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use QR decomposition and normalization for polyfit; normalization for polyval
author Ben Abbott <bpabbott@mac.com>
date Tue, 19 Feb 2008 07:28:40 -0500
parents 83a8781b529d
children 83cce070104f
comparison
equal deleted inserted replaced
7499:94d0cdd60dda 7500:2df882e69f13
16 ## You should have received a copy of the GNU General Public License 16 ## You should have received a copy of the GNU General Public License
17 ## along with Octave; see the file COPYING. If not, see 17 ## along with Octave; see the file COPYING. If not, see
18 ## <http://www.gnu.org/licenses/>. 18 ## <http://www.gnu.org/licenses/>.
19 19
20 ## -*- texinfo -*- 20 ## -*- texinfo -*-
21 ## @deftypefn {Function File} {} polyval (@var{c}, @var{x}) 21 ## @deftypefn {Function File} {@var{y}=} polyval (@var{p}, @var{x})
22 ## Evaluate a polynomial. 22 ## @deftypefnx {Function File} {@var{y}=} polyval (@var{p}, @var{x}, [], @var{mu})
23 ## 23 ## Evaluate the polynomial at of the specified values for @var{x}. When @var{mu}
24 ## @code{polyval (@var{c}, @var{x})} will evaluate the polynomial at the 24 ## is present evaluate the polynomial for (@var{x}-@var{mu}(1))/@var{mu}(2).
25 ## specified value of @var{x}. 25 ## If @var{x} is a vector or matrix, the polynomial is evaluated for each of
26 ##
27 ## If @var{x} is a vector or matrix, the polynomial is evaluated at each of
28 ## the elements of @var{x}. 26 ## the elements of @var{x}.
29 ## @seealso{polyvalm, poly, roots, conv, deconv, residue, filter, 27 ## @deftypefnx {Function File} {[@var{y}, @var{dy}] =} polyval (@var{p}, @var{x}, @var{S})
28 ## @deftypefnx {Function File} {[@var{y}, @var{dy}] =} polyval (@var{p}, @var{x}, @var{S}, @var{mu})
29 ## In addition to evaluating the polynomial, the second output
30 ## represents the prediction interval, @var{y} +/- @var{dy}, which
31 ## contains at least 50% of the future predictions. To calculate the
32 ## prediction interval, the structured variable @var{s}, originating
33 ## form `polyfit', must be present.
34 ## @seealso{polyfit, polyvalm, poly, roots, conv, deconv, residue, filter,
30 ## polyderiv, polyinteg} 35 ## polyderiv, polyinteg}
31 ## @end deftypefn 36 ## @end deftypefn
32 37
33 ## Author: Tony Richardson <arichard@stark.cc.oh.us> 38 ## Author: Tony Richardson <arichard@stark.cc.oh.us>
34 ## Created: June 1994 39 ## Created: June 1994
35 ## Adapted-By: jwe 40 ## Adapted-By: jwe
36 41
37 function y = polyval (c, x) 42 function [y, dy] = polyval (p, x, s, mu)
38 43
39 if (nargin != 2) 44 if (nargin < 2 || nargin > 4 || (nargout == 2 && nargin < 3))
40 print_usage (); 45 print_usage ();
41 endif 46 endif
42 47
43 if (! (isvector (c) || isempty (c))) 48 if (nargin < 3)
49 s = [];
50 endif
51
52 if (! (isvector (p) || isempty (p)))
44 error ("polyval: first argument must be a vector"); 53 error ("polyval: first argument must be a vector");
54 endif
55
56 if (nargin < 4)
57 mu = [0, 1];
45 endif 58 endif
46 59
47 if (isempty (x)) 60 if (isempty (x))
48 y = []; 61 y = [];
49 return; 62 return;
50 endif 63 endif
51 64
52 if (length (c) == 0) 65 if (length (p) == 0)
53 y = c; 66 y = p;
54 return; 67 return;
55 endif 68 endif
56 69
57 n = length (c); 70 n = length (p) - 1;
58 y = c (1) * ones (rows (x), columns (x)); 71 k = numel (x);
59 for index = 2:n 72 x = (x - mu(1)) / mu(2);
60 y = c (index) + x .* y; 73 A = (x(:) * ones (1, n+1)) .^ (ones (k, 1) * (n:-1:0));
61 endfor 74 y(:) = A * p(:);
75 y = reshape (y, size (x));
76
77 if (nargout == 2)
78 ## The line below is *not* the result of a conceptual grasp of statistics.
