Question

我有我在MATLAB中编写的用于存储矩阵的代码。我当时使用的是cells array，但是在Python中，我不知道该怎么做。

有人知道我该怎么做吗？

MATLAB代码为：

S =[0.5 0.7 0.9 1.1]; % distância entre tx e rx[m]
d = 0.2*ones(1,10); 
h = [ 0 0.1 0.2 0.3 0.4]

n_cam = length(d); % numero de camadas
n_alt = length(h); % numero de alturas
n_S = length(S); % numero de receptores
z = zeros(n_cam,n_alt); % profundidade
Rv_h = zeros(n_S,n_alt);
Rv_v = zeros(n_S,n_alt);
Rv = zeros(n_cam,n_alt);
Rh = zeros(n_cam,n_alt);
S_Rv = cell(1,n_S);
S_Rh = cell(1,n_S);
sigma = 0.3*ones(1,n_cam);
sigmaah = zeros(n_S,n_alt);


for i = 1:n_S
    for  j = 1:n_alt
        for k = 1:n_cam
            z(k,j)= (sum(d(1:k))+h(j))/S(i);
            Rv(k,j) = 1/((4*z(k,j)^2+1)^0.5);
            Rh(k,j) = ((4*z(k,j)^2+1)^0.5)-2*z(k,j);
        end 
        Rv_h(i,j) = 1/((4*(h(j)/S(i))^2+1)^0.5);
        Rh_h(i,j)=((4*(h(j)/S(i))^2+1)^0.5)-2*(h(j)/S(i));
    end
    S_Rv(:,i) = {Rv}; % z para cada camada em cada altura, para cada S
    S_Rh(:,i) = {Rh};
end

for i = 1:n_S
    for  j = 1:n_alt
        Rv = cell2mat(S_Rv(1,i));
        Rh = cell2mat(S_Rh(1,i));
        sigma_ah = sigma(1)*(Rh_h(i,j)-Rh(1,j));
        sigma_av = sigma(1)*(Rv_h(i,j)-Rv(1,j));
        for k = 2:(n_cam-1)
           sigma_ah_ant = sigma_ah;
           sigma_av_ant = sigma_av;
           sigma_ah = sigma_ah_ant + sigma(k)*(Rh(k-1,j)-Rh(k,j));
           sigma_av = sigma_av_ant + sigma(k)*(Rv(k-1,j)-Rv(k,j));
        end
        sigmaah (i,j)  = sigma_ah + sigma(end)*Rh(n_cam-1,j)
        sigmaav (i,j)  = sigma_av + sigma(end)*Rv(n_cam-1,j)
    end
end

我当时想在Python中我可以做类似的事情：

n_S = 4
n_alt = 9
n_cam = 6
Rv =[]
for i in range(1,n_S):
    for j in range(1,n_alt):
        for k in range(1,n_cam):
            z[k][j]= (sum(d[0:k])+h[j])/S[i]
            Rv[i][j][k] = 1/((4*z[k,j]**2+1)**0.5)

但是它不起作用，我收到的错误消息是“列表索引超出范围”。

Answer 1

我建议您使用broadcasting功能

然后可以替换循环以澄清代码：

from numpy import array, ones
S =array([0.5, 0.7, 0.9, 1.1])
d = 0.2*ones((10)); 
h = array([ 0, 0.1, 0.2, 0.3, 0.4])

z = ((d.cumsum()[:, None] + h).ravel() / S[:, None]).reshape((S.size, d.size, h.size))
Rv = 1 / (4 * z ** 2 + 1)** .5
# ...etc
print(Rv[-1])

哪个输出：

    [[0.93979342 0.87789557 0.80873608 0.73994007 0.67572463]
 [0.80873608 0.73994007 0.67572463 0.61782155 0.56652882]
 [0.67572463 0.61782155 0.56652882 0.52145001 0.48191875]
 [0.56652882 0.52145001 0.48191875 0.4472136  0.41665471]
 [0.48191875 0.4472136  0.41665471 0.38963999 0.36565237]
 [0.41665471 0.38963999 0.36565237 0.34425465 0.32507977]
 [0.36565237 0.34425465 0.32507977 0.30782029 0.29221854]
 [0.32507977 0.30782029 0.29221854 0.27805808 0.26515648]
 [0.29221854 0.27805808 0.26515648 0.25335939 0.24253563]
 [0.26515648 0.25335939 0.24253563 0.23257321 0.22337616]]

与八度/ matlab中的计算重叠：

Rv =

   0.93979   0.87790   0.80874   0.73994   0.67572
   0.80874   0.73994   0.67572   0.61782   0.56653
   0.67572   0.61782   0.56653   0.52145   0.48192
   0.56653   0.52145   0.48192   0.44721   0.41665
   0.48192   0.44721   0.41665   0.38964   0.36565
   0.41665   0.38964   0.36565   0.34425   0.32508
   0.36565   0.34425   0.32508   0.30782   0.29222
   0.32508   0.30782   0.29222   0.27806   0.26516
   0.29222   0.27806   0.26516   0.25336   0.24254
   0.26516   0.25336   0.24254   0.23257   0.22338

由于麻木的魔法，这减少了厄运的金字塔，并且可能比for循环更快。编辑：格式化 Edit2：给了支票

Answer 2

在执行此操作之前，需要将z和Rv定义为已知大小的2D和3D数组。

还要注意，python（和numpy）数组是从零开始的，所以直接使用range。

未测试：

import numpy as np

d = 0.2*np.arange(10)   ### not sure if this is what you meant
h = np.array([ 0 0.1 0.2 0.3 0.4])

n_S = 4
n_alt = 9
n_cam = 6
Rv = np.zeros((n_S, n_alt, n_cam))
z = np.zeros((n_cam, n_alt))
for i in range(n_S):
    for j in range(n_alt):
        for k in range(n_cam):
            z[k][j]= (sum(d[0:k])+h[j])/S[i]   ## not sure about 0:10
            Rv[i][j][k] = 1/((4*z[k,j]**2+1)**0.5)

但是，正如@GlobalTraveler指出的那样，执行此操作的最pythonic / numpyish方式是利用广播而不是完全不使用循环：

Rv = 1/np.sqrt(4 * z**2 + 1)

存储来自三重for循环的数据

2 个答案: