Question

我已经创建了两个地理数据框，并试图根据答案here找到一个地理数据框中每个点到另一个地理数据框中任何点之间的最短距离。

但是，即使我认为我已经创建了一个几何列，我也无法访问这些行。我收到此错误：

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

nameofgeodataframe.head（）返回：

   node  x_coord  y_coord   Coordinates1
0     0        0      258  POINT (0 258)
1     1        0      259  POINT (0 259)
2     2        0      260  POINT (0 260)
3     3        0      261  POINT (0 261)
4     4        0      262  POINT (0 262)

这是我非常优雅的剧本。

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的构造方式？

非常感谢您，任何帮助都会很棒！

Answer 1

您在列的新名称和名称几何之间进行混合，这是导致错误的原因：（名称翻译并非总是如此）

 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：

node  x_coord  y_coord   Coordinates1  Nearest
0     0        0      258  POINT (0 258)        3
1     1        0      259  POINT (0 259)        3
2     2        0      260  POINT (0 260)        3
3     3        0      261  POINT (0 261)        3
4     4        0      230  POINT (0 230)        4

之后，如果您有很多分数，建议您使用https://wp-kama.com/hook/get_terms_orderby：

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

以防其他任何人都在挣扎旧的Geopandas版本。这是解决方案：

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)

在地理系列中访问几何中的行

2 个答案: