# mapping numbers to another number

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bkshn on 19 Feb 2015
Edited: Jan on 4 Feb 2021
Hello
I want to map numbers 1,2,3,4,5,6,7,8,9 to 11,12,13,21,22,23,31,32,33. for example if I have number '1' I get '11'
if I have number '2' I get '12' . I don't want to use 'IF statement', because time and performance is important for me. In addition to 'switch case' is there any solution?
Thanks
##### 2 CommentsShowHide 1 older comment
Jan on 4 Feb 2021
In your code the variable full_file_name is not defined.
But this is a new question, so post it in a new thread instead of the section for comments to another question.

David Young on 19 Feb 2015
I don't see why switching case might be relevant to this problem.
Put the numbers you want to map to into an array
outputNumbers = [11,12,13,21,22,23,31,32,33];
and then perform the mapping using array indexing
x = 3; % input number for testing
y = outputNumbers(x); % output number
which will assign 13 to y. This will work for multiple numbers, for example
x = [3 5 9];
y = outputNumbers(x);
which will assign [13 22 33] to y.
If you want your numbers to be represented by strings rather than as binary, use a cell array of strings rather than an array of numerical values.

Billy St. John on 6 Feb 2020
Edited: Billy St. John on 7 Feb 2020
While Indexing and Linear Interpolation can be used with the data in the question, for anyone else (like me) that came here trying to solve the general problem of mapping certain values in any array to a different values (E.g. substituting values in a data array), consider using changem if you have the Mapping Toolbox or implement your own function as detailed here
Ex:
dataToConvert = [1 1 2 20 22 3 55];
from = [3 55 20];
to = [3000 1234 888];
convertedChangem = changem(dataToConvert, to, from)
% convertedChangem =
% [1 1 2 888 22 3000 1234]
convertedInterp1 = interp1(from, to, dataToConvert)
convertedIndexing = to(dataToConvert)
% convertedInterp1 =
% [NaN NaN NaN 888 907.8 3000 1234]
% convertedIndexing = to(dataToConvert) raises an error with "Index exceeds matrix dimensions."
Note that Linear Interpolation results in NaNs for certain values and Basic Indexing does not work unless the values in the dataToConvert array are all valid indices of the to array.
Billy St. John on 7 Feb 2020
Interesting, i didn't realize that. I'll add that note to my comment.

Chandrasekhar on 19 Feb 2015
Edited: Chandrasekhar on 19 Feb 2015
use
vq = interp1(x,v,xq)
where x is the first set of numbers and r is the second set of numbers.
x = [1,2,3,4,5,6,7,8,9]
v = [11,12,13,21,22,23,31,32,33]
xq is the input number
David Young on 19 Feb 2015
There's nothing in the question to suggest that interpolation is needed. Indexing - much simpler and more efficient - suffices.

Michael Haderlein on 19 Feb 2015
If you don't want to limit it to the first 9 numbers, you can also use this equation:
>> x=1:15;
>> y=mod(x-1,3)+1+(1+fix((x-1)/3))*10;
>> y
y =
11 12 13 21 22 23 31 32 33 41 42 43 51 52 53

Walter Roberson on 7 Feb 2020
One solution:
%this does not require that old or new be integers or positive
old = [1,2,3,4,5,6,7,8,9];
new = [11,12,13,21,22,23,31,32,33];
[found, idx] = ismember(TheNumbers, [1,2,3,4,5,6,7,8,9]);
replaced_numbers = TheNumbers;
replaced_numbers(found) = new(idx(found));
But in the case where the old series is consecutive integers:
oldmin = 1; oldmax = 9; %does not have to start at 1
new = [11,12,13,21,22,23,31,32,33];
mask = TheNumbers >= oldmin & TheNumbers <= oldmax;
replaced_numbers = TheNumbers;
Another approach, especially if you know the range of values ahead of time and they are integers:
LUT = cast(0 : maximum_possible_input, class(TheNumbers)); %e.g., cast(0:255,'uint8')
LUT(1+(1:10)) = [11,12,13,21,22,23,31,32,33];
replaced_numbers = LUT(double(TheNumbers)+1);
This version of the code assumes that TheNumbers might be an integer data type that includes 0, such as uint8 . The double(TheNumbers) is needed because uint8(255)+1 would saturate to 255 whereas we need 255 to map to 256 because we have to shift index by 1 to allow for 0.
In the special case where the inputs are integer but never include 0:
LUT = cast(1 : maximum_possible_input, class(TheNumbers)) %e.g., cast(1:255, 'uint8')
LUT(1:10) = [11,12,13,21,22,23,31,32,33];
replaced_numbers = LUT(TheNumbers);
Walter Roberson on 7 Feb 2020
If you are working with integers and you have the Image Processing Toolbox, you can also use https://www.mathworks.com/help/images/ref/intlut.html
LUT = uint8(0:255);
LUT(1+(1:10)) = [11,12,13,21,22,23,31,32,33];
replaced_numbers = imlut(TheNumbers, LUT);