IF statement not generating results?

Hello! I wrote a code for calculation of some geological parameters. The problem is that the code is not generating parameter dipdir which should be calculated using IF statement.

Can some point me in the right direction for solving this problem?

(I am attaching small part of the data set for testing.

Thank you all in advance

data=load('artificial_slope.txt') %loading data
% defining input values based of input txt file 
nx = data(:,4);
ny = data(:,5);
nz = data(:,6);
n = [nx,ny,nz];
n = n./vecnorm(n,2,2); % normalize to unit-vectors. People promise to only give you the unit-length
                      % normal-vector, if you normalize it you're sure it is. Big
                      % difference. Big.


cosA = n(:,1);
cosB = n(:,2);
cosG= n(:,3);

% dip calculation
dip_disc = acos(cosG)*(180/pi);
    

if n(:,1) >= 0 & n(:,2) >= 0
    dipdir = atan(n(:,2)./n(:,1));
elseif n(:,1) > 0 & n(:,2) < 0
    dipdir = 180 - atan(n(:,2)./n(:,1));
elseif n(:,1) <= 0 & n(:,2) <= 0
    dipdir = 180 + atan(n(:,2)./n(:,1));
elseif n(:,1) < 0 & n(:,2) > 0
    dipdir = 360 - atan(n(:,2)./n(:,1));

end


disc_plane = [dip_dir, dip_disc, nx, ny, nz]

Answer 1

[VBBV]

data=load('artificial_slope.txt') %loading data

data = 8×6

-0.7880 0.5238 -0.5724 -0.6947 -0.5819 0.4228 -0.3967 0.8456 0.5138 -0.6947 -0.5819 0.4228 -1.0000 1.3560 0.2264 -0.6947 -0.5819 0.4228 -1.1571 0.9290 -0.6202 -0.6947 -0.5819 0.4228 1.0021 0.8320 1.0846 0.0989 -0.1012 0.9899 1.9014 0.9879 1.0091 0.0989 -0.1012 0.9899 2.7729 0.7271 0.8938 0.0989 -0.1012 0.9899 1.4571 1.5036 1.1069 0.0989 -0.1012 0.9899

% defining input values based of input txt file 
nx = data(:,4);
ny = data(:,5);
nz = data(:,6);
n = [nx,ny,nz];
n = n./vecnorm(n,2,2); % normalize to unit-vectors. People promise to only give you the unit-length
                      % normal-vector, if you normalize it you're sure it is. Big
                      % difference. Big.


cosA = n(:,1);
cosB = n(:,2);
cosG= n(:,3);

% dip calculation
dip_disc = acos(cosG)*(180/pi);
dipdir = zeros(8,1);   
for k = 1:length(n)
    if n(k,1) >= 0 & n(k,2) >= 0
        dipdir(k) = atan(n(k,2)/n(k,1));
    elseif n(k,1) > 0 & n(k,2) < 0
        dipdir(k) = 180 - atan(n(k,2)/n(k,1));
    elseif n(k,1) <= 0 & n(k,2) <= 0
        dipdir(k) = 180 + atan(n(k,2)/n(k,1));
    elseif n(k,1) < 0 & n(k,2) > 0
        dipdir(k) = 360 - atan(n(k,2)/n(k,1));    
    end
end

disc_plane = [dipdir, dip_disc, nx, ny, nz] %  check the spelling of variable dipdir

disc_plane = 8×5

180.6973 64.9896 -0.6947 -0.5819 0.4228 180.6973 64.9896 -0.6947 -0.5819 0.4228 180.6973 64.9896 -0.6947 -0.5819 0.4228 180.6973 64.9896 -0.6947 -0.5819 0.4228 180.7970 8.1340 0.0989 -0.1012 0.9899 180.7970 8.1340 0.0989 -0.1012 0.9899 180.7970 8.1340 0.0989 -0.1012 0.9899 180.7970 8.1340 0.0989 -0.1012 0.9899

Check the spelling of variable dipdir as against dip_dir

[VBBV]

you have incorrect spelling in the line

disc_plane = [dip_dir, dip_disc, nx, ny, nz]

in place of

disc_plane = [dipdir, dip_disc, nx, ny, nz]

[Hrvoje Lukacic]

Thank you for your help, the core works. The key was in the for loop

Answer 2

[Davide Masiello]

data=load('artificial_slope_2.txt') %loading data

data = 8×6

-0.7880 0.5238 -0.5724 -0.6947 -0.5819 0.4228 -0.3967 0.8456 0.5138 -0.6947 -0.5819 0.4228 -1.0000 1.3560 0.2264 -0.6947 -0.5819 0.4228 -1.1571 0.9290 -0.6202 -0.6947 -0.5819 0.4228 1.0021 0.8320 1.0846 0.0989 -0.1012 0.9899 1.9014 0.9879 1.0091 0.0989 -0.1012 0.9899 2.7729 0.7271 0.8938 0.0989 -0.1012 0.9899 1.4571 1.5036 1.1069 0.0989 -0.1012 0.9899

