Converting Python to Matlab

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Zach Dunagan 2017-10-31

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/364178-converting-python-to-matlab

评论： Walter Roberson 2018-6-5

Here is my Python code...

import numpy as np
import matplotlib.pyplot as plt
import copy
def SourceDistOnPt(xc,yc,xp,yp,theta):
      ups=np.zeros((len(xc),len(xp)-1))
      wps=np.zeros((len(xc),len(xp)-1)) 
      for i in range(len(xc)):
          for j in range(len(xp)-1):
              r1=((xc[i]-xp[j])**2+(yc[i]-yp[j])**2)**(1/2)
              r2=((xc[i]-xp[j+1])**2+(yc[i]-yp[j+1])**2)**(1/2)
              nu2=np.arctan2(-(xc[i]-xp[j])*np.sin(theta[j])+(yc[i]-yp[j])*np.cos(theta[j])+(xp[j+1]-xp[j])*np.sin(theta[j])-(yp[j+1]-yp[j])*np.cos(theta[j]),\
                               (xc[i]-xp[j])*np.cos(theta[j])+(yc[i]-yp[j])*np.sin(theta[j])-(xp[j+1]-xp[j])*np.cos(theta[j])-(yp[j+1]-yp[j])*np.sin(theta[j]))
              nu1=np.arctan2(-(xc[i]-xp[j])*np.sin(theta[j])+(yc[i]-yp[j])*np.cos(theta[j]),(xc[i]-xp[j])*np.cos(theta[j])+(yc[i]-yp[j])*np.sin(theta[j]))
              ups[i,j]=1/(4*np.pi)*np.log(r1**2/r2**2)
              wps[i,j]=1/(2*np.pi)*(nu2-nu1)
      return ups,wps

I am trying to test the output of my Matlab code. I watched this video https://www.mathworks.com/videos/managing-code-in-matlab-functions-of-multiple-inputs-and-outputs-97211.html

I am trying to obtain a set of outputs for every input. I am not sure if I would add the for loops after the end for the function.

Matlab code...

function [r1, r2, nu2, nu1, ups, wps] = SourceDistOnPt(xc, yc, xp, yp, theta)
ups = zeros(length(xc), length(xp)-1);
wps = zeros(length(xc), length(xp)-1);
for i = 1:length(xp) - 1
      for j = 1:length(xc) - 1
      r1 = ((xc(i) - xp(j)).^2 + (yc(i) - yp(j)).^2)^(1/2.0);
      r2 = ((xc(i) - xp(j+1)).^2 + (yc(i) - yp(j+1)).^2).^(1/2.0);
      nu2 = atan2(-(xc(i)-xp(j)) * sin(theta(j)) + (yc(i) - yp(j)) * cos(theta(j))+(xp(j+1) - xp(j)) * sin(theta(j))-(yp(j+1) - yp(j)) * cos(theta(j)),...
                  (xc(i) - xp(j)) * cos(theta(j)) + (yc(i) - yp(j)) * sin(theta(j)) - (xp(j+1) - xp(j)) * cos(theta(j)) - (yp(j+1) - yp(j)) * sin(theta(j)));
      nu1 = atan2(-(xc(i)-xp(j)) * sin(theta(j)) + (yc(i) - yp(j)) * cos(theta(j)), (xc(i) - xp(j)) * cos(theta(j)) + (yc(i) - yp(j)) * sin(theta(j)));
      ups(i, j) = 1 /(4 * pi) * log(r1.^2/r2.^2);
      wps(i, j) = 1 /(2 * pi) * (nu2-nu1);
      end
end
      r1 = ((xc(i) - xp(j)).^2 + (yc(i) - yp(j)).^2)^(1/2.0);
      r2 = ((xc(i) - xp(j+1)).^2 + (yc(i) - yp(j+1)).^2).^(1/2.0);
      nu2 = atan2(-(xc(i)-xp(j)) * sin(theta(j)) + (yc(i) - yp(j)) * cos(theta(j))+(xp(j+1) - xp(j)) * sin(theta(j))-(yp(j+1) - yp(j)) * cos(theta(j)),...
                  (xc(i) - xp(j)) * cos(theta(j)) + (yc(i) - yp(j)) * sin(theta(j)) - (xp(j+1) - xp(j)) * cos(theta(j)) - (yp(j+1) - yp(j)) * sin(theta(j)));
      nu1 = atan2(-(xc(i)-xp(j)) * sin(theta(j)) + (yc(i) - yp(j)) * cos(theta(j)), (xc(i) - xp(j)) * cos(theta(j)) + (yc(i) - yp(j)) * sin(theta(j)));
      ups(i, j) = 1 /(4 * pi) * log(r1.^2/r2.^2);
      wps(i, j) = 1 /(2 * pi) * (nu2-nu1);

回答（2 个）

Haritha 2018-6-5

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/364178-converting-python-to-matlab#answer_323251

编辑：Walter Roberson 2018-6-5

在 MATLAB Online 中打开

I have the code in python

n=6;
for a, b in product(range(n), range(n))

i need this code in matlab. if anyone knows please let me know

1 个评论
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Walter Roberson 2018-6-5

在 MATLAB Online 中打开

[a, b] = ndgrid(0:n-1, 0:n-1);

请先登录，再进行评论。

Walter Roberson 2017-10-31

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/364178-converting-python-to-matlab#answer_288607

在 MATLAB Online 中打开

for i = 1:length(xp) - 1

Should not have the -1

Python range(n) gives 0 to n-1 but to use as a MATLAB index you need to add 1, so 1:n

Notice that the Python code subtracted 1 from the upper bound but that you coded the same as the one that did not subtract 1. The MATLAB translation of the second one does need to subtract one on the upper bound.

Perhaps it would help to define

PyIdx = @(K) K+1
PyRange = @(N) 0:N-1

And then copy the Python code using the appropriate wrapper, like

for i = PyRange(len(xc))
 for j = PyRange(len(xc) - 1)
    r1=((xc(PyIdx(i)) -xp(PyIdx(j))**2+(yc(PyIdx(i)) -yp(PyIdx(j))**2)**(1/2);

Get the code working first with something that is an obvious equivalent of the Python. You can always optimize later.

17 个评论
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Walter Roberson 2017-11-9

He wants data to be saved on every run?

The pathName line you showed from Python does not save data. It might be setting up a file name for a later command to save data.

The savepath() command from MATLAB does not save data, it saves the current MATLAB path -- the list of directories to search for code. See https://www.mathworks.com/help/matlab/data-import-and-export.html for more on saving data.

Zach Dunagan 2017-11-9