如何用pylab绘制心脏

Question

如何用pylab绘制心脏？我搜索谷歌的方式来绘制图片，但我想知道如何用pylab绘制它。有人可以帮忙吗？画面应该像this：

Answer 1

在其他溶液使​​用连接式：

import pylab 
x = scipy.linspace(-2,2,1000) 
y1 = scipy.sqrt(1-(abs(x)-1)**2) 
y2 = -3*scipy.sqrt(1-(abs(x)/2)**0.5) 
pylab.fill_between(x, y1, color='red') 
pylab.fill_between(x, y2, color='red') 
pylab.xlim([-2.5, 2.5]) 
pylab.text(0, -0.4, 'Stack Overflow', fontsize=24, fontweight='bold', 
      color='white', horizontalalignment='center') 
pylab.savefig('heart.png') 

Answer 2

提示：3D Love Heart：从鼠尾草看看例如

x, y, z = var('x, y, z') 
f(x, y, z) = (x^2+(9/4)*y^2+z^2-1)^3-x^2*z^3-(9/80)*y^2*z^3 
P = implicit_plot3d(f, (x, -3, 3), (y, -3, 3), (z, -3, 3), 
        frame=False, axes=True, figsize=6,color="red") 
P.show(viewer='tachyon') 

Answer 3

你可以看到here，你怎么绘制一个3D炉膛。

文章的作者已经把隐函数绘图can be found here和炉膛的隐函数，并得到了下面的代码：

#!/usr/bin/env python3 
from mpl_toolkits.mplot3d import Axes3D 
from matplotlib import cm 
from matplotlib.ticker import LinearLocator, FormatStrFormatter 
import matplotlib.pyplot as plt 
import numpy as np 


def heart_3d(x,y,z): 
    return (x**2+(9/4)*y**2+z**2-1)**3-x**2*z**3-(9/80)*y**2*z**3 


def plot_implicit(fn, bbox=(-1.5, 1.5)): 
    ''' create a plot of an implicit function 
    fn ...implicit function (plot where fn==0) 
    bbox ..the x,y,and z limits of plotted interval''' 
    xmin, xmax, ymin, ymax, zmin, zmax = bbox*3 
    fig = plt.figure() 
    ax = fig.add_subplot(111, projection='3d') 
    A = np.linspace(xmin, xmax, 100) # resolution of the contour 
    B = np.linspace(xmin, xmax, 40) # number of slices 
    A1, A2 = np.meshgrid(A, A) # grid on which the contour is plotted 

    for z in B: # plot contours in the XY plane 
     X, Y = A1, A2 
     Z = fn(X, Y, z) 
     cset = ax.contour(X, Y, Z+z, [z], zdir='z', colors=('r',)) 
     # [z] defines the only level to plot 
     # for this contour for this value of z 

    for y in B: # plot contours in the XZ plane 
     X, Z = A1, A2 
     Y = fn(X, y, Z) 
     cset = ax.contour(X, Y+y, Z, [y], zdir='y', colors=('red',)) 

    for x in B: # plot contours in the YZ plane 
     Y, Z = A1, A2 
     X = fn(x, Y, Z) 
     cset = ax.contour(X+x, Y, Z, [x], zdir='x',colors=('red',)) 

    # must set plot limits because the contour will likely extend 
    # way beyond the displayed level. Otherwise matplotlib extends the plot limits 
    # to encompass all values in the contour. 
    ax.set_zlim3d(zmin, zmax) 
    ax.set_xlim3d(xmin, xmax) 
    ax.set_ylim3d(ymin, ymax) 

    plt.show() 

if __name__ == '__main__': 
    plot_implicit(heart_3d) 

我已经改变了蟒蛇的第一行中的python3。如果您使用Python 2，则需要将其设置回来。