how to get variable of function in for loop

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Amron Rosyada
Amron Rosyada 2024-5-29
回答： Steven Lord 2024-5-29
Ran in:
Hello, someone please help me.
I created a polynomial function f(x,y) using two variables. I varied the values of both variables to get the maximum f(x,y) using for loop and i got it. But, i dont know how to get or display the values of the x and y that made it. can you tell me the syntax to get them? thanks!!
x = 1:20;
y = 1:13;
f = [];
for i = 1:length(x), j = 1:length(j);
f(i,j) = x(i).^2 - x(i) -2*y(j).^2 + y(j) -25;
end
fmax = max(max(f));
display(fmax)
fmax = 354
  2 个评论
VBBV
VBBV 2024-5-29
Ran in:
Do you mean display x and y values only or f values ?
x = 1:20 % delete the semi colon
x = 1x20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
y = 1:13 %
y = 1x13
1 2 3 4 5 6 7 8 9 10 11 12 13
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
disp(x)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
disp(y)
1 2 3 4 5 6 7 8 9 10 11 12 13
Amron Rosyada
Amron Rosyada 2024-5-29
编辑：Amron Rosyada 2024-5-29
you didnt get my question. i got the fmax. but at which variable of x and y that excactly subtituted in that f function. i mean, when (the value of x and y subtituted) fmax happens?

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回答（2 个）

Steven Lord
Steven Lord 2024-5-29
Ran in:
Don't call max twice. Call max once with two output arguments and specify 'all' as the dimension over which to operate. That second output will be the linear index of the location where the maximum value returned as the first output was located. Then use ind2sub to get the row and column indices.
x = 1:20;
y = 1:13;
I've written your function as a function handle so I can call it from multiple places without retyping it.
fh = @(x, y) x.^2-x-2*y.^2+y-25;
Note that as originally written, this for loop doesn't do what you think it does. You can't have one for loop iterate over two arrays like this.
% f = [];
% for i = 1:length(x), j = 1:length(j);
% f(i,j) = fh(x(i), y(j));
% end
You need two for loops. Or you could use the vectorized approach others have suggested. But there's no need to grid the data explicitly as others have done; you can take advantage of the implicit expansion behavior of the basic operators in MATLAB like plus.
f = fh(x, y.')
f = 13x20
-26 -24 -20 -14 -6 4 16 30 46 64 84 106 130 156 184 214 246 280 316 354 -31 -29 -25 -19 -11 -1 11 25 41 59 79 101 125 151 179 209 241 275 311 349 -40 -38 -34 -28 -20 -10 2 16 32 50 70 92 116 142 170 200 232 266 302 340 -53 -51 -47 -41 -33 -23 -11 3 19 37 57 79 103 129 157 187 219 253 289 327 -70 -68 -64 -58 -50 -40 -28 -14 2 20 40 62 86 112 140 170 202 236 272 310 -91 -89 -85 -79 -71 -61 -49 -35 -19 -1 19 41 65 91 119 149 181 215 251 289 -116 -114 -110 -104 -96 -86 -74 -60 -44 -26 -6 16 40 66 94 124 156 190 226 264 -145 -143 -139 -133 -125 -115 -103 -89 -73 -55 -35 -13 11 37 65 95 127 161 197 235 -178 -176 -172 -166 -158 -148 -136 -122 -106 -88 -68 -46 -22 4 32 62 94 128 164 202 -215 -213 -209 -203 -195 -185 -173 -159 -143 -125 -105 -83 -59 -33 -5 25 57 91 127 165
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[fmaxval, fmaxloc] = max(f, [], 'all')
fmaxval = 354
fmaxloc = 248
What does that location of 248 correspond to in terms of x and y?
[row, col] = ind2sub(size(f), fmaxloc)
row = 1
col = 20
fprintf("Maximum of f is %d at x = %d, y = %d.\n", fmaxval, col, row)
Maximum of f is 354 at x = 20, y = 1.
Note that the index into x is the col index and the index into y is the row index. Let's check by evaluating the function once more at those specific values of x and y.
fh(x(col), y(row))
ans = 354
That matches the maximum value found by max (and by inspection of the displayed f array.)
VBBV
VBBV 2024-5-29
Ran in:
x = 1:20;
y = 1:13;
f = [];
for i = 1:length(x),
for j = 1:length(y);
f(i,j) = x(i).^2 - x(i) -2*y(j).^2 + y(j) -25;
end
end
f
f = 20x13
-26 -31 -40 -53 -70 -91 -116 -145 -178 -215 -256 -301 -350 -24 -29 -38 -51 -68 -89 -114 -143 -176 -213 -254 -299 -348 -20 -25 -34 -47 -64 -85 -110 -139 -172 -209 -250 -295 -344 -14 -19 -28 -41 -58 -79 -104 -133 -166 -203 -244 -289 -338 -6 -11 -20 -33 -50 -71 -96 -125 -158 -195 -236 -281 -330 4 -1 -10 -23 -40 -61 -86 -115 -148 -185 -226 -271 -320 16 11 2 -11 -28 -49 -74 -103 -136 -173 -214 -259 -308 30 25 16 3 -14 -35 -60 -89 -122 -159 -200 -245 -294 46 41 32 19 2 -19 -44 -73 -106 -143 -184 -229 -278 64 59 50 37 20 -1 -26 -55 -88 -125 -166 -211 -260
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
fmax = max(max(f));
display(fmax)
fmax = 354
  6 个评论
显示 4更早的评论
Akira Agata
Akira Agata 2024-5-29
Ran in:
+1
By using meshgrid, you can avoid for-loop, like:
x = 1:20;
y = 1:13;
% Create x- and y-grid
[xGrid, yGrid] = meshgrid(x, y);
% Calculate f
f = xGrid.^2 - xGrid - 2*(yGrid.^2) + yGrid -25;
% Find max(f) and it's linear index
[fmax, idx] = max(f(:));
% Show the result
sprintf('max(f) = %d', fmax)
ans = 'max(f) = 354'
sprintf('(x,y) = (%d, %d)', xGrid(idx), yGrid(idx))
ans = '(x,y) = (20, 1)'
% Visualize f
figure
surf(xGrid, yGrid, f)

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