Changing Double to String Automatically in a Dataset Array

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Hi there,
I have a dataset array called 'data' of 2738 observations with 9 columns(2738*9) in size.
The 3rd column header is 'Surf' and contains 1's, 10's, 100's, 1000's, 10000's and 100000's.
I intend to write a for loop that goes through the data.surf column and wherever it sees a 1 it replaces it with 'F' and similarly replaces 10's with 'R' and so forth.
It seems as though I can replace the whole column with a 'F' by just typing in data.surf = 'F' but I am not quite sure how to write a mini script that automates this process. Any help or guidance will be appreciated.
Thanks in advance. Nj

Accepted Answer

Teja Muppirala
Teja Muppirala on 27 Dec 2012
Here are two ways to do it:
%% Method 1, using "containers.Map"
ds = dataset([10;100;10;1000;100000;1;1;10000],'VarNames','Surf');
C = containers.Map([1 10 100 1000 10000 100000],{'F' 'R' 'A' 'B' 'C' 'D'});
ds.Surf = C.values(num2cell(ds.Surf))
%% Method 2, using ISMEMBER
ds = dataset([10;100;10;1000;100000;1;1;10000],'VarNames','Surf');
C = {[1 10 100 1000 10000 100000]; {'F'; 'R'; 'A'; 'B'; 'C'; 'D'}};
[~,loc] = ismember(ds.Surf,C{1});
ds.Surf = C{2}(loc)
  6 Comments
Teja Muppirala
Teja Muppirala on 27 Dec 2012
If you need to do be able to reverse the operation also, then I think the ISMEMBER way might be simpler. You could still do it either way though:
%% Using containers.Map
ds = dataset([10;100;10;1000;100000;1;1;10000],'VarNames','Surf');
C = containers.Map([1 10 100 1000 10000 100000],{'F' 'R' 'A' 'B' 'C' 'D'});
ds.Surf = C.values(num2cell(ds.Surf))
Cinv = containers.Map({'F' 'R' 'A' 'B' 'C' 'D'},[1 10 100 1000 10000 100000]);
ds.Surf = cell2mat(Cinv.values(ds.Surf))
%% Using ISMEMBER with an appropriate cell array
ds = dataset([10;100;10;1000;100000;1;1;10000],'VarNames','Surf');
C = {[1; 10; 100; 1000; 10000; 100000]; {'F'; 'R'; 'A'; 'B'; 'C'; 'D'}};
[~,loc] = ismember(ds.Surf,C{1});
ds.Surf = C{2}(loc)
[~,loc] = ismember(ds.Surf,C{2});
ds.Surf = C{1}(loc)
Niraj Poudel
Niraj Poudel on 9 Jan 2013
Thank you so much Teja Muppirala. I really appreciate your answers, they helped me out a lot!

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More Answers (3)

Sean de Wolski
Sean de Wolski on 26 Dec 2012
So something like this?
ds = dataset(rand(10,1)>0.5,'VarNames','Surf');
idx = ds.Surf;
ds.Surf = repmat('F',numel(idx),1);
ds.Surf(idx) = 'T';
ds.Surf
  4 Comments
Niraj Poudel
Niraj Poudel on 27 Dec 2012
I think you are absolutely right. Matlab probably does not allow for me to incrementally change from a double to a string. I might have to create a new column and have matlab input string values to the corresponding double values. Which leads me to my question, How can I write a for loop to index each individual cell in the column 'Surf' and simultaneously write the corresponding 'string' value in the new column? I guess my issue is with the for loop indexing in a dataset array. Thank you so much for taking the time to look into this and respond to me. I really appreciate it. Nj
Walter Roberson
Walter Roberson on 27 Dec 2012
If you are looking at data.surfs(K) then the new thing to write would be something like data.surfcodes(K)

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Walter Roberson
Walter Roberson on 26 Dec 2012
letteridx = 1 + round(log10(data.surf);
letters = 'FRXUH';
data.surf = letters(letteridx);
  2 Comments
Niraj Poudel
Niraj Poudel on 27 Dec 2012
Unfortunately when I try running this code, it keeps telling me that the index exceeds the matrix dimensions. I am hoping there is another way of indexing each element in the "Surf" column and changing the numbers (1,10,100) one by one to a strings (f, R and S) respectively?
Thank you so much for looking into this for me. Nj
Walter Roberson
Walter Roberson on 27 Dec 2012
Could you show max(data.surf) before this code?
I notice that I should have allowed one more letter,
letters = 'fRSXUH';
Change the XUH to the appropriate codes for 1000, 10000, and 100000

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Peter Perkins
Peter Perkins on 9 Jan 2013
Nj, you might also consider using an ordinal variable here:
>> ds = dataset([10;100;10;1000;100000;1;1;10000],'VarNames','Surf')
ds =
Surf
10
100
10
1000
1e+05
1
1
10000
>> ds.Surf = ordinal(ds.Surf,{'F' 'R' 'A' 'B' 'C' 'D'},[1 10 100 1000 10000 100000])
ds =
Surf
R
A
R
B
D
F
F
C
You can now do things like
>> ds(ds.Surf < 'A',:)
ans =
Surf
R
R
F
F
where Surf is created so that F<R<A<B<C<D (which may or my not be what you want).

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