python interpolate.interp1d_将MATLAB的interp1转换为Python interp1d

我正在将MATLAB代码转换为Python代码。在

代码使用MATLAB中的函数interp1。我发现scipy函数interp1d应该是我想要的，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，但是代码已经在2010年左右实现了]。在

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:

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