python - for 循环中的嵌套字典向所有嵌套键添加相同的值

Question

我有带有多边形的地址数据和形状文件，并尝试确定每个地址与每个多边形的最近距离(以英里为单位)，然后创建一个包含所有信息的嵌套字典，格式如下:

nested_dict = {poly_1: {address1: distance, address2 : distance}, 
               poly2: {address1: distance, address2: distance}, etc}

我正在使用的完整的、适用的代码是:

import pandas as pd
from shapely.geometry import mapping, Polygon, LinearRing, Point
import geopandas as gpd
from math import radians, cos, sin, asin, sqrt

address_dict = {k: [] for k in addresses_geo.input_string}
sludge_dtc = {k: [] for k in sf_geo.unique_name}

def haversine(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

    # haversine formula
    dlon = lon2 - lon1
    dlat = lat2 - lat1
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a))
    r = 3956 # Radius of earth in miles. Use 6371 for kilometers
    return c * r

# Here's the key loop that isn't working correctly
for unique_name, i in zip(sf_geo.unique_name, sf_geo.index):
    for address, pt in zip(addresses_geo.input_string, addresses_geo.index):
        pol_ext = LinearRing(sf_geo.iloc[i].geometry.exterior.coords)
        d = pol_ext.project(addresses_geo.iloc[pt].geometry)
        p = pol_ext.interpolate(d)
        closest_point_coords = list(p.coords)[0]
        # print(closest_point_coords)
        dist = haversine(addresses_geo.iloc[pt].geometry.x,
                         addresses_geo.iloc[pt].geometry.y,
                         closest_point_coords[0], closest_point_coords[1])
        address_dict[address] = dist
    sludge_dtc[unique_name] = address_dict
# Test results on a single address
addresses_with_sludge_distance = pd.DataFrame(sludge_dtc)
print(addresses_with_sludge_distance.iloc[[1]].T)

如果我打破这段代码并尝试计算单个多边形的距离，它似乎工作得很好。但是，当我创建数据帧并检查地址时，它列出了每个多边形的相同距离。

因此，内部字典键“123 Main Street”对于外部字典中的每个多边形键将具有 5.25 英里，而“456 South Street”对于外部字典中的每个多边形键将具有 6.13 英里。 (编造例子。)

我意识到我在设置 for 循环的方式中一定做了一些愚蠢的事情，但我无法弄清楚。我颠倒了 for 语句的顺序，加上缩进——结果都是一样的。

为了明确起见，我想要发生的是:

• 取一个多边形，然后
• 对于地址数据中的每个地址，找到距该多边形的距离，并将其添加到以地址为键、距离为值的 address_dict 字典中
• 计算完所有地址后，将整个地址字典添加为sludge_dtc中多边形键的值
• 移至下一个多边形并继续

我缺少什么想法吗？

Answer 1

问题很简单，您始终使用相同的 address_dict 实例。您只需在每个按键循环内重新创建它即可。

import pandas as pd
from shapely.geometry import mapping, Polygon, LinearRing, Point
import geopandas as gpd
from math import radians, cos, sin, asin, sqrt

def haversine(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

    # haversine formula
    dlon = lon2 - lon1
    dlat = lat2 - lat1
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a))
    r = 3956 # Radius of earth in miles. Use 6371 for kilometers
    return c * r

sludge_dtc = {k: [] for k in sf_geo.unique_name}

# Here's the key loop that isn't working correctly
for unique_name, i in zip(sf_geo.unique_name, sf_geo.index):

    address_dict = {k: [] for k in addresses_geo.input_string}

    for address, pt in zip(addresses_geo.input_string, addresses_geo.index):
        pol_ext = LinearRing(sf_geo.iloc[i].geometry.exterior.coords)
        d = pol_ext.project(addresses_geo.iloc[pt].geometry)
        p = pol_ext.interpolate(d)
        closest_point_coords = list(p.coords)[0]
        # print(closest_point_coords)
        dist = haversine(addresses_geo.iloc[pt].geometry.x,
                         addresses_geo.iloc[pt].geometry.y,
                         closest_point_coords[0], closest_point_coords[1])
        address_dict[address] = dist
    sludge_dtc[unique_name] = address_dict
# Test results on a single address
addresses_with_sludge_distance = pd.DataFrame(sludge_dtc)
print(addresses_with_sludge_distance.iloc[[1]].T)

另一个考虑因素:

您正在创建空字典，其中空列表作为值，但在您直接设置值之后(空列表将被替换)。如果您需要收集值列表，您应该附加值到现有列表，例如:

address_dict[address].append(dist)

和

sludge_dtc[unique_name].append(address_dict)