% defining input values based of input txt file 
nx = data(:,4);
ny = data(:,5);
nz = data(:,6);
n = [nx,ny,nz];
n = n./vecnorm(n,2,2); % normalize to unit-vectors. People promise to only give you the unit-length
                      % normal-vector, if you normalize it you're sure it is. Big
                      % difference. Big.
cosA = n(:,1);
cosB = n(:,2);
cosG= n(:,3);
% dip calculation
dip_disc = acos(cosG)*(180/pi);
    
disp(n)

   -0.6947   -0.5819    0.4228
   -0.6947   -0.5819    0.4228
   -0.6947   -0.5819    0.4228
   -0.6947   -0.5819    0.4228
    0.0989   -0.1012    0.9899
    0.0989   -0.1012    0.9899
    0.0989   -0.1012    0.9899
    0.0989   -0.1012    0.9899

The problem as I see it is that you are passing vectors to the conditional statement.

In order for the command to be executed, you need the condition to be true for all elements of the vector.

The very first instruction in each statement, i.e.

n(:,1) >= 0

ans = 8×1 logical array
   0
   0
   0
   0
   1
   1
   1
   1

n(:,1) > 0

ans = 8×1 logical array
   0
   0
   0
   0
   1
   1
   1
   1

n(:,1) <= 0

ans = 8×1 logical array
   1
   1
   1
   1
   0
   0
   0
   0

n(:,1) < 0

ans = 8×1 logical array
   1
   1
   1
   1
   0
   0
   0
   0

is never verified for all elements of the first column of n, therefore none of the commands in the conditional statement are executed and dip_dir is not defined in the first place.

You can probably get away without conditional statement.

If you put your four conditions in a matrix, you will see that the conditions being fulfilled are number 3 for the first 4 rows and number 2 for the second 4 rows.

cond    = [n(:,1) >= 0 & n(:,2) >= 0,n(:,1) > 0 & n(:,2) < 0,n(:,1) <= 0 & n(:,2) <= 0,n(:,1) < 0 & n(:,2) > 0]

cond = 8×4 logical array
   0   0   1   0
   0   0   1   0
   0   0   1   0
   0   0   1   0
   0   1   0   0
   0   1   0   0
   0   1   0   0
   0   1   0   0

You can also calculate all the possible occurences for dip_dir

dip_dir0 = [atan(n(:,2)./n(:,1)),180 - atan(n(:,2)./n(:,1)),80 + atan(n(:,2)./n(:,1)),360 - atan(n(:,2)./n(:,1))];

And finally get only those which fulfill the conditions given by the matrix cond.

dip_dir = dip_dir0(cond)

dip_dir = 8×1

180.7970 180.7970 180.7970 180.7970 80.6973 80.6973 80.6973 80.6973

disc_plane = [dip_dir, dip_disc, nx, ny, nz]

disc_plane = 8×5

180.7970 64.9896 -0.6947 -0.5819 0.4228 180.7970 64.9896 -0.6947 -0.5819 0.4228 180.7970 64.9896 -0.6947 -0.5819 0.4228 180.7970 64.9896 -0.6947 -0.5819 0.4228 80.6973 8.1340 0.0989 -0.1012 0.9899 80.6973 8.1340 0.0989 -0.1012 0.9899 80.6973 8.1340 0.0989 -0.1012 0.9899 80.6973 8.1340 0.0989 -0.1012 0.9899

Hope this helps you.

[Hrvoje Lukacic]

Thank you, I understand what are you saying. But, I need to apply the certain formula for dipdip calculation according to the values of n. So for example if you look the first row in n with values -0.6947 -0.5819 0.4228 it should calculte dipdir according to the formula dipdir = 180 + atan(n(:,2)./n(:,1));.... because n1 and n2 are less than 0 ... how do I achive that?

[Davide Masiello]

I think I now understand what you need to do.

I have modified my answer to accommodate your need, please let me know if I am right.

[Hrvoje Lukacic]

The code works thank you but for some reasno results of dip direction are not correct. I have to figure out why

[Hrvoje Lukacic]

Formulas are correct but matlab calculates it wrong. FOr example, for the first row the result should be 230 and not 180

[Davide Masiello]

What formula should be used to obtain the value of 230?

[Hrvoje Lukacic]

Sorry values are correct. The code written by VBBV is correct. Thank you for your help