79 ## Instead, after reading the links below and comparing to the output of Matlab's polyval.m,
80 ## http://www.mathworks.com/access/helpdesk/help/toolbox/stats/index.html?/access/helpdesk/help/toolbox/stats/finv.html
81 ## http://www.mathworks.com/access/helpdesk/help/toolbox/curvefit/index.html?/access/helpdesk/help/toolbox/curvefit/bq_5ka6-1_1.html
82 ## Note: the F-Distribution is generally considered to be single-sided.
83 ## http://www.itl.nist.gov/div898/handbook/eda/section3/eda3673.htm
84 ## t = finv (1-alpha, s.df, s.df);
85 ## dy = t * sqrt (1 + sumsq (A/s.R, 2)) * s.normr / sqrt (s.df)
86 ## If my inference is correct, then t must equal 1 for polyval.
87 ## This is because finv (0.5, n, n) = 1.0 for any n.
88 dy = sqrt (1 + sumsq (A/s.R, 2)) * s.normr / sqrt (s.df);
89 dy = reshape (dy, size (x));
90 endif
62 91
63 endfunction 92 endfunction
64 93
65 %!assert(polyval ([1, 1, 1], 2) == 7); 94 %!test
95 %! fail("polyval([1,0;0,1],0:10)");
66 96
67 %!assert(all (all (polyval ([1, 1, 1], [0; 1; 2]) == [1; 3; 7]))); 97 %!test
98 %! r = 0:10:50;
99 %! p = poly (r);
100 %! p = p / max(abs(p));
101 %! x = linspace(0,50,11);
102 %! y = polyval(p,x) + 0.25*sin(100*x);
103 %! [pf, s] = polyfit (x, y, numel(r));
104 %! [y1, delta] = polyval (pf, x, s);
105 %! expected = [0.37235, 0.35854, 0.32231, 0.32448, 0.31328, ...
106 %! 0.32036, 0.31328, 0.32448, 0.32231, 0.35854, 0.37235];
107 %! assert (delta, expected, 0.00001)
68 108
69 %!assert(isempty (polyval ([1, 1, 1], []))); 109 %!test
110 %! x = 10 + (-2:2);
111 %! y = [0, 0, 1, 0, 2];
112 %! p = polyfit (x, y, numel (x) - 1);
113 %! [pn, s, mu] = polyfit (x, y, numel (x) - 1);
114 %! y1 = polyval (p, x);
115 %! yn = polyval (pn, x, [], mu);
116 %! assert (y1, y, sqrt(eps))
117 %! assert (yn, y, sqrt(eps))
70 118
71 %!assert(all (all (polyval ([1, 1, 1], [-1, 0; 1, 2]) == [1, 1; 3, 7]))); 119 %!test
120 %! p = [0, 1, 0];
121 %! x = 1:10;
122 %! assert (x, polyval(p,x), eps)
123 %! x = x(:);
124 %! assert (x, polyval(p,x), eps)
125 %! x = reshape(x, [2, 5]);
126 %! assert (x, polyval(p,x), eps)
127 %! x = reshape(x, [5, 2]);
128 %! assert (x, polyval(p,x), eps)
129 %! x = reshape(x, [1, 1, 5, 2]);
130 %! assert (x, polyval(p,x), eps)
72 131
73 %!error polyval ([1, 2; 3, 4], [-1, 0; 1, 2]);
74
75 %!assert(isempty (polyval ([], [-1, 0; 1, 2])));
76