# mapping numbers to another number

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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 Comments

Jan
on 4 Feb 2021

Edited: 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.

It is recommended not to publish your email address in this forum, because this will increase the number of spam mails you get.

### Answers (5)

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.

##### 0 Comments

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.

##### 3 Comments

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

##### 1 Comment

David Young
on 19 Feb 2015

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

##### 0 Comments

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;

replaced_numbers(mask) = new(TheNumbers(mask)-oldmin+1); %requires that old is integers

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);

##### 1 Comment

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);

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