I have a data 5x148992 , i have to extract data column wise and check if there is any changes wrt previous column and display when there are chnages?

Idont know where i am going wrong it also displays I when there are no changes ?

counter=0;
for i = (data( : , : ))
    if diff(i)~=0
        counter=counter+1;
        disp(i);
    else
        continue;
    end
    disp(counter)
end

[Dyuman Joshi]

(Assuming data is a numeric array)

It's not clear what you want to check for -

If you want to check that there are no two consecutive element with same values, use all()

if all(diff(i)~=0)

If you want to check something else, please mention it.

[dpb]

The above code is rife with issues...and probably doesn't do what you think it does...but as @Dyuman Joshi says, it's not clear just what you are looking for and what you want to do with it in the end...

But, to answer the actual Q? asked; there's a disp statement in all possible paths the code can take whether the test is true or not -- the disp(counter) line is executed every iteration through the loop whether the if clause is satisfied or not.

However, in MATLAB, if is TRUE if and only if ALL elements of the quantity in the test are true. diff() returns a vector or an array so If there are more than two elements in the argument, then there are at least two results returned and a single false comparison amongst them all results in an overall false result.

To make things more of an issue, for on an array returns the elements of the array by column; thus what you have in each pass through the loop are the values for each column in data in turn; thus you're actually comparing rows in each column, not columns.

See what a simple sample program shows you...

data=rand(3)            % a 3x3 array to look at

data = 3×3

0.5194 0.9479 0.1395 0.6629 0.3225 0.3490 0.4716 0.4489 0.9537

j=0;                    % a loop counter to see how many times loop operates
for i=data              % your loop construct -- the (:,:) is superfluous
  j=j+1; disp(j)        % increment loop counter, display it
  disp(i)               % what is "i" each pass?
  diff(i)               % and the difference to match...
end

     1

    0.5194
    0.6629
    0.4716

ans = 2×1

0.1435 -0.1913

     2

    0.9479
    0.3225
    0.4489

ans = 2×1

-0.6253 0.1264

     3

    0.1395
    0.3490
    0.9537

ans = 2×1

0.2096 0.6047

So, you see there are three passes through the loop because there are three columns in the array and for works by rows. Then 'i" is expanded to be each column in turn and so the difference is the two-vector of those differences.

Now, tell us what you're really after here...

[Celestie Gladys]

yes the data is numeric , even if i did all(diff(i)~=0) the disp(i) chould only show if the two consecutive column are not same , but insted it displays all the consecutive column .

11.500183 11.500183

-0.50000000 -0.50000000

975.51489 975.51489

-28.500427 -28.500427

211.50323 211.50323

[Celestie Gladys]

@dpb, i want to print only if the consequtive colmns are different , like in cpp it is, print only if the consequtive colmns are different .

data[i][j]
 j=j+1
    if [j]`~=[j+1]
        printf([j])
end

Answer 1

[Walter Roberson]

for i = (data( : , : ))

data appears to be a 2D array.

When you have something of the form

for VARIABLE = EXPRESSION; BODY; end

then that is equivalent to

HIDDEN_INTERNAL_VARIABLE_ORIGINAL = EXPRESSION;
HIDDEN_INTERNAL_VARIABLE_COLUMNS = size(EXPRESSION,2);   %COLUMNS
HIDDEN_INTERNAL_VARIABLE_COUNTER = 1;
while HIDDEN_INTERNAL_VARIABLE_COUNTER <= HIDDEN_INTERNAL_VARIABLE_COUNTER
    VARIABLE = HIDDEN_INTERNAL_VARIABLE_ORIGINAL(:,HIDDEN_INTERNAL_VARIABLE_COUNTER,:);
    HIDDEN_INTERNAL_VARIABLE_COUNTER = HIDDEN_INTERNAL_VARIABLE_COUNTER + 1;
    BODY
end
clear HIDDEN_INTERNAL_VARIABLE_ORIGINAL HIDDEN_INTERNAL_VARIABLE_COLUMNS HIDDEN_INTERNAL_VARIABLE_COUNTER

That is, the loop control variable is assigned the columns of the array.

if diff(i)~=0

Your i would contain an entire column out of data . diff() applied to the column column would (generally) be a column vector. You are then testing whether the column vector == 0, which will calculate a column vector of results.

When if is applied to a non-scalar, the condition is considered to be true if all of the values are non-zero, so that test would be equivalent to

if all(diff(i)~=0)

which is probably not going to be the case very often...

[Celestie Gladys]

Thank you !

Answer 2

[dpb]

" i want to print only if the conse[c]utive col[u]mns are different ..."

Well, then there should not be a disp statement outside the test condition; as noted above, you've got the other one where it is executed every pass regardless the condition test.

As @Walter Roberson and I both explained (and as is documented at <for>, the MATLAB vectorized loop construct passes the variable by column as the loop index variable. This is uniquely different behavior than other non-vectorized languages with which you may be familiar.

What you're looking for here in loop construction would then be something more like

nc=size(data,2);                    % number of data array columns
for i=2:nc                          % iterate over the columns beginning w/ second
  if all(data(:,i)==data(:,i-1))    % see if column i==column i-1
    disp("Column "+i+" matches previous column")
  end
end

However, "the MATLAB way" to do the above would use the vector capabilities of MATLAB something more like

ix=find(~all(diff(data,1,2)))+1;                    % find all columns match previous
disp("Column "+ix(:)+" matches previous column")    % disp() all those to screen

where ix would contain the index to all columns in the original array for which that column matched the previous.

NB: Neither of the above finds repeated columns throughout the array, only subsequent columns that are identical.

NB2nd: There's implied issue in above comparison of floating point values that even one rounded bit of precision will return FALSE -- if values of these columns are computed by alternate routes, even simply the order of numeric operations, then two values that are nominally the same may fail in the test. If this can possibly be an issue here, then see ismembertol for comparisons with tolerances.

[Celestie Gladys]

Thank you !