a=0.222833 a=sprintf('%.6f',a)

You can either use string(arr) or num2str(arr), depending on what you desire. string(arr) will create a string array by converting each element in the array to string (for example, [ "1", "2", "3" ]) whereas num2str(arr) will convert the whole array into a string (example, [ '1 2 3' ]).

How do you want it displayed? v=[0,0,2736,1824]; d=num2str(v);%once it is a character array you can manipulate it as you desire use sprintf() or regexp()

sprintf gives you control over the formatting: http://www.mathworks.com/help/matlab/ref/sprintf.html#input_argument_formatspec

One way to do this is to use sprintf to convert the array to a long string of digits. You can then reshape this string into the appropriate shape. Then you can use cellstr to convert each row of the reshaped string into a separate cell array element.

out = cellstr(reshape(sprintf('%d', A), [], size(A,2)));

Explanation

First convert the matrix into a long string of digits.

s = sprintf('%d', A)
%// 1234123411112222111111112222222211111111 

Then we want to reshape this so that each row of numbers in the original is a row of numbers in the output

s = reshape(s, [], size(A,2))
%// 11121
%// 21121
%// 31121
%// 41121
%// 12121
%// 22121
%// 32121
%// 42121

Then we can use cellstr to convert each row of this into it's own cell array

out = cellstr(s);
%// '11121'
%// '21121'
%// '31121'
%// '41121'
%// '12121'
%// '22121'
%// '32121'
%// '42121'

A different approach

Another way that you could accomplish this is to treat each column of A as a place value (i.e. 10000's place, 1000's, place, 100's place, etc.) and convert each row to an integer knowing that. This can easily be done by multiplying each row with an array of 10^(N-1:-1:0) and summing the elements. This will yield a digit for each row that combines all of the columns. We can then use num2str to convert this to a cell array of strings.

%// Then convert each number to a string in a cell array
out = arrayfun(@num2str, A * (10.^(size(A, 2)-1:-1:0)).', 'uni', 0);

Or to shorten this even more, we can borrow a page out of @rayryeng's book and use sprintfc to convert this array of integers into a cell array of strings:

out = sprintfc('%d', A * (10.^(size(A, 2)-1:-1:0)).');

Benchmark

I was curious about the performance of the methods presented here and in @rayryeng's answer and Dev-iL's answer when you increase the number of rows. I wrote up a quick test script.

function tests()
    % Test the number of rows between 100 and 10000
    nRows = round(linspace(100, 10000, 100));

    times1 = zeros(numel(nRows), 1);
    times2 = zeros(numel(nRows), 1);
    times3 = zeros(numel(nRows), 1);
    times4 = zeros(numel(nRows), 1);
    times5 = zeros(numel(nRows), 1);

    %// Generate a random matrix of N x 5
    getRandom = @(n)randi([0, 9], [n, 5]);

    for k = 1:numel(nRows)
        A = getRandom(nRows(k));
        times1(k) = timeit(@()string_reshape_method(A));
        A = getRandom(nRows(k));
        times2(k) = timeit(@()base10_method(A));
        A = getRandom(nRows(k));
        times3(k) = timeit(@()sprintfc_method(A));
        A = getRandom(nRows(k));
        times4(k) = timeit(@()addition_method(A));
    end

    %// Plot the results
    plot(nRows, cat(2, times1, times2, times3, times4)*1000);
    legend({'String Reshape', 'Base-10 Conversion', 'sprintfc', 'addition of "0"'})

    xlabel('Number of Rows in A')
    ylabel('Execution Time (ms)');
end

function out = string_reshape_method(A)
    out = cellstr(reshape(sprintf('%d', A), [], size(A,2)));
end

function out = base10_method(A)
    out = sprintfc('%d', A * (10.^(size(A, 2)-1:-1:0)).');
end

function B = sprintfc_method(A)
    B = sprintfc(repmat('%d', 1, size(A,2)), A);
end

function B = addition_method(A)
    B = cellstr(char(A + '0'));
end

Here are the results.

B=arrayfun(@num2str,A,'un',0)

Here are two simpler solutions to convert a single double value to a string and back without loss.

I want the string to be a human-readable representation of the number

Use num2str to obtain 17 decimal digits in string form, and str2double to convert back:

>> s = mat2str(x,17) 
s =
    '2.2204460492503131e-16'
>> y = str2double(s);
>> y==x
ans =
  logical
   1

Note that 17 digits are always enough to represent any IEEE double-precision floating-point number.

I want a more compact string representation of the number

Use matlab.net.base64encode to encode the 8 bytes of the number. Unfortunately you can only encode strings and integer arrays, so we type cast to some integer array (we use uint8 here, but uint64 would work too). We reverse the process to get the same double value back:

>> s = matlab.net.base64encode(typecast(x,'uint8'))
s =
    'AAAAAAAAsDw='
>> y = typecast(matlab.net.base64decode(s),'double');
>> x==y
ans =
  logical
   1

Base64 encodes every 3 bytes in 4 characters, this is the most compact representation you can easily create. A more complex algorithm could likely convert into a smaller UTF-8-encoded string (which uses more than 6 bytes per displayable character).

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)

Is the source of the data an input text file? Why not do the conversion when you're reading in the data in a pre-processing step rather than doing it during runtime? Use fscanf or another file reading function to do the conversion to your desired data type. However, this doesn't seem really slow to me. 17 seconds to process 2M+ records seems reasonable for MATLAB. MATLAB is not a performance language. How fast do you need to get this function to run?