如何对经度和纬度进行分箱并绘制分箱的密度？

Question

我正在测试下面的代码。

import numpy as np
import pandas as pd
#import matplotlib.pyplot as plt
#plt.style.use('seaborn-white')


df = pd.read_csv('C:\\Users\\ryans\\OneDrive\\Desktop\\business.csv')
X = df[['latitude','longitude','address']].copy()

X['latitude'].value_counts()
X['longitude'].value_counts()

结果：

-115.123695    168
-111.940325    167
-115.171130    158
-111.821087    157
-115.224485    156

-82.032188       1
-89.383229       1
-89.533178       1
-81.475399       1
-111.857103      1

下一步...

X['lat'] = pd.cut(df['latitude'], bins=10)
X['lon'] = pd.cut(df['longitude'], bins=10)
print(X)

结果：

         latitude   longitude  ...               lat                   lon
0       33.522143 -112.018481  ...  (33.187, 35.014]  (-115.536, -111.235]
1       43.605499  -79.652289  ...  (42.252, 44.062]     (-81.428, -77.17]
2       35.092564  -80.859132  ...  (35.014, 36.824]     (-81.428, -77.17]
3       33.455613 -112.395596  ...  (33.187, 35.014]  (-115.536, -111.235]
4       35.190012  -80.887223  ...  (35.014, 36.824]     (-81.428, -77.17]
          ...         ...  ...               ...                   ...
192604  36.213732 -115.177059  ...  (35.014, 36.824]  (-115.536, -111.235]
192605  44.052658  -79.481850  ...  (42.252, 44.062]     (-81.428, -77.17]
192606  33.679992 -112.035569  ...  (33.187, 35.014]  (-115.536, -111.235]
192607  33.416137 -111.735743  ...  (33.187, 35.014]  (-115.536, -111.235]
192608  36.107267 -115.171920  ...  (35.014, 36.824]  (-115.536, -111.235]

现在，我正在尝试可视化这些坐标箱，并绘制箱的密度。因此，计数越高，颜色越强烈。这可能吗？

我在网上找到了几个示例，它们展示了如何创建经度和纬度数据的热图。这是唯一的做法，还是可以对这些数据点进行分箱？

Answer 1

下面的代码做得很好！

import pandas as pd 
import folium
from folium.plugins import HeatMap


df = pd.read_csv('C:\\your_path\\business.csv')
df.head(3)

max_amount = float(df['review_count'].max())

hmap = folium.Map(location=[42.5, -75.5], zoom_start=7, )

hm_wide = HeatMap( list(zip(df.latitude.values, df.longitude.values, df.review_count.values)),
                   min_opacity=0.2,
                   max_val=max_amount,
                   radius=17, blur=15, 
                   max_zoom=1, 
                 )

hmap.add_child(hm_wide)

我需要处理其中一些颜色的不透明度和强度，但这个概念绝对是有道理的。