如何绘制周围的散点图点圆环图?

问题描述 投票:0回答:1

我有,我可以很容易地绘制足够几点散点图。我想补充周围各点的圆环图,指示哪些类弥补了这一点。我看到nested donut charts的例子,但我想为多个点分散/甜甜圈情节。

这是我迄今为止制作散点图和圆环图的代码。它将绘制所有3个数据点和第一点一个圆环图。

import numpy as np
import matplotlib.pyplot as plt

# Fixing random state for reproducibility
np.random.seed(19680801)

## Scatter
# create three data points with three random class makeups
N = 3
N_class = 5
x = np.random.rand(N)
y = np.random.rand(N)
vals = [np.random.randint(2, size=N_class) for _ in range(N)]

plt.scatter(x, y, s=500)
plt.show()

## Donut plot
# Create 5 equal sized wedges
size_of_groups = np.ones(5)

# Create a pieplot
plt.pie(size_of_groups, colors=["grey" if val == 0 else "red" for val in vals[0]])
#plt.show()

# add a circle at the center
my_circle=plt.Circle( (0,0), 0.7, color='white')
p = plt.gcf()
p.gca().add_artist(my_circle)

plt.show()

同样的事情也以此为每个点(不考虑饼图中心,只是一个散点)

Desired output like this (disregarding the pie chart center for now, I've figured that)

python matplotlib scatter-plot donut-chart
1个回答
4
投票

适应Scatter plot with pie chart markers example,一个可以只在中间,使馅饼变成甜甜圈添加白色标记。

import numpy as np
import matplotlib.pyplot as plt

# first define the ratios
r1 = 0.2       # 20%
r2 = r1 + 0.4  # 40%

# define some sizes of the scatter marker
sizes = np.array([60, 80, 120])*4
center_sizes = sizes/3.

# calculate the points of the first pie marker
#
# these are just the origin (0,0) +
# some points on a circle cos,sin
x = [0] + np.cos(np.linspace(0, 2 * np.pi * r1, 10)).tolist()
y = [0] + np.sin(np.linspace(0, 2 * np.pi * r1, 10)).tolist()
xy1 = np.column_stack([x, y])
s1 = np.abs(xy1).max()

x = [0] + np.cos(np.linspace(2 * np.pi * r1, 2 * np.pi * r2, 10)).tolist()
y = [0] + np.sin(np.linspace(2 * np.pi * r1, 2 * np.pi * r2, 10)).tolist()
xy2 = np.column_stack([x, y])
s2 = np.abs(xy2).max()

x = [0] + np.cos(np.linspace(2 * np.pi * r2, 2 * np.pi, 10)).tolist()
y = [0] + np.sin(np.linspace(2 * np.pi * r2, 2 * np.pi, 10)).tolist()
xy3 = np.column_stack([x, y])
s3 = np.abs(xy3).max()

fig, ax = plt.subplots()
ax.scatter(range(3), range(3), marker=xy1,
           s=s1 ** 2 * sizes, facecolor='indigo')
ax.scatter(range(3), range(3), marker=xy2,
           s=s2 ** 2 * sizes, facecolor='gold')
ax.scatter(range(3), range(3), marker=xy3,
           s=s3 ** 2 * sizes, facecolor='crimson')
# centers
ax.scatter(range(3), range(3), s=center_sizes, marker="o", color="w")
plt.show()

enter image description here

相反,如果一个真正的pie图表需要,你可以使用参数centerradius对轴定位的几个馅饼。

import matplotlib.pyplot as plt

# first define the ratios
r1 = 0.2       # 20%
r2 = r1 + 0.4  # 40%

x = list(range(3))
y = list(range(3))

fig, ax = plt.subplots()

for xi,yi in zip(x,y):
    ax.pie([r1,r2,r2], colors=['indigo', "gold", 'crimson'],
           center=(xi, yi), radius=0.2+xi/4, 
           wedgeprops=dict(width=(0.2+xi/4)/2), frame=True)
ax.autoscale()

plt.show()

enter image description here

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