Pandas自定义函数报错：无法将Series转为float，求多城市距离计算方案

我创建了如下Pandas DataFrame：

import pandas as pd
import numpy as np
from math import sin, cos, sqrt, atan2, radians


ds1 = {'Longitude':[-46.6736,-46.50926,-46.75166,-46.54743], "Latitude" : [-23.69057,-23.41165,-23.51482,-23.42598]}
df1 = pd.DataFrame(data=ds1)

输出内容：

print(df1)

   Longitude  Latitude
0  -46.67360 -23.69057
1  -46.50926 -23.41165
2  -46.75166 -23.51482
3  -46.54743 -23.42598

需要计算该DataFrame中每条记录到多个巴西城市的距离（单位：公里），这些城市的经纬度如下：

coordinates = {
    "rio" : [-23.02,-43.474889],
    "curitiba" : [-25.38792,-49.27741],
    "portoAlegre" : [-29.98115,-51.19597],
    "salvador" : [-12.97369,-38.43908],
    "manaus" :[-3.012972,-59.926802],
    "campoGrande" : [-20.52243,-54.58743],
    "beloHorizonte" : [-19.79722,-43.95691],
    "portoVelho" : [-8.774148,-63.851237],
    "recife" : [-8.12673,-34.90491],
    "boaVista" : [2.844999,-60.718089],
    "fortaleza" : [-3.76489,-38.51496],
    "rioBranco" : [-9.972341,-67.801294],
    "palmas" : [-10.165953,-48.880833],
    "natal" : [-5.79861,-35.18398],
    "aracaju" : [-10.972717,-37.068985],
    "teresina" : [-5.10247,-42.79552]
}

距离计算函数如下：

def radius(latitude1, longitude1, latitude2, longitude2):
    R = 6373.0

    lat1 = radians(latitude1)
    lon1 = radians(longitude1)
    lat2 = radians(latitude2)
    lon2 = radians(longitude2)
    dlon = lon2 - lon1
    dlat = lat2 - lat1
    a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
    c = 2 * atan2(sqrt(a), sqrt(1 - a))
    distance = R * c
    return distance

尝试计算到里约热内卢的距离时：

df1['distanceFromRio'] = radius(coordinates["rio"][0],coordinates["rio"][1],df1['Latitude'],df1['Longitude'])

触发错误：

TypeError: cannot convert the series to <class 'float'>

希望避免使用数组/列表，直接为coordinates里的所有城市生成对应距离列（如distanceFromCuritiba、distanceFromPortoAlegre等），该怎么实现？

1. 错误根源 #

原来用的math模块函数只支持单个浮点数，而Pandas Series是一组数值集合，直接传入会导致类型不匹配报错。需要改用支持向量化运算的numpy对应函数，这样就能直接处理整个Series。

2. 修改距离计算函数 #

把math的函数替换成numpy版本，让函数兼容Series输入：

def radius(latitude1, longitude1, latitude2, longitude2):
    R = 6373.0

    lat1 = np.radians(latitude1)
    lon1 = np.radians(longitude1)
    lat2 = np.radians(latitude2)
    lon2 = np.radians(longitude2)
    dlon = lon2 - lon1
    dlat = lat2 - lat1
    a = np.sin(dlat / 2)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon / 2)**2
    c = 2 * np.arctan2(np.sqrt(a), np.sqrt(1 - a))
    distance = R * c
    return distance

3. 批量生成所有城市的距离列 #

遍历coordinates字典，对每个城市调用修改后的函数，直接生成对应列：

for city, (lat, lon) in coordinates.items():
    # 格式化列名，比如rio -> distanceFromRio
    col_name = f"distanceFrom{city.capitalize()}"
    df1[col_name] = radius(lat, lon, df1['Latitude'], df1['Longitude'])

4. 查看结果 #

执行后，df1会自动新增16个距离列，每个列对应DataFrame中所有点到目标城市的公里数。可以用print(df1)查看完整结果。

内容的提问来源于stack exchange，提问作者Giampaolo Levorato