How to create string of text-objects

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okoth ochola
okoth ochola 2023-4-26
评论： chicken vector 2023-4-26
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

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采纳的回答

dpb
dpb 2023-4-26
编辑：dpb 2023-4-26
Ran in:
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

更多回答（2 个）

chicken vector
chicken vector 2023-4-26
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
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dpb
dpb 2023-4-26
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.

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Dyuman Joshi
Dyuman Joshi 2023-4-26
编辑：Dyuman Joshi 2023-4-26
Ran in:
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

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