how to get variable of function in for loop

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

[VBBV]

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

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y = 1:13 %

y = 1x13

1 2 3 4 5 6 7 8 9 10 11 12 13

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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]

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?

Answer 1

[Steven Lord]

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

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[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.)

Answer 2

[VBBV]

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

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fmax = max(max(f));
display(fmax)

fmax = 354

[VBBV]

You can follow the for loop syntax given below

https://in.mathworks.com/help/matlab/ref/for.html

[Amron Rosyada]

thanks, but my question is how to get the variables ( x and y) that build the fmax?

[VBBV]

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
disp(f)

   -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
    84    79    70    57    40    19    -6   -35   -68  -105  -146  -191  -240
   106   101    92    79    62    41    16   -13   -46   -83  -124  -169  -218
   130   125   116   103    86    65    40    11   -22   -59  -100  -145  -194
   156   151   142   129   112    91    66    37     4   -33   -74  -119  -168
   184   179   170   157   140   119    94    65    32    -5   -46   -91  -140
   214   209   200   187   170   149   124    95    62    25   -16   -61  -110
   246   241   232   219   202   181   156   127    94    57    16   -29   -78
   280   275   266   253   236   215   190   161   128    91    50     5   -44
   316   311   302   289   272   251   226   197   164   127    86    41    -8
   354   349   340   327   310   289   264   235   202   165   124    79    30

fmax = max(max(f))

fmax = 354

[r,c] = find(f==fmax)

r = 20

c = 1

display(x(r))

    20

disp(y(c))

     1

[VBBV]

The x and y values are 20 and 1 that produce max value in f

[Akira Agata]

+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)

[VBBV]

@akira Agata what does "+1' mean ?