Reverse calculate a table of values

Hi,

I'm looking for a way to determine the Y axis from a given Z value in a 3D table. Basically reverse interpolate the table.

Is there a way to determine it?

X/Y 0 100 175 250 325 400 475 575

60 0 5 13 21 29 40 51 68

80 0 7 16 24 32 40 49 63

90 0 7 16 25 34 42 49 61

100 0 6 15 25 33 41 49 59

110 0 6 15 23 32 40 48 58

120 0 5 14 22 30 38 47 57

130 0 5 13 21 29 37 45 56

140 0 5 13 21 29 37 44 55

150 0 4 12 21 29 37 44 54

160 0 3 12 20 29 37 44 53

Say, Z = 45, X = 125, Y = ?

Thanks in advance!

Answer 1

[Star Strider]

I am not certain what a ‘3D table’ is. Perhaps you intend a three column table or array.

If that is true, then the interp1 function could be appropriate.

The code for that, assuming ‘Y’ and ‘Z’ are vectors, would be:

Yi - interp1(Z, Y, Zi);

and ‘Zi’ is the scalar or vector you want to find the ‘Yi’ values for.

[Ferd]

Hi Star,

Thanks for the quick response!

Actually, the X axis is the first column,

X =

60

80

90

100

110

120

130

140

150

160

Y = 0 100 175 250 325 400 475 575

Z is the table,

Z =

0 5 13 21 29 40 51 68

0 7 16 24 32 40 49 63

0 7 16 25 34 42 49 61

0 6 15 25 33 41 49 59

0 6 15 23 32 40 48 58

0 5 14 22 30 38 47 57

0 5 13 21 29 37 45 56

0 5 13 21 29 37 44 55

0 4 12 21 29 37 44 54

0 3 12 20 29 37 44 53

[Star Strider]

This is an indirect way of going about this, however it appears to work:

X = [60 
     80 
     90 
    100 
    110 
    120 
    130 
    140 
    150 
    160];
Y = [0  100  175  250  325  400  475  575];
Z = [0     5    13    21    29    40    51    68
     0     7    16    24    32    40    49    63
     0     7    16    25    34    42    49    61
     0     6    15    25    33    41    49    59
     0     6    15    23    32    40    48    58
     0     5    14    22    30    38    47    57
     0     5    13    21    29    37    45    56
     0     5    13    21    29    37    44    55
     0     4    12    21    29    37    44    54
     0     3    12    20    29    37    44    53];
 
[Xm,Ym] = ndgrid(X(:)',Y(:)');
F = griddedInterpolant(Xm,Ym,Z);
Yfcn = @(Yq) norm(F(Xq,Yq)-Zq);

Zq = 45;
Xq = 125;

Yi = fsolve(Yfcn, 10)

figure
mesh(Xm, Ym, Z)
hold on
stem3(Xq, Yi, Zq, '^r', 'filled')
hold off
grid on
xlabel('X')
ylabel('Y')
view(-125,40)

produces:

Yi =
          466.176470534387

and:

[Ferd]

Works perfectly! Thanks!

[Star Strider]

As always, my pleasure!

Answer 2

[Image Analyst]

Edited according to your new info in your comment to Star:

If the desired Z value is definitely in your Z matrix, then you can find them by

zMap = Z == desiredZValue; % Produces a logical matrix where it's 1 if the matrix has that Z value.
% To get the rows and columns of all the locations, do this:
[Y, X] = find(zMap);

[Ferd]

Yeah, this works!

What if the X value is between two cells and the Z value is in-between two neigboring cells?

Find closest two cells with logical 1's if condition is met?