python - 将 MATLAB 的 interp1 转换为 Python interp1d

Question

我正在将 MATLAB 代码转换为 Python 代码。

该代码使用 MATLAB 中的函数 interp1。我发现 scipy 函数 interp1d 应该是我所追求的，但我不确定。你能告诉我我实现的代码是否正确吗？ 我的Python版本是3.4.1，MATLAB版本是R2013a。然而，该代码已在 2010 年左右实现。

MATLAB:

S_T = [0.0, 2.181716948, 4.363766232, 6.546480392, 8.730192373, ...
      10.91523573, 13.10194482, 15.29065504, 17.48170299, 19.67542671, ...
      21.87216588, 24.07226205, 26.27605882, 28.48390208; ...
     1.0, 1.000382662968538, 1.0020234819906781, 1.0040560245904753, ...
       1.0055690037530718, 1.0046180687475195, 1.000824223678225, ...
       0.9954866694014762, 0.9891408937764872, 0.9822543350571298, ...
       0.97480163751874, 0.9666158376141503, 0.9571711322843011, ...
       0.9460998105962408; ...
     1.0, 0.9992731388936672, 0.9995093132493109, 0.9997021748479805, ...
       0.9982835412406582, 0.9926319477117723, 0.9833685776596993, ...
       0.9730725288209638, 0.9626092685176822, 0.9525234896714959, ...
       0.9426698515488858, 0.9326788630704709, 0.9218100196936996, ...
       0.9095717918978693];

S = transpose(S_T);

dist = 0.00137;
old = 15.61;
ll = 125;
ref = 250;
start = 225;
high = 7500;
low = 2;

U = zeros(low,low,high);

for ii=1:high
    g0= start-ref*dist*ii;
    g1= g0+ll;
    if(g0 <=0.0 && g1 >= 0.0)
        temp= old/2*(1-cos(2*pi*g0/ll));
        for jj=1:low
            U(jj,jj,ii)= temp;
        end
    end
end

for ii=1:low
 S_mod(ii,1,:)=interp1(S(:,1),S(:,ii+1),U(ii,ii,:),'linear');
end

python :

import numpy
import os
from scipy import interpolate

S = [[0.0, 2.181716948, 4.363766232, 6.546480392, 8.730192373, 10.91523573, 13.10194482, 15.29065504, \
      17.48170299, 19.67542671, 21.87216588, 24.07226205, 26.27605882, 28.48390208], \
     [1.0, 1.000382662968538, 1.0020234819906781, 1.0040560245904753, 1.0055690037530718, 1.0046180687475195, \
      1.000824223678225, 0.9954866694014762, 0.9891408937764872, 0.9822543350571298, 0.97480163751874, \
      0.9666158376141503, 0.9571711322843011, 0.9460998105962408], \
     [1.0, 0.9992731388936672, 0.9995093132493109, 0.9997021748479805, 0.9982835412406582, 0.9926319477117723, \
      0.9833685776596993, 0.9730725288209638, 0.9626092685176822, 0.9525234896714959, 0.9426698515488858, \
      0.9326788630704709, 0.9218100196936996, 0.9095717918978693]]

dist = 0.00137
old = 15.61
ll = 125
ref = 250
start = 225
high = 7500
low = 2

U = [numpy.zeros( [low, low] ) for _ in range(high)]

for ii in range(high):
    g0 = start - ref * dist * (ii+1)   
    g1 = g0 + ll
    if g0 <=0.0 and g1 >= 0.0:
        for jj in range(low):
            U[ii][jj,jj] = old / 2 * (1 - numpy.cos( 2 * numpy.pi * g0 / ll) )

S_mod = []

for jj in range(high):
    temp = []
    for ii in range(low):
        temp.append(interpolate.interp1d( S[0], S[ii+1], U[jj][ii,ii]))

    S_mod.append(temp)

Answer 1

好的，我已经解决了我自己的问题(感谢 Alex 对 MATLAB interp1 的解释!)。

python interp1d 本身没有查询点，而是创建一个函数，然后您可以使用该函数来获取新数据点。因此，它应该是:

        f = interpolate.interp1d( S[0], S[ii+1])
        temp.append(f(U[jj][ii,ii]))