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我有一些Python代碼用於創建隨機遊走的情節。步行將反映[-a,a]的障礙。序列中的隨後的值是由如何繪製matplotlib中一組點周圍的恆定斜率的虛線「圓錐體」?
r[n] = r[n-1] + Uni[-R, R]
,然後將其反映爲必要生成。我想要做的是繪製每個點周圍的「不確定性錐」,[-R, R]
。
這裏是Python代碼到目前爲止我有:
import matplotlib.pyplot as plt
import random
uni = random.uniform
t = []
r = []
r0 = .15 # Seed for our random walk. Can be between -a and a
a = .2 # Distance of barriers from 0. Should be in (0, 1]
R = .04 # Height of half-cone in r-direction
dt = 20 # Sample period
N = 20 # Number of samples
cone_ls = ':'
cone_clr = 'blue'#[0, .5, .5]
for i in range(N):
t.append(i*dt)
if i == 0:
r.append(r0)
else:
'''
When our cone of uncertainty outpaces out barriers,
simply sample uniformly inside the barriers.
'''
if(R > 2*a):
r.append(uni(-a, a))
continue
rn = r[i - 1] + uni(-R, R)
'''
If the sampled value comes above the upper barrier,
reflect it back below.
'''
if(rn > a):
r.append(2*a - rn)
continue
'''
If the sampled value comes below the lower barrier,
reflect it back above.
'''
if(rn < -a):
r.append(-2*a - rn)
continue
'''
Otherwise just append the sampled value.
'''
r.append(rn)
# Plot cones
for i, pt in enumerate(r):
plt.plot([t[i], t[i] + dt], [pt, pt + R], linestyle=cone_ls, color=cone_clr, linewidth=2)
plt.plot([t[i], t[i] + dt], [pt, pt - R], linestyle=cone_ls, color=cone_clr, linewidth=2)
plt.plot(t, r, 'ro')
plt.plot(t, [a]*N)
plt.plot(t, [-a]*N)
plt.axis([min(t), max(t), -2*a, 2*a])
plt.xlabel('Time (min)')
plt.ylabel('Relative Difference, r')
plt.show()
我想情節看起來像這樣添加錐後:
我也將包括在一個文件中,所以任何美化技巧值得讚賞。
編輯:解決,實現我只需要繪製錐形部分單獨。
在我意識到和你一樣意識到這只是另一組情節之後,我最終做了一些稍微複雜的事情。這一行很簡潔,謝謝! – ijustlovemath