How to create string of text-objects

Hi, I have the following matrix as an output. I wouldlike to make a third column such that the rows corresponding to 1, i would like to write "paid" and the rows corresponding to 0 I would like to write "not paid" so that the resulting table should have three columns, the first two columns are columns of the matrix D and the third should contain either "paid" or "not paid". I have inserted the expected output. The second matrix I have inesrted the column mannualy. Please help.

D =

13 0

4 1

5 1

1 1

2 1

3 1

14 0

5 1

16 0

7 1

8 1

9 1

3 1

4 1

2 1

1 1

3 1

98 0

100 0

what i expect is;

D =

13 0 Notpaid

4 1 paid

5 1 paid

1 1 paid

2 1 paid

3 1 paid

14 0 Notpaid

5 1 paid

16 0 Notpaid

7 1 paid

8 1 paid

9 1 paid

3 1 paid

4 1 paid

2 1 paid

1 1 paid

3 1 paid

98 0 Notpaid

100 0 Notpaid

Answer 1

[dpb]

Holding disparate data types would be a good place to use a table as alternative to the cell array...

D = [randi(100,10,1),rand(10,1)>=0.5];
tD=array2table(D,'VariableNames',{'Amount','PayStatus'});   % convert to table
CODE=categorical({'Notpaid';'Paid'});                       % the lookup values
tD=addvars(tD,CODE(tD.PayStatus+1),'NewVariableNames',{'BillStatus'}) % add the verbal status

tD = 10×3 table
    Amount    PayStatus    BillStatus
    ______    _________    __________

      71          0         Notpaid  
      65          0         Notpaid  
      77          0         Notpaid  
      49          1         Paid     
      80          1         Paid     
      73          1         Paid     
      41          1         Paid     
       8          0         Notpaid  
      74          0         Notpaid  
      26          0         Notpaid  

Answer 2

[chicken vector]

You can't concatenate doubles and string in one array, because the array can be only one of them.

If you want to only store the information you can use cells:

D = [randi(100,20,1),rand(20,1)>=0.5];
result = cell(2,1);
result{1} = D;
result{2} = strings(length(D),1);
for j = 1 : length(D)
    if ~D(j,2)
        result{2}(j) = "Notpaid";
    else
        result{2}(j) = "Paid";
    end
end

If it's for display purposes you can transform everything into a string:

D = [randi(100,20,1),rand(20,1)>=0.5];
result = strings(length(D),3);
result(:,1:2) = string(D);
for j = 1 : length(D)
    if ~D(j,2)
        result(j,3) = "Notpaid";
    else
        result(j,3) = "Paid";
    end
end

[dpb]

To eliminate the loop construct and explicit if...else clause, "the MATLAB way" would be to use a lookup table and vectorize...

CODE={'Notpaid';'Paid'};        % the lookup values
D=num2cell(D);                  % convert original to cell array to add disparate type
D(:,3)=CODE([D{:,2}]+1)         % add the third column based on second [0,1]==>[1,2] index

D = 10×3 cell array
    {[86]}    {[1]}    {'Paid'   }
    {[43]}    {[0]}    {'Notpaid'}
    {[45]}    {[1]}    {'Paid'   }
    {[17]}    {[1]}    {'Paid'   }
    {[51]}    {[0]}    {'Notpaid'}
    {[42]}    {[1]}    {'Paid'   }
    {[66]}    {[0]}    {'Notpaid'}
    {[ 5]}    {[1]}    {'Paid'   }
    {[93]}    {[0]}    {'Notpaid'}
    {[95]}    {[0]}    {'Notpaid'}

[chicken vector]

My usual way of doing it would be to:

paidIdx = D(:,2) == 1;
D(paidIdx,3) = "Paid";
D(~paidIdx,3) = "Notpaid";

But your approach is much better.

[dpb]

I don't know that it's "much" better, but does illustrate using the lookup table approach that many novices may have not seen...it trades one temporary variable for another so isn't a tremendous advantage one way or another other than, perhaps, putting the constants into one array so can just adjust it if number of choices changes.

[chicken vector]

Aesthetically much better*

Answer 3

[Dyuman Joshi]

Given the format of your final output, you will either have to use string arrays, cell arrays or categorical arrays.

%Categorical array method
D =[13    0  
    4     1
    5     1
    1     1
    2     1
    3     1
    14    0
    5     1
    16    0
    7     1
    8     1
    9     1
    3     1
    4     1
    2     1
    1     1
    3     1
    98    0
    100   0];
%Values corresponding to paid
idx = D(:,2);
%Convert D to a categorical array
D = categorical(D);
str = categorical(["Notpaid", "paid"]);

D(:,3) = str(idx+1)

D = 19×3 categorical array
     13       0      Notpaid 
     4        1      paid    
     5        1      paid    
     1        1      paid    
     2        1      paid    
     3        1      paid    
     14       0      Notpaid 
     5        1      paid    
     16       0      Notpaid 
     7        1      paid    
     8        1      paid    
     9        1      paid    
     3        1      paid    
     4        1      paid    
     2        1      paid    
     1        1      paid    
     3        1      paid    
     98       0      Notpaid 
     100      0      Notpaid 

[dpb]

Actually, the categorical is better choice than the string... +1