MATLAB code requires a loop explanation

I am very new in MATLAB programming, I am trying several examples from google, I understood several codes, however, a code example I need to understand for starting my further, analysis. can you please explain the equation or meaning of this written code in a loop? what is the mathematics or meaning of this? below in command I get T is data that we took from excel. secondly, n is the length of data, next yt= n/12 fine. What is there in for loop? I will appreciate your answer. thank you so much. I run data examples and found output from this code but I am unable to understand the mathematics, equation or meaning of this that use in for loop for generating the data. For example, my data length is 732, then yt=n/12=61, in ys=(fix((i-1)/12))+1; what code want to get, why it define while k>12 and so on,

T=xlsread('ali.xls','temp','C:C');
n=length(T);
yt=n/12;
for i=n:-1:1
    ys=(fix((i-1)/12))+1;
    k=i;
    while k>12
        k=k-12;
    end
    m=k;
    X(m,ys)=T(i);
end

Answer 1

[Walter Roberson]

    k=i;
    while k>12
        k=k-12;
    end
    m=k;

That code is equiavelent to

    m = mod(k-1, 12) + 1; 

Another way of writing the code would be:

T = xlsread('ali.xls','temp','C:C');
n = length(T);
padding_needing = mod(-n, 12);
T = [T; nan(padding_needed, 1)];
X = reshape(T, 12, []).';

This says:

• read in the data
• find out how many entries we got
• we want to take the data and put it across 12 rows
• but what happens if the original data was not an exact multiple of 12 entries?
• we figure out how many more entries would be needed to take us to a multiple of 12
• then we pad the data with that many nan
• so now T is an exact multiple of 12 items long
• take consecutive entries from T, and arrange them in 12 rows with as many columns as needed
• once we have that, transpose the data so we now have 12 columns and as many rows as needed

In the special case where we were sure that the data had an exact multiple of 12 entries, then this could all be simplified to

T = xlsread('ali.xls','temp','C:C');
X = reshape(T, 12, []).';

[Rizwan Niaz]

This means what ? ys=(fix((i-1)/12))+1 ???? please exlain whole meaning of this for i=n:-1:1

ys=(fix((i-1)/12))+1;

k=i;

while k>12

k=k-12;

end

m=k;

X(m,ys)=T(i);

end

[Voss]

@Rizwan Niaz Please see my explanation of the for loop, in my answer.

[Walter Roberson]

i = [1;11;12;13;23;24;25];
[i, fix((i-1)/12)+1]

ans = 7×2

1 1 11 1 12 1 13 2 23 2 24 2 25 3

That is, the first full cycle of 12 gets assigned the value 1, the second full cycle gets assigned the value 2, ...

[Walter Roberson]

for i=n:-1:1

In context, taking into account the fact that some unusual behaviour is not coded in that loop, that code is equivalent to:

i = [];
if n >= 1
   hidden_temporary_variable = n;
   while hidden_temporary_variable >= 1
       i = hidden_temporary_variable;
       execute the code for the loop body
       hidden_temporary_variable = hidden_tempory_variable - 1;
   end
end

Or if you just want the summary: count i backwards from n to 1, leaving i as 1 afterwards.

[Walter Roberson]

The rest, other than the assignment statement, has to do with figuring out how many full groups of 12, and how far you are into that group.

The logic could have been written like

i = [1;2;3;11;12;13;23;24;25];

m = mod(i-1, 12) + 1;
ys = (i - m)/12 + 1;

[i, m, ys]

ans = 9×3

1 1 1 2 2 1 3 3 1 11 11 1 12 12 1 13 1 2 23 11 2 24 12 2 25 1 3

So m is how many months into the year you are, and ys is how many years in you are.

Answer 2

[Voss]

% T=xlsread('ali.xls','temp','C:C');
T = (1:100).'

T = 100×1

1 2 3 4 5 6 7 8 9 10

n=length(T);
yt=n/12;
for i=n:-1:1 % this for loop populates the matrix X with elements 
             % from T (starting with the last element of T and ending with
             % the first, but the order doesn't matter in this algorithm).
    
    % fix() rounds toward 0.
    % fix((i-1)/12) rounds (i-1)/12 toward 0.
    % since i is positive in this case, this is equivalent to
    % floor((i-1)/12), which rounds to the largest integer smaller than or
    % equal to (i-1)/12:
    % i    (i-1)/12   floor((i-1)/12)
    % -------------------------------
    % 1      0/12           0  [ 12 zeros ]
    % 2      1/12           0
    % 3      2/12           0
    % ...    ...            0
    % 12     11/12          0
    % 13     12/12          1  [ 12 ones ]
    % ...    ...            1
    % 24     23/12          1
    % 25     24/12          2  [ 12 twos ]
    % ...    ...            2
    %
    % so fix((i-1)/12)+1 is one more than the values in the right-hand
    % column in the table above: twelve 1's followed by twelve 2's and so
    % on, as i increases from 1.
    %
    % this number will be the column in X that T(i) will go to.
    ys=(fix((i-1)/12))+1;
    
    % find the integer m (1 <= m <= 12) that is congruent to the given integer i, modulo 12.
    % [i mod 12 is the remainder when some number i is divided by 12. for example:
    %  0, 12, 24, 36, ... are all congruent to 0 mod 12 (these are multiples of 12), and 
    %  1, 13, 25, 37, ... are all congruent to 1 mod 12 (1 more than a multiple of 12) and 
    %  2, 14, 26, 38, ... are all congruent to 2 mod 12 (2 more than a multiple of 12), 
    % and so on, up to 
    % 11, 23, 35, 47, ... are all congruent to 11 mod 12 (11 more than a multiple of 12).
    
    k=i; % start with i, an integer
    while k>12 % and as long as it is more than 12
        k=k-12; % keep subtracting 12
    end
    m=k; % now you have an integer between 1 and 12, inclusive.
         % this number is the row in X that T(i) will go to.
    
    % set the element in row m, column ys of matrix X to T(i) (the ith 
    % element of T):
    X(m,ys)=T(i);
    
end
disp(X);

     1    13    25    37    49    61    73    85    97
     2    14    26    38    50    62    74    86    98
     3    15    27    39    51    63    75    87    99
     4    16    28    40    52    64    76    88   100
     5    17    29    41    53    65    77    89     0
     6    18    30    42    54    66    78    90     0
     7    19    31    43    55    67    79    91     0
     8    20    32    44    56    68    80    92     0
     9    21    33    45    57    69    81    93     0
    10    22    34    46    58    70    82    94     0
    11    23    35    47    59    71    83    95     0
    12    24    36    48    60    72    84    96     0

[Rizwan Niaz]

wonderful explaination

Answer 3

[Image Analyst]

Take the semicolons off the lines and step through it one line at a time in the debugger and see how the variables change.

Debugging in MATLAB | Doug's MATLAB Video Tutorials

[Rizwan Niaz]

while k>12

k=k-12;