How can i encode strings in a array to numbers?

Question

arda has 2024-3-2

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回答： Steven Lord 2024-3-2

I have different strings in a matrix and some of the coloums of this matrix is strings. they includes city names etc. so i would like to convert this city names into number i did below code. my matrix is made of string so firstly i try to learn how many unique string are exist in a coloumn for example in 9th coloumn.

c=unique(matrix(:,9));

then for example i find 3400 elements and i give numbers to each one. (first coloumn of the c is city names and second coloumn is numbers 1 to 3400)

c(:,2)=1:3400;

then i have 1 million rows and i try to match everyone of them with numbers and create a new matrix that includes only numbers

for i=1:1000000

for j=1:3400

if matrix(i,9)==c(j,1)

numbermatrix(i,9)=str2double(c(j,2));

end

so this code works well but it takes a lot of time to compare every variable with all possibilities.

Is there any easy and fast working method to do this job?

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MarKf 2024-3-2

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Ok, maybe you can make that a little more clear.

>"I have different strings in a matrix and some of the coloums of this matrix is strings"

So you have a string matrix. Which is unlikely, so you have either a string array or more likely a cell. You could just have an actual matrix array if you have the same number of characters and have the shorter city names just padded with spaces, but ugh hopefully not. Or maybe city codes of the same lenght. Anyway it does not matter since it seems you are only interested in column 9 and unique works on all of the above. So maybe:

matcol9 = {'New York', 'London', 'Paris', 'San Francisco', 'Toronto', 'Sydney', 'Singapore', 'London', 'Paris'}'; % example cell; ["London", "Paris" etc] in case of string array
[c,ia,ic]=unique(matcol9); %If A is a vector, then C = A(ia) and A = C(ic).
c = 7×1 cell array
    {'London'       }
    {'New York'     }
    {'Paris'        }
    {'San Francisco'}
    {'Singapore'    }
    {'Sydney'       }
    {'Toronto'      }

So unique has already a built-in way to assign an index to the original elements (the variable ic). And that may already be what you need. Assuming that the 1 million rows is the same "matrix" (why are they rows now tho?). Otherwise ismember might be more helpful than comparing each elemnt to each.

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Answer 1

Hassaan 2024-3-2

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Using containers.Map

% Assume 'matrix' is your string matrix
uniqueCities = unique(matrix(:,9));
% Create a map object where key-value pairs correspond to city names and their unique integer identifiers
cityMap = containers.Map(uniqueCities, 1:numel(uniqueCities));
% Preallocate the number matrix for efficiency
numberMatrix = zeros(size(matrix,1), 1); % Adjust the size according to your needs
% Now, iterate through each city name in your matrix and replace it with its corresponding number using the map
for i = 1:size(matrix,1)
    city = matrix(i,9);
    if isKey(cityMap, city)
        numberMatrix(i,9) = cityMap(city);
    end
end

Using ismember

[uniqueCities, ~, idx] = unique(matrix(:,9));
% The 'idx' array contains the indices of the unique values in 'uniqueCities' that correspond to each entry in 'matrix(:,9)'
% Now, directly use these indices as your numeric representation
% If 'numberMatrix' should retain the original 'matrix' structure but with numbers in the 9th column
numberMatrix = matrix; % Assuming 'matrix' is a numeric matrix and you want to keep other data unchanged
numberMatrix(:,9) = idx; % This replaces the 9th column with numerical identifiers

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arda has 2024-3-2

thank you

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Answer 2

Steven Lord 2024-3-2

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If you have data where some of the columns are text and some are numbers, consider storing your data in a table array rather than trying to use numbers as proxies for the text data. In a table array, each variable has to have the same type of data but different variables can contain different types of data.

names = ["Walter Roberson"; "Image Analyst"; "Star Strider"];
reputation = [135067; 77260; 65504];
contributors = table(names, reputation)
contributors = 3×2 table
          names          reputation
    _________________    __________

    "Walter Roberson"    1.3507e+05
    "Image Analyst"           77260
    "Star Strider"            65504

Here the names variable is text while the reputation variable is numeric.

starStriderReputation = contributors{3, 'reputation'}
starStriderReputation = 65504

Alternately, instead of treating the names as data I could treat them as row names and use them to index into the table.

contributors2 = table(reputation, RowNames=names)
contributors2 = 3×1 table
                       reputation
                       __________

    Walter Roberson    1.3507e+05
    Image Analyst           77260
    Star Strider            65504
walterRobersonReputation = contributors2{"Walter Roberson", 1}
walterRobersonReputation = 135067

If your data is time-based a timetable may be more useful than a table because it provides capabilities to change the time basis of your data or perform time-based analysis (compute daily averages from data collected more frequently, for example.)