Question

我写了这样的代码

clc;
clear;
clf;
A = NaN(141, 7);
ind = [1, 41, 141];
nind = setdiff(1:141, ind);
A(ind, :) = [1.46, 1.117, 0.945, 0.805, 0.675, 0.613, 0.555;
             0.877, 0.687, 0.605, 0.535, 0.463, 0.423, 0.385; 
             0.52, 0.425, 0.405, 0.37, 0.325, 0.315, 0.28];

for k = 1:7;
  A(nind, k) = interp1(ind, A(ind, k), nind, 'pchip');

end


  B = NaN(141, 61);
  inde = [1, 11, 21, 31, 41, 51, 61];
  ninde = setdiff(1:61, inde);
  B(:, inde) = A;
  for n = 1:141;
  B(n, ninde) = interp1(inde, B(n, inde), ninde, 'pchip');
  end

  for x = 30:90
  xi = x./100;
  XM = x.-29;
  for P = 1:0.1:15
   PM = P.*10-9;
   ASU = (68.070476/xi)*((xi-0.23)*(0.11+0.6998*log(P))+(1.23-xi)*0.7734*log(P));
   CSU = B(PM,XM);
   RCV = (0.00004*P.^5-0.004*P.^4+0.146*P.^3-2.5525*P.^2+21.396*P+806.87)*1000; 
   R = (RCV-ASU-CSU)/874590*100;
   fprintf('Pressure %f bar Concentration %f percent Efficiency %f percent\n', P, x, R)
 end
end

并且它显示＆＃39;下标索引的错误必须是小于2 ^ 31的正整数或逻辑＆＃39;。 Octave表示问题出在这里。

CSU = B(PM,XM);

我该如何解决这个问题？

为什么我的下标索引必须是小于2 ^ 31的正整数或逻辑＆＃39;这里？

0 个答案: