Plotly (python) 上的地图：在哪一层创建带有城市名称的卫星 map？

Question

I want to create a satellite map like these: https://plotly.com/python/mapbox-layers/ , but i would like to display the names of the countries and cities on the map like on Google Maps (not in hover boxes like on this example, but written on the map).

Do you know how to do?

Answer 1

Simply use the text parameter, as explained in the scatter_mapbox documentation :

import pandas as pd
us_cities = pd.read_csv("https://raw.githubusercontent.com/plotly/datasets/master/us-cities-top-1k.csv")

fig = px.scatter_mapbox(us_cities, lat="lat", lon="lon", text="City", hover_name="City", hover_data=["State", "Population"],
                        color_discrete_sequence=["fuchsia"], zoom=3, height=300)
fig.update_layout(mapbox_style="open-street-map")
fig.update_layout(margin={"r":0,"t":0,"l":0,"b":0})
fig.show()

This will display the city names as shown in the map below:

Please note that all city names will not be displayed at low zoom levels: they will gradually start appearing as you zoom in.

Answer 2

Obtaining a Mapbox API token will enable the display of place names on satellite maps. The Mapbox API can be obtained here . (It is free).

import plotly.express as px
import plotly.graph_objects as go
import pandas as pd

us_cities = pd.read_csv("https://raw.githubusercontent.com/plotly/datasets/master/us-cities-top-1k.csv")

token = open("mapbox_api_key.txt").read()

fig = px.scatter_mapbox(us_cities,
                        lat="lat",
                        lon="lon",
                        hover_name="City",
                        hover_data=["State", "Population"],
                        color_discrete_sequence=["fuchsia"],
                        zoom=3,
                        height=300
                       )
fig.update_layout(
    mapbox=dict(
    style="satellite-streets", 
    accesstoken=token,
    center=go.layout.mapbox.Center(
        lat=42.5,
        lon=-76
    ),
    pitch=0,
    zoom=5
    )
)

fig.update_layout(margin={"r":0,"t":0,"l":0,"b":0})
fig.show()