Plotly绘制县级FIPS数据Choropleth地图遇MultiPolygon不可迭代错误

问题重现 #

运行官方示例代码绘制佛罗里达州县级人口 choropleth 地图时，抛出错误：

TypeError: 'MultiPolygon' object is not iterable

原代码如下：

import plotly.figure_factory as ff
import numpy as np
import pandas as pd

df_sample = pd.read_csv('https://raw.githubusercontent.com/plotly/datasets/master/minoritymajority.csv')
df_sample_r = df_sample[df_sample['STNAME'] == 'Florida']

values = df_sample_r['TOT_POP'].tolist()
fips = df_sample_r['FIPS'].tolist()

endpts = list(np.mgrid[min(values):max(values):4j])
colorscale = ["#030512","#1d1d3b","#323268","#3d4b94","#3e6ab0",
              "#4989bc","#60a7c7","#85c5d3","#b7e0e4","#eafcfd"]
fig = ff.create_choropleth(
    fips=fips, values=values, scope=['Florida'], show_state_data=True,
    colorscale=colorscale, binning_endpoints=endpts, round_legend_values=True,
    plot_bgcolor='rgb(229,229,229)',
    paper_bgcolor='rgb(229,229,229)',
    legend_title='Population by County',
    county_outline={'color': 'rgb(255,255,255)', 'width': 0.5},
    exponent_format=True,
)
fig.layout.template = None
fig.show()

问题原因 #

这个错误源于Shapely 2.x+版本的API变更：旧版Shapely允许直接迭代MultiPolygon对象，新版则需要通过geoms属性访问其包含的多边形。而plotly.figure_factory.create_choropleth的底层逻辑还未适配该变更，导致迭代MultiPolygon时出错。

解决方案 #

方案1：临时补丁适配Shapely 2.x #

在代码开头添加一段补丁，让MultiPolygon对象支持旧版迭代方式，无需修改原绘图逻辑：

# 先添加补丁代码
import shapely.geometry.multipolygon
def _multipolygon_iter(self):
    yield from self.geoms
shapely.geometry.multipolygon.MultiPolygon.__iter__ = _multipolygon_iter

# 然后是原有的绘图代码
import plotly.figure_factory as ff
import numpy as np
import pandas as pd

df_sample = pd.read_csv('https://raw.githubusercontent.com/plotly/datasets/master/minoritymajority.csv')
df_sample_r = df_sample[df_sample['STNAME'] == 'Florida']

values = df_sample_r['TOT_POP'].tolist()
fips = df_sample_r['FIPS'].tolist()

endpts = list(np.mgrid[min(values):max(values):4j])
colorscale = ["#030512","#1d1d3b","#323268","#3d4b94","#3e6ab0",
              "#4989bc","#60a7c7","#85c5d3","#b7e0e4","#eafcfd"]
fig = ff.create_choropleth(
    fips=fips, values=values, scope=['Florida'], show_state_data=True,
    colorscale=colorscale, binning_endpoints=endpts, round_legend_values=True,
    plot_bgcolor='rgb(229,229,229)',
    paper_bgcolor='rgb(229,229,229)',
    legend_title='Population by County',
    county_outline={'color': 'rgb(255,255,255)', 'width': 0.5},
    exponent_format=True,
)
fig.layout.template = None
fig.show()

方案2：改用Plotly Express（推荐） #

Plotly官方现在更推荐使用plotly.express的choropleth_mapbox绘制地理地图，它对新版Shapely兼容性更好，代码也更简洁：

import plotly.express as px
import pandas as pd

df_sample = pd.read_csv('https://raw.githubusercontent.com/plotly/datasets/master/minoritymajority.csv')
df_sample_r = df_sample[df_sample['STNAME'] == 'Florida']

fig = px.choropleth_mapbox(
    df_sample_r,
    geojson="https://raw.githubusercontent.com/plotly/datasets/master/geojson-counties-fips.json",
    locations='FIPS',
    color='TOT_POP',
    color_continuous_scale=["#030512","#1d1d3b","#323268","#3d4b94","#3e6ab0",
                            "#4989bc","#60a7c7","#85c5d3","#b7e0e4","#eafcfd"],
    range_color=(df_sample_r['TOT_POP'].min(), df_sample_r['TOT_POP'].max()),
    mapbox_style="carto-positron",
    zoom=6,
    center={"lat": 27.9944, "lon": -81.7603},  # 佛罗里达州中心坐标
    labels={'TOT_POP':'Population'},
    title='Population by County (Florida)'
)
fig.update_layout(margin={"r":0,"t":40,"l":0,"b":0})
fig.show()

该方案无需依赖旧版Shapely，同时支持交互性更强的Mapbox地图，更适合长期使用。

内容的提问来源于stack exchange，提问作者Tyler