需要帮助为直方图的每个区间加权（缩放）不同的因子

Question

我正在尝试制作一个粒子圆形散布的径向分布直方图，并且想要将这个直方图缩放，使得径向分布表示为每单位面积的粒子数量。

声明: 如果你对我所说的数学内容不感兴趣，可以跳过这一部分：

我将径向分布分成宽度相等的环形区域，从中心向外扩展。所以，在中心，我会有一个半径为a的圆。这个最内层区域的面积是$\pi a^{2}$。

现在，如果我们想知道从半径a到2a的环形区域的面积，我们可以计算 $$ \int_{a}^{2a} 2 \pi r \ dr = 3 \pi a^{2} $$

继续以类似的方式（从2a到3a，3a到4a，等等），我们可以看到面积是这样增加的: $$ 面积 = \pi a^{2}, 3 \pi a^{2}, 5 \pi a^{2}, 7 \pi a^{2}, ... $$

所以，当我为散布的径向分布加权直方图时，从中心向外，每个区间的计数需要加权：第一个区间的计数保持不变，第二个区间的计数要除以3，第三个区间的计数要除以5，依此类推。

所以：这是我尝试写的代码：

import numpy as np
import matplotlib.pyplot as plt

# making random sample of 100000 points between -2.5 and 2.5
y_vec = 5*np.random.random(100000) - 2.5
z_vec = 5*np.random.random(100000) - 2.5

# blank canvasses for the y, z, and radial arrays
y_vec2 = []
z_vec2 = []
R_vec =  []

# number of bins I want in the ending histogram
bns = 40

# cutting out the random samplings that aren't in a circular distribution
# and making the radial array
for i in range(0, 100000):
        if np.sqrt((y_vec[i]*y_vec[i] + z_vec[i]*z_vec[i])) <= 2.5:
                y_vec2.append(y_vec[i])
                z_vec2.append(z_vec[i])
                R_vec.append(np.sqrt(y_vec[i]*y_vec[i] + z_vec[i]*z_vec[i]))

# setting up the figures and plots
fig, ax = plt.subplots()
fig2, hst = plt.subplots()

# creating a weighting array for the histogram
wghts = []
i = 0
c = 1

# making the weighting array so that each of the bins will be weighted correctly
# (splitting the radial array up evenly in to groups of the size the bins will be
# and weighting them appropriately). I assumed the because the documentation says
# the "weights" array has to be the same size as the "x" initial input, that the 
# weights act on each point individually...
while i < bns:
        wghts.extend((1/c)*np.ones(len(R_vec)/bns))
        c = c + 2
        i = i + 1

# Making the plots
ax.scatter(y_vec2, z_vec2)
hst.hist(R_vec, bins = bns, weights = wghts)

# plotting
plt.show()

散点图看起来很好：