python+matplotlib绘制简单的海豚(顶点和节点的操作)

Ines ·
更新时间:2024-11-15
· 767 次阅读

海豚

本文例子主要展示了如何使用补丁、路径和转换类绘制和操作给定的顶点和节点的形状。

测试可用。

import matplotlib.cm as cm import matplotlib.pyplot as plt from matplotlib.patches import Circle, PathPatch from matplotlib.path import Path from matplotlib.transforms import Affine2D import numpy as np # Fixing random state for reproducibility np.random.seed(19680801) r = np.random.rand(50) t = np.random.rand(50) * np.pi * 2.0 x = r * np.cos(t) y = r * np.sin(t) fig, ax = plt.subplots(figsize=(6, 6)) circle = Circle((0, 0), 1, facecolor='none', edgecolor=(0, 0.8, 0.8), linewidth=3, alpha=0.5) ax.add_patch(circle) im = plt.imshow(np.random.random((100, 100)), origin='lower', cmap=cm.winter, interpolation='spline36', extent=([-1, 1, -1, 1])) im.set_clip_path(circle) plt.plot(x, y, 'o', color=(0.9, 0.9, 1.0), alpha=0.8) # Dolphin from OpenClipart library by Andy Fitzsimon # <cc:License rdf:about="http://web.resource.org/cc/PublicDomain"> # <cc:permits rdf:resource="http://web.resource.org/cc/Reproduction"/> # <cc:permits rdf:resource="http://web.resource.org/cc/Distribution"/> # <cc:permits rdf:resource="http://web.resource.org/cc/DerivativeWorks"/> # </cc:License> dolphin = """ M -0.59739425,160.18173 C -0.62740401,160.18885 -0.57867129,160.11183 -0.57867129,160.11183 C -0.57867129,160.11183 -0.5438361,159.89315 -0.39514638,159.81496 C -0.24645668,159.73678 -0.18316813,159.71981 -0.18316813,159.71981 C -0.18316813,159.71981 -0.10322971,159.58124 -0.057804323,159.58725 C -0.029723983,159.58913 -0.061841603,159.60356 -0.071265813,159.62815 C -0.080250183,159.65325 -0.082918513,159.70554 -0.061841203,159.71248 C -0.040763903,159.7194 -0.0066711426,159.71091 0.077336307,159.73612 C 0.16879567,159.76377 0.28380306,159.86448 0.31516668,159.91533 C 0.3465303,159.96618 0.5011127,160.1771 0.5011127,160.1771 C 0.63668998,160.19238 0.67763022,160.31259 0.66556395,160.32668 C 0.65339985,160.34212 0.66350443,160.33642 0.64907098,160.33088 C 0.63463742,160.32533 0.61309688,160.297 0.5789627,160.29339 C 0.54348657,160.28968 0.52329693,160.27674 0.50728856,160.27737 C 0.49060916,160.27795 0.48965803,160.31565 0.46114204,160.33673 C 0.43329696,160.35786 0.4570711,160.39871 0.43309565,160.40685 C 0.4105108,160.41442 0.39416631,160.33027 0.3954995,160.2935 C 0.39683269,160.25672 0.43807996,160.21522 0.44567915,160.19734 C 0.45327833,160.17946 0.27946869,159.9424 -0.061852613,159.99845 C -0.083965233,160.0427 -0.26176109,160.06683 -0.26176109,160.06683 C -0.30127962,160.07028 -0.21167141,160.09731 -0.24649368,160.1011 C -0.32642366,160.11569 -0.34521187,160.06895 -0.40622293,160.0819 C -0.467234,160.09485 -0.56738444,160.17461 -0.59739425,160.18173 """ vertices = [] codes = [] parts = dolphin.split() i = 0 code_map = { 'M': (Path.MOVETO, 1), 'C': (Path.CURVE4, 3), 'L': (Path.LINETO, 1)} while i < len(parts): code = parts[i] path_code, npoints = code_map[code] codes.extend([path_code] * npoints) vertices.extend([[float(x) for x in y.split(',')] for y in parts[i + 1:i + npoints + 1]]) i += npoints + 1 vertices = np.array(vertices, float) vertices[:, 1] -= 160 dolphin_path = Path(vertices, codes) dolphin_patch = PathPatch(dolphin_path, facecolor=(0.6, 0.6, 0.6), edgecolor=(0.0, 0.0, 0.0)) ax.add_patch(dolphin_patch) vertices = Affine2D().rotate_deg(60).transform(vertices) dolphin_path2 = Path(vertices, codes) dolphin_patch2 = PathPatch(dolphin_path2, facecolor=(0.5, 0.5, 0.5), edgecolor=(0.0, 0.0, 0.0)) ax.add_patch(dolphin_patch2) plt.show()

效果如下:

总结

以上就是本文关于python+matplotlib绘制简单的海豚(顶点和节点的操作)的全部内容,希望对大家有所帮助。感兴趣的朋友可以继续参阅本站其他相关专题,如有不足之处,欢迎留言指出。感谢朋友们对本站的支持!

您可能感兴趣的文章:Python实现针对给定单链表删除指定节点的方法Python基于lxml模块解析html获取页面内所有叶子节点xpath路径功能示例python 通过xml获取测试节点和属性的实例Python算法之求n个节点不同二叉树个数python xml.etree.ElementTree遍历xml所有节点实例详解Python selenium 父子、兄弟、相邻节点定位方式详解Python获取任意xml节点值的方法python实现单链表中删除倒数第K个节点的方法



海豚 matplotlib Python

需要 登录 后方可回复, 如果你还没有账号请 注册新账号