Question

我正在尝试创建一种经典的近似Pi的方法，该方法是将飞镖随机“扔”在带有内切圆的正方形内，然后计算落在圆上的飞镖与抛出的飞镖的比例

我的代码有效，但是我想增加被抛出的“飞镖”的数量。问题是我的代码运行非常慢。为什么会这样，我如何加快速度？

请参阅下面的代码。

import matplotlib.pyplot as plt
import matplotlib.patches as patches
import numpy as np
from decimal import Decimal

darts = 2000  # number of "darts" to be thrown

r = 10  # radius/sidelength

plt.cla()

# Make a box with an inscribed circle
box = patches.Rectangle((-r,-r), 2*r, 2*r, 
                        linewidth=2, edgecolor='k', facecolor='none')

circle = patches.Circle((0,0), radius=r,
                        linewidth=2, edgecolor='k', facecolor='none')

inCircle = 0  # number of points that land in the circle

# Plotting
ax = plt.gca()
ax.add_patch(box)
ax.add_patch(circle)
ax.axis('equal')
plt.axis('off')

for i in range(darts):
    x = np.random.uniform(-r,r)
    y = np.random.uniform(-r,r)
    
    dist = x * x + y * y
    
    if dist <= r * r:
        inCircle += 1
        
    plt.scatter(x,y, marker='o', s=2, color = '#1f77b4')

plt.title(f'Pi Approximation with {darts} "Darts"')

plt.tight_layout()

plt.savefig(f'PiApproximationwith{darts}darts.png', dpi = 600)
plt.show()

# Computes pi
print(Decimal(4 * inCircle/darts))

感谢您的帮助。

Answer 1

代码的大部分时间都花在运行for循环上，因此您应该将注意力集中在那里。可以将Numpy.random.uniform设置为返回给定大小的随机数数组。通过分配2个数组类型以分别容纳2000个变量，您可以一次获得所有点。运行for循环以查看圆中有多少，然后将这两个数组一次全部传递到散点图，如下所示：

import matplotlib.pyplot as plt
import matplotlib.patches as patches
import numpy as np
from decimal import Decimal

darts = 2000  # number of "darts" to be thrown

r = 10  # radius/sidelength

plt.cla()

# Make a box with an inscribed circle
box = patches.Rectangle((-r, -r), 2 * r, 2 * r,
                        linewidth=2, edgecolor='k', facecolor='none')

circle = patches.Circle((0, 0), radius=r,
                        linewidth=2, edgecolor='k', facecolor='none')

inCircle = 0  # number of points that land in the circle

# Plotting
ax = plt.gca()
ax.add_patch(box)
ax.add_patch(circle)
ax.axis('equal')
plt.axis('off')

array_of_rand_x = np.random.uniform(-r, r, 2000)
array_of_rand_y = np.random.uniform(-r, r, 2000)

for i in range(darts):
    x = array_of_rand_x[i]
    y = array_of_rand_y[i]

    dist = x * x + y * y

    if dist <= r * r:
        inCircle += 1

plt.scatter(array_of_rand_x, array_of_rand_y, marker='o', s=2, color='#1f77b4')

plt.title(f'Pi Approximation with {darts} "Darts"')

plt.tight_layout()

plt.savefig(f'PiApproximationwith{darts}darts.png', dpi=600)
plt.show()

# Computes pi
print(Decimal(4 * inCircle / darts))

Answer 2

我在这里看到2名嫌疑犯：

plt.scatter(x,y, marker='o', s=2, color = '#1f77b4')-每个数字的速度可能会变慢
plt.savefig(f'PiApproximationwith{darts}darts.png', dpi = 600); plt.show()渲染图。（更多点数=更多时间）

尝试禁用第2点的线条，并使用1Milion飞镖运行。如果运行很短，您将知道其显示/渲染

为什么运行这么慢，如何加快速度？

2 个答案: