change output from row to column and display one cell value instead of multiple

Hi I have this code:

[locAll,typeAll]=find(mVocs==1) 

which is getting index from 1-30 from a matrix (mVocs size x*32) of mostly NaNs and outputting into a 1170*1 matrix x*32 in ascending order. So 1 (for column 1) appears 200 times and then 2, 3, 4 and so on, e.g.

1

1

1

The cell values are either 1,2 or 3. What I want is an output x*33 matrix with the cell value displayed under each column. How do I do this?

Answer 1

[dpb]

[locAll,typeAll]=find(mVocs==1) 

will return two vectors of the row,column indices (that you've inexplicably called loc/typeAll, respectively, for some unknownst reason).

Oh! Now it dawns on me what it is you want...

RC=accumarray(locAll,typeAll,[],@(v){v.'});

RC will be a cell array of numel(unique(locAll)) x 1 where each cell will have the numel() column indices satisfying the condition for the given row. That could be from empty to size(mVocs,2)

To put into a regular array you would have to allocate the array and augment each cell array to the full length with NaN or other missing indicator. This is fairly straightforward w/ a looping construct or might even be done in a cellfun call.

ADDENDUM/ERRATUM:

I realized I had forgotten to transpose the column vector from column to row to accumulate cell array content being row vectors instead of column vectors. Fixed up above (the ".'" inside the "{}" is the array transpose operator in MATLAB)

[Emu]

Thank you. I realise I've explained this badly - am very new to this. locAll is fine as it is - it is giving me the location of all the datapoints (as opposed to some other analyses where I am looking at segmented chunks) within a time series (1 sec per row). typeAll is giving me the column names which correspond to a variable called type.

I want the output matrix to look simething like :

1 2 3 4 5 6 7 8 9 10 11 ....

1 1 1 1 1 1

1 1 3 3 2

2 1 3

3

[dpb]

Doesn't matter what you call them; find has defined outputs when called with two output arguments, the row and column locations that satisfy the given logical condition in the argument expression.

There is no such thing in MATLAB as an array as the above -- arrays are and must be rectangular; no empty "holes" in them. MATLAB is NOT a spreadsheet.

The only way to have something approximating the above is as shown above, a cell array, or alternatively, a square array where the "don't care" elements are some missing value indicator like NaN -- but they can't be "not there"

Whatever you think locAll is giving you, what it is is the list of rows that contain a 1 in them, regardless of which column -- the corresponding column is the other variable and is the column, nothing more, nothing less. If the values range only between 1 and 3 and the size of the array is large, then it's highly probable all rows will contain at least one "1" and so, indeed, the returned vector will be all rows -- but just because it happens to be all rows doesn't imply that it always will be. Needs must be careful in drawing improper conclusions based on what numeric results may be or you would like them to be -- they are what is defined by the function outputs and only that.

If you want, instead, the value at those locations, then it would be mVocs(locAll,typeAll), but we already know what those values are; they're all identically == 1 because that's the condition you placed on them in calling find. From the data available to us, it's certainly not clear where any 2's or 3's would be coming from -- not from any of this code for certain.

I think you need to go back to the beginning and explain just what it is you're actually trying to do...

Attach a (smallish) sample dataset that's easy enough to look at on the screen without panning (5x7 or somesuch would be plenty big enough to illustrate with) and then show us what you would have the result be (although recognize the limitations of arrays being rectangular)

[dpb]

Here's an illustration of what you've coded -- it may not be what you're looking for, but it's what you've done so far --

>> M=randi(3,5,7)           % best can tell an array that matches what you described
M =
     1     2     3     1     2     1     2
     3     2     1     3     2     2     1
     2     1     1     1     1     3     1
     2     3     2     1     3     1     1
     3     3     3     2     1     1     1
>> [r,c]=find(M==1);        % what you computed whatever you used for variable names
>> [r c]
ans =
     1     1
     3     2
     2     3
     3     3
     1     4
     3     4
     4     4
     3     5
     5     5
     1     6
     4     6
     5     6
     2     7
     3     7
     4     7
     5     7
>> 

You can see how each "1" in M is at the location specified by the r,c pair; "r" is row, "c" is column. Remember MATLAB stores arrays in column-major order.

Now, to consolidate row positions by column (as near as I could tell as to what were asking for),

>> C=accumarray(r,c,[],@(v){v.'});  % accumulate for each row a cell array of columns
>> C{:}                             % see what each cell contains
ans =
     1     4     6
ans =
     3     7
ans =
     2     3     4     5     7
ans =
     4     6     7
ans =
     5     6     7
>> 

The other non-1-valued array locations are those not in the list above; there's no way directly to use what has been calculated so far to get those so we're still stuck with the Q? of what it is that you're expecting to see here.

[Emu]

I see. Thank you for the explanation. Maybe it will be ok as it is, and I could add to extract 2s and 3s - does that work with find?

I'm trying to find all the column values where there is a number; and find the different numbers (from 1-3) for each columb value. Have only got round to looking for 1s so far as current dataset doesn't have 2s or 3s.

Sample attached.

[dpb]

"I'm trying to find all the column values where there is a number..."

Well, that's a different question than the one you wrote code for --

C=any(isfinite(mvocsDemo));     % logical true anywhere there's a finite element in the array
% results in
>> disp([1:size(mvocsDemo,2);C])
     1     2     3     4     5     6     7
     1     1     0     1     0     1     1
>> 

This is global; if you are interested by row, then it is simply

C=isfinite(movcsDemo);  % logical true any location that is not NaN in array

"and find the different numbers (from 1-3) for each colum[n] value."

Well, that I'm not sure I know what you want -- the values of each number simply the values in the original array...create a dataset that is fully representative of the overall condition and then what would you expect the result to be.

[Emu]

This is incredibly helpful thank you ,worked ! The second bit I am figuring out.

[dpb]

OK, if you have some other indexing operation needed, come back when can identify just what it is you're trying to accomplish.

Oftentimes we can help more if we are told the final/overall problem trying to be solved instead of just being given a particular stumbling block; the tack being taken may be going down the wrong road and approaching the end result from another direction can be all the difference. The easiest problem to solve is the one avoided in the first place! :)

[Emu]

Ok yes I am slowiy learning this :)