使用地gepandas的空间饼图

Question

我正在尝试制作一个饼图，看起来像下面的 -

我正在使用geopandas进行此 -

us_states = gpd.read_file("conus_state.shp")
data = gpd.read_file("data_file.shp")

fig, ax = plt.subplots(figsize= (10,10))
us_states.plot(color = "None", ax = ax)
data.plot(column = ["Column1","Column2"], ax= ax, kind = "pie",subplots=True)

这给我以下错误 -

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
C:\Users\LSRATH~1.STU\AppData\Local\Temp/ipykernel_17992/1047905594.py in <module>
      1 fig, ax = plt.subplots(figsize= (10,10))
      2 us_states.plot(color = "None", ax = ax)
----> 3 diff_env.plot(column = ["WS_MON1","WS_MON2"], ax= ax, kind = "pie")

c:\python38\lib\site-packages\geopandas\plotting.py in __call__(self, *args, **kwargs)
    951         if kind in self._pandas_kinds:
    952             # Access pandas plots
--> 953             return PlotAccessor(data)(kind=kind, **kwargs)
    954         else:
    955             # raise error

c:\python38\lib\site-packages\pandas\plotting\_core.py in __call__(self, *args, **kwargs)
    921             if isinstance(data, ABCDataFrame):
    922                 if y is None and kwargs.get("subplots") is False:
--> 923                     raise ValueError(
    924                         f"{kind} requires either y column or 'subplots=True'"
    925                     )

ValueError: pie requires either y column or 'subplots=True'

即使在指定后，subplots = true，它也无法正常工作。

如何使用2列数据框制作饼图？

以下是相关列的前五行 -

diff_env[["Column1", "Column2", "geometry"]].head().to_dict()

{'Column1': {0: 2, 1: 0, 2: 0, 3: 1, 4: 12},
 'Column2': {0: 2, 1: 0, 2: 0, 3: 1, 4: 12},
 'geometry': {0: <shapely.geometry.point.Point at 0x2c94e07f190>,
  1: <shapely.geometry.point.Point at 0x2c94e07f130>,
  2: <shapely.geometry.point.Point at 0x2c94e07f0d0>,
  3: <shapely.geometry.point.Point at 0x2c94bb86d30>,
  4: <shapely.geometry.point.Point at 0x2c94e07f310>}}

Answer 1

• 您尚未提供任何可用的示例数据。已经随机生成了一些
• 这是受如何使用matplotlib绘制散布派遣饼图的启发。

样本数据

	值	1几何	大小	0
5	3	点	（-105.96116535117056 31.014979344448164）	1
2	3	点	） 36728093645）	3
7	9	点	）	303
（ -79.7060924414147155	312	（	-118.12344177257525 31.909303946488293	363
4	2	7	-102.1001252173913 28.57591221070234）	326
5	3	3点	（-96.88772103678929 47.76324024025083612） 7 7	8
3	1	-95.15025297658862	46.787613401337794	421
）	）	（	594	6
点（	Point	Point	（-100.36265715719063	44.59245298996656
9	2	4	点（-81.82966451505015 35.1613934444816056）	401

完整代码

import geopandas as gpd
import numpy as np
import shapely
import matplotlib.pyplot as plt

states = (
    gpd.read_file(
        "https://raw.githubusercontent.com/nvkelso/natural-earth-vector/master/geojson/ne_110m_admin_1_states_provinces.geojson"
    )
    .loc[lambda d: d["iso_3166_2"].ne("US-AK"), "geometry"]
    .exterior
)

# geodataframe of points where pies are to be plotted
n = 10
pies = gpd.GeoDataFrame(
    geometry=[
        shapely.geometry.Point(xy)
        for xy in zip(
            np.random.choice(np.linspace(*states.total_bounds[[0, 2]], 300), n),
            np.random.choice(np.linspace(*states.total_bounds[[1, 3]], 300), n),
        )
    ],
    data={f"value{c}": np.random.randint(1, 10, n) for c in range(2)},
    crs=states.crs,
).pipe(lambda d: d.assign(size=np.random.randint(300, 600, n)))

# utility function inspired by https://stackoverflow.com/questions/56337732/how-to-plot-scatter-pie-chart-using-matplotlib
def draw_pie(dist, xpos, ypos, size, ax):
    # for incremental pie slices
    cumsum = np.cumsum(dist)
    cumsum = cumsum / cumsum[-1]
    pie = [0] + cumsum.tolist()

    colors = ["blue", "red", "yellow"]
    for i, (r1, r2) in enumerate(zip(pie[:-1], pie[1:])):
        angles = np.linspace(2 * np.pi * r1, 2 * np.pi * r2)
        x = [0] + np.cos(angles).tolist()
        y = [0] + np.sin(angles).tolist()

        xy = np.column_stack([x, y])
        ax.scatter([xpos], [ypos], marker=xy, s=size, color=colors[i], alpha=1)

    return ax


fig, ax = plt.subplots()

ax = states.plot(ax=ax, edgecolor="black", linewidth=0.5)
for _, r in pies.iterrows():
    ax = draw_pie([r.value0, r.value1], r.geometry.x, r.geometry.y, r["size"], ax)

输出