在过去的几天里,我一直在尝试用python绘制循环数据,通过构建一个从0到2pi的循环直方图并拟合von mises分布。我真正想要实现的是:
拟合von mises分布的定向数据。此绘图由matplotlib、scipy和numpy构建,可在以下位置找到:http://jpktd.blogspot.com/2012/11/polar-histogram.html
这个图是用r生成的,但是给出了我想要绘制的内容。可以在这里找到:https://www.zeileis.org/news/circtree/
到目前为止我所做的:
from scipy.special import i0
import numpy as np
import matplotlib.pyploy as plt
# From my data I fitted a Von-Mises distribution, calculating Mu and Kappa.
mu = -0.343
kappa = 10.432
# Construct random Von-Mises distribution based on Mu and Kappa values
r = np.random.vonmises(mu, kappa, 1000)
# Adjust Von-Mises curve from fitted data
x = np.linspace(-np.pi, np.pi, num=501)
y = np.exp(kappa*np.cos(x-mu))/(2*np.pi*i0(kappa))
# Adjuste x limits and labels
plt.xlim(-np.pi, np.pi)
plt.xticks([-np.pi, -np.pi/2, 0, np.pi/2, np.pi],
labels=[r'$-\pi$ (0º)', r'$-\frac{\pi}{2}$ (90º)', '0 (180º)', r'$\frac{\pi}{2}$ (270º)', r'$\pi$'])
# Plot adjusted Von-Mises function as line
plt.plot(x, y, linewidth=2, color='red', zorder=3
# Plot distribution
plt.hist(r, density=True, bins=20, alpha=1, edgecolor='white');
plt.title('Slaty Cleavage Strike', fontweight='bold', fontsize=14)
我试图根据诸如python中的圆形/极性直方图之类的问题来绘制圆形直方图
# From the data above (mu, kappa, x and y):
theta = np.linspace(-np.pi, np.pi, num=50, endpoint=False)
radii = np.exp(kappa*np.cos(theta-mu))/(2*np.pi*i0(kappa))
# Bin width?
width = (2*np.pi) / 50
# Construct ax with polar projection
ax = plt.subplot(111, polar=True)
# Set Zero to North
ax.set_theta_zero_location('N')
ax.set_theta_direction(-1)
# Plot bars:
bars = ax.bar(x = theta, height = radii, width=width)
# Plot Line:
line = ax.plot(x, y, linewidth=1, color='red', zorder=3)
# Grid settings
ax.set_rgrids(np.arange(1, 1.6, 0.5), angle=0, weight= 'black');
笔记:
我的圆形直方图将我的数据绘制在错误的方向上,相差180度:比较两个直方图。请参见编辑1
我相信这可能与scipy.stats.vonmises有关,它是在[-pi,pi]上定义的https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.vonmises.htmlsee 编辑1
我的数据最初在[0,2pi]之间变化。我转换成[-pi,pi]来拟合von mises并计算mu和kappa。请参见编辑1
我真的很想把我的数据作为第一个绘图之一。我的数据是地质方向数据(方位角)。有人有什么想法吗?另外,很抱歉发了这么长的邮件。我希望这至少有帮助
编辑1:
通过浏览这些评论,我意识到有一些人对数据是否从 [0,2pi]
或者 [-pi,pi]
. 我意识到在圆形直方图中绘制的错误方向来自以下几点:
我的原始数据(地质数据)介于 [0,2pi]
,即0至360度;
但是,scipy.stats.vonmises在 [-pi, pi]
;
我从数据中减去pi,以便使用scipy.stats.vonmises my_data - pi
;
一次 Mu
以及 Kappa
我补充说,这是经过计算的(正确的) pi
到 Mu
值,恢复原来的方向,现在再次介于 [0,2pi]
.
现在,我的数据正确地指向东南:
# Add pi to fitted Mu.
mu = - 0.343 + np.pi
kappa = 10.432
x = np.linspace(-np.pi, np.pi, num=501)
y = np.exp(kappa*np.cos(x-mu))/(2*np.pi*i0(kappa))
theta = np.linspace(-np.pi, np.pi, num=50, endpoint=False)
radii = np.exp(kappa*np.cos(theta-mu))/(2*np.pi*i0(kappa))
# Bin width?
width = (2*np.pi) / 50
ax = plt.subplot(111, polar=True)
# Angles increase clockwise from North
ax.set_theta_zero_location('N')
ax.set_theta_direction(-1)
bars = ax.bar(x = theta, height = radii, width=width)
line = ax.plot(x, y, linewidth=1, color='red', zorder=3)
ax.set_rgrids(np.arange(1, 1.6, 0.5), angle=0, weight= 'black');
编辑2
正如在接受答案的评论中所暗示的,诀窍正在改变 y_lim
,如下所示:
# SE DIRECTION
mu = - 0.343 + np.pi
kappa = 10.432
x = np.linspace(-np.pi, np.pi, num=501)
y = np.exp(kappa*np.cos(x-mu))/(2*np.pi*i0(kappa))
theta = np.linspace(-np.pi, np.pi, num=50, endpoint=False)
radii = np.exp(kappa*np.cos(theta-mu))/(2*np.pi*i0(kappa))
# PLOT
plt.figure(figsize=(5,5))
ax = plt.subplot(111, polar=True)
# Bin width?
width = (2*np.pi) / 50
# Angles increase clockwise from North
ax.set_theta_zero_location('N'); ax.set_theta_direction(-1);
bars = ax.bar(x=theta, height = radii, width=width, bottom=0)
# Plot Line
line = ax.plot(x, y, linewidth=2, color='firebrick', zorder=3 )
# 'Trick': This will display Zero as a circle. Fitted Von-Mises function will lie along zero.
ax.set_ylim(-0.5, 1.5);
ax.set_rgrids(np.arange(0, 1.6, 0.5), angle=60, weight= 'bold',
labels=np.arange(0,1.6,0.5));
要做的工作:
要么规范化直方图,要么将von mises分布乘以样本的数量,如注解所示。
1条答案
按热度按时间vwkv1x7d1#
这就是我的成就:
我不太确定你是否希望x的范围是
[-pi,pi]
或者[0,2pi]
. 如果你想要这个范围[0,2pi]
相反,只需注解掉这些行ax.set_xlim
以及ax.set_xticks
.