Return the maximum value of every subgroup in a table

I have a section of table like this:

For each group of rows having the same X, I need to select only the row having the largest Value. How can I generate a table like this?

The code I have so far produces just one row:

X = [1; 1; 1; 2; 2; 2; 3; 3; 3];
Y = [2; 3; 4; 1; 3; 4; 1; 2; 4];
Value = [6.9; 6.8; 8.1;7.2;11.7;16;22.6;20.5;18.1];
T = table(X,Y,Value);
[~,maxidx] = max(Value);
T(maxidx,:)
%{
ans =
  1×3 table
    X    Y    Value
    _    _    _____
    3    1    22.6 
%}

If you're using R2015b or newer, you could use splitapply:

function T2 = q56413455()
% Define some example inputs:
X = [1; 1; 1; 2; 2; 2; 3; 3; 3];
Y = [2; 3; 4; 1; 3; 4; 1; 2; 4];
Value = [6.9; 6.8; 8.1;7.2;11.7;16;22.6;20.5;18.1];
T = table(X,Y,Value);
% Call the "business logic" and assign the output:
T2 = getMaxRows(T);

function out = getMaxRows(T)
GROUPING_VAR = 1; % We assume that the 1st column contains the grouping variable
varnames = T.Properties.VariableNames;
tmp = splitapply(@maxrow, T, T.(varnames{ GROUPING_VAR }));
out = array2table(tmp, 'VariableNames', varnames );

function outrow = maxrow(varargin)
COL_WITH_MAX = 3; % We assume that the 3rd columns is the one that matters for max()
subArr = cell2mat(varargin);
[~,I] = max(subArr(:,COL_WITH_MAX));
outrow = subArr(I,:);

Calling this produces the desired result:

>> q56413455
ans =
  3×3 table
    X    Y    Value
    _    _    _____
    1    4     8.1 
    2    4      16 
    3    1    22.6 

Another variation uses the 2^nd output of splitapply(@max, ...) which is the index of the maximum within the group. We then need to add the amount of elements in previous groups to this (this is done using diff):

X = [1; 1; 1; 2; 2; 2; 3; 3; 3];
Y = [2; 3; 4; 1; 3; 4; 1; 2; 4];
Value = [6.9; 6.8; 8.1;7.2;11.7;16;22.6;20.5;18.1];
T = table(X,Y,Value);

% Get the position of the maximum Value in every group
[~,I] = splitapply(@max, T.Value, T.X); % I == [3; 3; 1]

% Get beginnings of every group
lastGroupEnd = find([1; diff(X)])-1; % lastGroupEnd == [0; 3; 6]

% Offset the maximum positions by group starts to get row indices in the original table
T2 = T(I + lastGroupEnd, :);