您可以使用splitapply，每行有一个id。请查看评论了解详细信息...

% Assign unique ID to each row
tbl.id = (1:size(tbl,1))';
% Get groups of the different strings
g = findgroups(tbl.String);
% create function which gets id of max within each group 
% f must take arguments corresponding to each splitapply table column
f = @(num,id) id(find(num == max(num), 1));
% Use splitapply to apply the function f to all different groups
idx = splitapply( f, tbl(:,{'Number','id'}), g );
% Collect rows
outTbl = tbl(idx, {'Location', 'String', 'Number'});

>> outTbl = 
   Location   String    Number
      3        'a'        78   
      1        'b'       361   
      2        'c'        28

或者只是一个简单的循环。这个循环只在String的唯一值上，所以应该很快。

u = unique(tbl.String);
c = cell(numel(u), size(tbl,2));
for ii = 1:numel(u)
    temp = tbl(strcmp(tbl.String, u{ii}),:);
    [~, idx] = max(temp.Number);
    c(ii,:) = table2cell(temp(idx,:));
end
outTbl = cell2table(c, 'VariableNames', tbl.Properties.VariableNames);

找到每个字符串的最大值我的想法是。
为所有字符串创建一个vector，并只包含一次。类似于：
strs=['a','b','c'];
然后创建一个vector来存储每个字符串的最大值：

n=length(strs);
max_values=zeros(1,n);

现在创建一个循环，将当前的max_value与当前值进行比较，并替换if current_value>max_value：

for i=1:your_table_size
   m=find(strs==current_table_string);   % This finds the index of max_values
   if max_values(m)<current_table_Number   % This the the i_th row table_number
      max_values(m)=current_table_Number;
   end
end