates'].apply(Point)
inner = geopandas.GeoDataFrame(df2, geometry='Coordinates')

from shapely.ops import nearest_points
pts3 = inner.geometry.unary_union
def near(point, pts=pts3):
     nearest = inner.geometry == nearest_points(point, pts)[1]
     return inner[nearest].node.get_values()[0]
outer['Nearest'] = outer.apply(lambda row: near(row.geometry), axis=1)

我是否误解了地理数据框是如何构造的？在

非常感谢，任何帮助都会很棒！在

Answer 1

您将列的新名称与名称geometry混合使用，这是错误的原因：（名称转换并不总是完成）

 data1 = """
 node  x_coord  y_coord 
0        0      258  
1        0      259  
2        0      260  
3        0      261  
4        0      230  
 """
data2 = """
  node  x_coord  y_coord 
0        0      288  
1        0      249  
2        0      210  
3        0      259  
4        0      232  
"""
df1 = pd.read_csv(pd.compat.StringIO(data1), sep='\s+')
df2 = pd.read_csv(pd.compat.StringIO(data2), sep='\s+')
df1['Coordinates1'] = list(zip(df1.x_coord, df1.y_coord))
df1['Coordinates1'] = df1['Coordinates1'].apply(Point)
df2['Coordinates2'] = list(zip(df2.x_coord, df2.y_coord))
df2['Coordinates2'] = df2['Coordinates2'].apply(Point)

outer = gpd.GeoDataFrame(df1, geometry='Coordinates1')
inner = gpd.GeoDataFrame(df2, geometry='Coordinates2')

from shapely.ops import nearest_points
pts3 = inner.geometry.unary_union #you could use inner.Coordinates2.unary_union

def near(point, pts=pts3):
    #you could use inner.Coordinates2
    nearest = inner.geometry == nearest_points(point, pts)[1]
    return inner[nearest].node.get_values()[0]

# apply or lambda doesnt translate geometry to Coordinates1
outer['Nearest'] = outer.apply(lambda row: near(row.Coordinates1), axis=1)
print(outer)

输出：

^{pr2}$

之后，如果您有大量的要点，我建议您使用cKDTree：

from scipy.spatial import cKDTree
def ckdnearest(gdA, gdB, bcol):
    nA = np.array(list(zip(gdA.geometry.x, gdA.geometry.y)))
    nB = np.array(list(zip(gdB.geometry.x, gdB.geometry.y)))
    btree = cKDTree(nB)
    dist, idx = btree.query(nA, k=1)
    df = pd.DataFrame.from_dict({'distance': dist.astype(int),
                                 'bcol': gdB.loc[idx, bcol].values})
    return df


df = ckdnearest(outer, inner, 'node')
print(df)

输出：

    distance  bcol            (bcol equal node of inner
0         1     3
1         0     3
2         1     3
3         2     3
4         2     4

Answer 2

以防有人在为一个古老的geopandas版本而挣扎。解决方法如下：

from scipy.spatial import cKDTree
def ckdnearest(gdA, gdB, bcol):
    nA = np.array(list(zip(gdA.geometry.map(lambda val: val.x), 
    gdA.geometry.map(lambda val: val.y))))
    nB = np.array(list(zip(gdB.geometry.map(lambda val: val.x), 
    gdB.geometry.map(lambda val: val.y))))
    btree = cKDTree(nB)
    dist, idx = btree.query(nA, k=1)
    df = pd.DataFrame.from_dict({'distance': dist.astype(int),
                                 'bcol': gdB.loc[idx, bcol].values})
    return df


df = ckdnearest(outer, inner, 'node')
print(df)