存储来自三重 for 循环的数据

Question

我有这段代码是我在 MATLAB 中编写的，用于存储矩阵。我使用的是单元格数组，但在 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

我在想 Python 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)

但它不起作用，我收到的错误消息是

list index out of range.

Answer 1

在执行此操作之前，您需要将 z 和 Rv 定义为已知大小的 2D 和 3D 数组。

另请注意，python（和 numpy）数组是从零开始的，因此只需直接使用 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)

然而，正如@GlobalTraveler 指出的那样，最pythonic/numpyish 的方法是利用广播而不使用循环：

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

Answer 2

我建议使用 broadcasting 功能

然后可以替换循环以阐明代码：

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])

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octave/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

这减少了厄运金字塔，并且由于 numpy 的魔法，它可能比 for 循环更快。编辑：格式化 Edit2：检查

存储来自三重 for 循环的数据

Store data from a triple for loop

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storing-data