Remove columns from cell array when several cells in such columns have zero-values

Hi all,

I use R2019b. I have a cell array (11 x 400) in which the first row and first column contain words (e.g. 'Person#001', 'Person#002', ... 'Person#399' for the first row and 'ITEM1', 'ITEM2', ... 'ITEM10' for the first column). The remainder of the cells (2:end,2:end) are occupied by numerical values that range from 0 to 1. This dataset represents the amounts of each of the 10 items belonging to each of the 399 persons.

Some of these persons have only 0 values for all of their respective numerical values. I am trying to remove the columns from the cell array that contain only zero numerical values in rows 2 through 11 (i.e. for all 10 items). In the example below (that represents a condensed version of my data), I want the columns with 'Person#004' and 'Person#006' at the top to be removed entirely. There are several of these all-zero columns throughout the entire dataset and ideally this script would remove all of them.

Thanks in advance for your help!

 BEFORE
             'Person#001' 'Person#002' 'Person#003' 'Person#004' 'Person#005' 'Person#006'     . . .        'Person#399'
'ITEM1'           0            0.2         0.4           0           0.1           0                            0.4
'ITEM2'           0             0          0.2           0            0            0                            0.1
'ITEM3'          0.5            0          0.1           0            0            0                             0
'ITEM4'           0            0.1          0            0            0            0                            0.3

AFTER
            'Person#001' 'Person#002' 'Person#003' 'Person#005'     . . .        'Person#399'
'ITEM1'           0            0.2         0.4          0.1                           0.4
'ITEM2'           0             0          0.2           0                            0.1
'ITEM3'          0.5            0          0.1           0                             0
'ITEM4'           0            0.1          0            0                            0.3

Answer 1

[Adam Danz]

cellArray is your cell array.

threshold is the percent of 0-values per column that defines "several".

See inline comments for details.

% Define a threshold that defines "several". 
% For example. 0.5 means 50% of the values in a column is 0 (ignoring 1st row)
% 1.0 means 100% of the values in a column is 0 (ignoring 1st row)
threshold = 0.5;

% Compute the percent of 0-values per column in rows 2:end (pZero)
isZero = cellfun(@(x) isnumeric(x) &&  x==0, cellArray(2:end,:));   % <--- UPDATED FROM ORIGINAL ANSWER
pZero = mean(isZero,1);  

% Remove columns with "several" zeros
cellArray(:,pZero >= threshold) = [];

If you want to retain the original cell array, copy it to a different variable before the final line.

[flameohotman]

Thank you Adam for helping me out - I really appreciate it! I am finding that this script removes the column that precedes the all-zero column in cellArray. pZero stores the values in columns that are 1 integer less than the cellArray column containing all-zeros. e.g. person#145 has all zeros and is found in column 146 of cellArray, pZero stores it in column 145, and then column 145 (representing person#144) gets eliminated in cellArray. Do you have any suggestions to fix this? Thank you.

[Adam Danz]

I see that your first column contains strings which would cause an error when applied to my original answer.

I'm guessing that you fixed that error by changing

isZero = cellfun(@(x)  x==0, cellArray(2:end,:)); 
%                            ^^^^^^^^^^^^^^^^^^

to

isZero = cellfun(@(x)  x==0, cellArray(2:end,2:end)); 
%                            ^^^^^^^^^^^^^^^^^^^^^^

which would produce the results you're describing.

Instead, to fix the error, change that line to

isZero = cellfun(@(x) isnumeric(x) &&  x==0, cellArray(2:end,:)); 

I'll update my answer to reflect this improvement.

[flameohotman]

Thank you so much Adam. I'm now getting what I was hoping for. I am still getting a grasp of MATLAB so I am grateful for the support!

[Adam Danz]

Glad I could help!