我不知道这是否是 matplotlib/python 的错误,但从 emacs 运行以下命令会消除我通过单击图形窗口上的 [x] 来终止 matplotlib 进程的能力,它没有任何效果。是否有用于终止 emacs 启动的特定进程的命令(我用谷歌搜索过,没有运气)?我可以通过找到缓冲区并执行来终止进程,C-x k
但这有点麻烦,有什么方法可以杀死所有正在运行的 python 进程?
#Simple circular box simulator, part of part_sim
#Restructure to import into gravity() or coloumb () or wind() or pressure()
#Or to use all forces: sim_full()
#Note: Implement crashing as backbone to all forces
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial.distance import pdist, squareform
N = 100 #Number of particles
R = 10000 #Box width
pR= 5 #Particle radius
r = np.random.randint(0, R, (N,2)) #Position vector
v = np.random.randint(-R/100,R/100,(N,2)) #velocity vector
a = np.array([0,-10]) #Forces
v_limit = R/2 #Speedlimit
plt.ion()
line, = plt.plot([],'o')
line2, = plt.plot([],'o') #Track a particle
plt.axis([0,R+pR,0,R+pR]
while True:
v=v+a #Advance
r=r+v
#Collision tests
r_hit_x0 = np.where(r[:,0]<0) #Hit floor?
r_hit_x1 = np.where(r[:,0]>R) #Hit roof?
r_hit_LR = np.where(r[:,1]<0) #Left wall?
r_hit_RR = np.where(r[:,1]>R) #Right wall?
#Stop at walls
r[r_hit_x0,0] = 0
r[r_hit_x1,0] = R
r[r_hit_LR,1] = 0
r[r_hit_RR,1] = R
#Reverse velocities
v[r_hit_x0,0] = -0.9*v[r_hit_x0,0]
v[r_hit_x1,0] = -v[r_hit_x1,0]
v[r_hit_LR,1] = -0.95*v[r_hit_LR,1]
v[r_hit_RR,1] = -0.99*v[r_hit_RR,1]
#Collisions
D = squareform(pdist(r))
ind1, ind2 = np.where(D < pR)
unique = (ind1 < ind2)
ind1 = ind1[unique]
ind2 = ind2[unique]
for i1, i2 in zip(ind1, ind2):
eps = np.random.rand()
vtot= v[i1,:]+v[i2,:]
v[i1,:] = -(1-eps)*vtot
v[i2,:] = -eps*vtot
line.set_ydata(r[:,1])
line.set_xdata(r[:,0])
line2.set_ydata(r[:N/5,1])
line2.set_xdata(r[:N/5,0])
plt.draw()