How to obtain length of number of columns which has data

Hi,

I need to loop though the columns in " mTot " and select the columns which has data (for example column 1 and 2).

If I do

length(mTot)

ans
51681

But, I need to obtain the answer as 2 (becuase only column 1 and 2 has data). Can somebody suggest me a method?

Thanks in advance.

[per isakson]

Do you know beforehand that "only column 1 and 2 has data"?

Try

    size( mTot, 2 )

and

    all( mTot == 0, 1 )

[Damith]

yes i know the columns beforehand. But, if I can get this through a code I can automate a process otherwise I have run upto a certain point and then need to see how many columns has data.

all( mTot == 0, 1 ) gives me

ans =

     0     0     1     1     1     1     1     1     1     1     1     1

[James Tursa]

sum(~all( mTot == 0, 1 )) ?

[dpb]

See below; looping sometimes is the right answer. How have you ensured there's nothing past the first N months and it's not the missing values problem we were discussing in the original thread? Simply iterate over the columns and test for each; if you have somehow got it where they are all contiguous then after the first false result use break to terminate the loop.

Answer 1

[Image Analyst]

Sum the columns and look for zero

columnSums = sum(mTot, 1);
% Find column numbers that are not all zeros
colIndexesWithData = find(columnsSums ~= 0);

This can even handle columns with zeros in between other columns, in case that ever happens.

If you want to find the very last column that has data in it, get the last element.

lastColWithData = colIndexesWithData (end);

[Image Analyst]

By the way, length is the max of the two dimensions, i.e. the larger of the number of rows or the number of columns.

[Damith]

Thanks.

Answer 2

[dpb]

IA has the right idea but coming in from the cold doesn't know the full picture... :)

As I pointed out in the discussion accompanying the code that I wrote from whence the above comes, the NaN initialization option gives you the columns that are missing data or it's simply remove the zero columns from the array (albeit you do need to keep track of which is which since the columns represent month of year and if there's a case where there's a missing other than trailing your index otherwise will be off).

Probably the most generic thing is to simply loop over the full array and check that any(:,colIdx) is True, if so, then write that column, if False skip it. This keeps the indices and the month in synch.

[Damith]

Thanks a lot again.

[dpb]

I added an additional thought on the overall process at the other thread if you're back, Damith, on how to "have your cake and eat it, too!"