Find the first value of each groups.

Hello everybody,

I would like to find the first value of each groups. Is there a way to find the each group's first value?

the expecting answer is [1 5 2 1] in this case.

I tried with

GroupFirst1 = arrayfun(@(k) result(find(result(ic==k), 1, 'first')),1:length(GroupId))';

but it doest not work. Could anyone give me some helps?

x = [3 3 3 4 4 5 5 5 5 3 11]';

y = [1 2 4 5 7 2 1 8 10 10 1]';

result = [x,y];

[GroupId,~,ic] = unique(result(:,1));

% Calculating the averages of different groups of values

GroupMean1 = arrayfun(@(k) mean(result(ic==k,2)),1:length(GroupId))';

% How to find the first y value of according to x

% I need to get 1st value of Ys, corresponds to similar groups of Xs separately.

% first y value when x = 3 is.. 1, and first y value when x = 4 is.. 5, and

% first y value when x = 5 is.. 2, and first y value when x = 11 is.. 1.

% so the expecting answer is [1 5 2 1]

% GroupFirst1 = arrayfun(@(k) result(find(result(ic==k), 1, 'first')),1:length(GroupId))'; % it doest not work.

Answer 1

[Raghava S N]

x = [3 3 3 4 4 5 5 5 5 3 11]';
y = [1 2 4 5 7 2 1 8 10 10 1]';
result = [x,y];
[GroupId,~,ic] = unique(result(:,1));
GroupFirst1 = arrayfun(@(k) result(find(result(:,1)==GroupId(k),1),2),1:length(GroupId))'

GroupFirst1 = 4×1

1 5 2 1

This should work. I think you made the mistake of not searching for the 'GroupId's, but their indices with result(ic==k).

[Smithy]

Thank you very much for your answer. I really really really appreciate with it. It works really well now.

Answer 2

[Luca Ferro]

It's not the most efficient way since it uses loops and find instead of indexing but this works:

x = [3 3 3 4 4 5 5 5 5 3 11]'; 
y = [1 2 4 5 7 2 1 8 10 10 1]';
result = [x,y];
[GroupId,~,ic] = unique(result(:,1));
for gg=1:size(GroupId,1)
    allCorr=find(x==GroupId(gg));
    firstCorr(gg)=y(allCorr(1));
end
firstCorr

firstCorr = 1×4

1 5 2 1

[Smithy]

Thank you very much for your huge helps. It works for me reall well.

Answer 3

[Edoardo Mattia Piccio]

Hi Smithy, here there is an attempt to solve your problem. I hope it will be all clear, this is my first answer here

x = [3 3 3 4 4 5 5 5 5 3 11]'; 
y = [1 2 4 5 7 2 1 8 10 10 1]';
[newX,idx]= sort(x,'ascend'); % Sort x so the equal numbers are consecutive
newY= y(idx); % sort y in the same way of x
idx2= [1; diff(newX)~=0];% with diff(newX) I find when the number changes; adding 1 as first element, to taking account the first group of equal numbers
temp= newY.*idx2; % take only the numbers of y that corresponds to the first value of x
GroupFirst1= temp(temp>0);

[Smithy]

Really thank you very much for your helps. It works for me reall well.