2021-05-12 14:32:11
如何解決Matlab錯誤Too many output arguments.
最佳化方法與理論這本書的程式碼中有很多印刷錯誤的地方,大家平時要留意下
複製最佳化理論書上的原始碼,用Matlab編譯時出現如下錯誤:
??? Error using ==> qmin Too many output arguments.
1
書上的程式碼如下:
function [s, phis, ds, dphi, S] = qmin (phi, a, b, delta, epsilon)
s0 = 1; maxj = 20; maxk = 30; big = 1e6; err = 1; k = 1;
S(k) = s0; cond = 0; h = 1; ds = 0.00001;
if (abs (s0) > 1e4), h = abs(s0) * (1e-4); end
while (k < maxk && err > epsilon && cond ~= 5)
f1 = (feval (phi, s0 + ds) - feval(phi, s0 - ds)) / (2 * ds)
if(f1 > 0), h = -abs(h); end
s1 = s0 + h; s2 = s0 + 2 * h; bars = s0;
phi0 = feval(phi, s0); phi1 = feval (phi, s1);
phi2 = feval(phi, s2); barphi = phi0; cond = 0;
j = 0;
while (j < maxj && abs (h) > delta && cond == 0)
if (phi0 <= phi1),
s2 = s1; phi2 = phi1; h = 0.5 * h;
s1 = s0 + h; phi1 = feval (phi, s1);
else if (phi2 < phi1),
s1 = s2; phi1 = phi2; h = 2 * h;
s2 = s0 + 2 * h; phi2 = feval (phi, 2);
else
cond = -1;
end
end
j = j + 1;
if (abs (h) > big || abs (s0) > big), cond = 5; end
end
if (cond == 5),
bars = s1; barphi = feval (phi, s1);
else
d = 2 * (2 * phi1 - phi0 - phi2);
if (d < 0),
barh = h * (4 * phi1 - 3 * phi0 - phi2)/d;
else
barh = h / 3; cond = 4;
end
bars = s0 + barh; barphi = feval (phi, bars);
h = abs (h); h0 = abs (barh);
h1 = abs (barh - h); h2 = abs (barh - 2 * h);
if (h0 < h), h = h0; end
if (h1 < h), h = h1; end
if (h2 < h), h = h2; end
if(h == 0), h = barh; end
if (h < delta), cond = 1; end
if (abs (h) > big || abs (bars) > big), cond = 5; end
err = abs (phi1 - barphi);
s0 = bars; k = k+1; S(k) = s0;
end
if (cond == 2 && h < delta), cond = 3; end
end
s = s0; phis = feval (phi, s);
ds = h; dphi = err;
end
2
在Matlab中新建一個M檔案,名稱為qmin
3
編譯時出現如下錯誤提示:
大意是輸出引數太多
4
仔細檢查一下程式碼,發現命令語句的輸出引數中多了一個k引數
5
將命令語句中的K引數刪除,再次執行便可成功輸出結果了
6
注意:最佳化方法與理論這本書的程式碼中有很多印刷錯誤的地方,大家平時要留意下
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