how to ignore a value and jumps to next row for calculation?

in the matrix i have an strange number 999, I need to perform an averaging between two numbers to get a new value of u(i) u(i)=(u(i+1)-u(i-1))/2 but if u(i),or u(i+1) or u(i-1) comes 999 then it should skip this row and jump to next value in the table u(i). so that when i will get the new value of u(i), the value 999 remain undisturbed in it position and other value should not get affected by this value.

[Guillaume]

ellora, could you please give a short example of input and expected output.

Are you sure you want (u(i+1) - u(i-1))/2 and not (u(i+1) + u(i-1))/2, note the + instead of -?

[ellora]

yeah sure, it is (u(i+1) - u(i-1))/2

Answer 1

[Guillaume]

Probably the simplest thing, is to store the location of the 999, remove them, perform your averaging (which is then very straightforward as you don't have special cases anymore) and finally put back the 999:

v = [1 2 3 999 5 6 7 10 999 20 40 80 160] %example data
isspecial = v == 999; %logical array indicating the position of 999
filteredv = v(~isspecial);  %remove 999 from vector
%do your averaging any way you want. Not sure what you want to do at the edges
filteredv = conv(filteredv, [0.5 0 0.5], 'same'); %does u(i)=(u(i+1)-u(i-1))/2, use zero-padding at the edges
%now create a vector with filtered values and 999 in their original location
finalv = zeros(size(v));
finalv(~isspecial) = filteredv;
finalv(isspecial) = v(isspecial)

[Shameer Parmar]

your code fails for ellora's requirement..

as per your code..

if input

v =

     1     2     3   999     5     6     7    10   999    20    40    80   160

output =

     1     2     4   999     5     6     8    14   999    25    50   100    40

the second value should be should be '1' (i.e. (3-1)/2 = 1)and not '2' and so on for other values.....

[ellora]

yeah it is wrong, in 2nd place it should be 1, in 3rd place it should be 3.5 and 4th place will be same as 999 and 5th place will be 5 and so on

[Guillaume]

@Shameer, I missed the negative sign in the (u(i+1) - u(i-1))/2 expression (as it makes more sense to me to take the mean). If half the difference between u(i+1) and u(i-1) is really needed, then simply replace the [0.5 0 0.5] filter in the convolution by [0.5 0 -0.5].

@ellora, can you explain how you obtain 1 in 2nd place and 3.5 in 4th and 5 in 5th?

[Guillaume]

Ellora asked in another comment: Guillaume i need a help, your solution is correct for single column, but how it can be done for 891x3 matrix?can u please explain.

Operate on each column one at a time:

%m: matrix 
newm = m;
for column = 1 : size(newm, 2)
   include = newm(:, column) ~= 999;
   filteredcol = newm(include, column);    
   newm(include, column) = conv(filteredcol, [0.5; 0; 0.5], 'same');
end

I'm still not clear on what operation you want to perform (half difference?, mean ?) but this is the idea.

If you're uncomfortable with convolutions, this does the same:

newm(include, column) = ([0; filteredcol(1:end-1)] + [filteredcol(2:end); 0]) / 2;

[ellora]

whatever you are saying is correct,but there are some problems persist. As my columns do not have equal number of 999 values. so while performing the run, it is giving only the last column values. if you will say i can attach my matrix file here

[Guillaume]

" my columns do not have equal number of 999 values". I assumed as much, the code above does not assume that there is the same number of 999 in each column.

Yes, please attach your matrix (as a mat file please), and show what output you expect for the first few values.

[ellora]

here is the file. the procedure you have used is correct, but i am unable to do it for all the columns at a time.

Answer 2

[KSSV]

You can get the indices of 999 using find..

idx = find(mymatrix==999)

Now you can do averaging by excluding idx..or make idx to zero and get avarage...

[Guillaume]

find is not required most of the time and is usually a waste of time. You can use the logical array directly.

Answer 3

[Shameer Parmar]

Let us consider, you have values as follows as example:

u = [112, 54, 86, 547, 999, 458, 657, 999, 56, 422, 100];

for i = 2:(length(u)-1)

    if (u(i) ~= 999)
        u(i) = (u(i+1)-u(i-1))/2;
    else
        u(i) = 999;
    end
end

[Guillaume]

Your algorithm fails on that original requirement by ellora: "and other value should not get affected by this value".

There are much better ways than a for loop for performing the filtering required:

option 1: plain matrix operation:

u = (u(1:end-2) - u(3:end)) / 2;

option 2: convolution with [0.5 0 0.5]

u = conv(u, [0.5 0 -0.5], 'valid');

edit: replaced + by - and convolution coefficients, as I missed that ellora was asking for the half difference.

[Shameer Parmar]

@Guillaume:

How do you know, my algorithm fails ?

What is mean by "and other value should not get affected by this value" ?

Do you know the expected output ? if input matrix is...

u = [112, 54, 86, 547, 999, 458, 657, 999, 56, 422, 100];

[Guillaume]

I know it fails because for i = 4 for example you're averaging 86 with 999 which is not "other value should not get affected by this value"

[ellora]

Guillaume i need a help, your solution is correct for single column, but how it can be done for 891x3 matrix?can u please explain.