allocate values avoiding loop

I have the following matrix [t k p]

    1.0000    1.0000   -1.1471
    1.0000    2.0000   -1.0689
    2.0000    1.0000   -0.8095
    2.0000    2.0000   -2.9443
    3.0000    1.0000    1.4384
    3.0000    2.0000    0.3252

and I want an additional column with the mean of p for every t, hence

    1.0000    1.0000   -1.1471   -1.1080
    1.0000    2.0000   -1.0689   -1.1080
    2.0000    1.0000   -0.8095   -1.8769
    2.0000    2.0000   -2.9443   -1.8769
    3.0000    1.0000    1.4384    0.8818
    3.0000    2.0000    0.3252    0.8818

I can do it with the following code

if true
%Calulate the mean
A=[t p_tk];
p_t= accumarray(A(:,[1]), A(:,2), [], @nanmean, NaN);
% allocate it to long form
p_t_long= NaN(size(t));
for d = 1:max(t)
  ind= t ==d; 
  p_t_long(ind)= p_t(d);
end
end

However, I want to avoid loops since I have a big dataset. Can anybody help?

Answer 1

[Stephen23]

Some indexing using the first column does what you want, more efficiently than a loop or unique:

>> M = [1,1,-1.1471;1,2,-1.0689;2,1,-0.8095;2,2,-2.9443;3,1,1.4384;3,2,0.3252]
M =
   1.00000   1.00000  -1.14710
   1.00000   2.00000  -1.06890
   2.00000   1.00000  -0.80950
   2.00000   2.00000  -2.94430
   3.00000   1.00000   1.43840
   3.00000   2.00000   0.32520
>> V = accumarray(M(:,1),M(:,3),[],@mean)
V =
  -1.10800
  -1.87690
   0.88180
>> M(:,4) = V(M(:,1))
M =
   1.00000   1.00000  -1.14710  -1.10800
   1.00000   2.00000  -1.06890  -1.10800
   2.00000   1.00000  -0.80950  -1.87690
   2.00000   2.00000  -2.94430  -1.87690
   3.00000   1.00000   1.43840   0.88180
   3.00000   2.00000   0.32520   0.88180

[Rahel Braun]

Thank you for your quick answer, it works perfectly, however, I abbreviated my question, because I thought it would work the same way. What if I have 3 or more groups and want to calculate the mean for each group and reassign it?

M =

    1.0000    1.0000    1.0000   -5.2359
    1.0000    1.0000    2.0000  -12.7706
    1.0000    2.0000    1.0000    8.0797
    1.0000    2.0000    2.0000    5.8198
    2.0000    1.0000    1.0000    3.3348
    2.0000    1.0000    2.0000  -10.3368
    2.0000    2.0000    1.0000   14.2749
    2.0000    2.0000    2.0000    6.5018
    3.0000    1.0000    1.0000    0.0093
    3.0000    1.0000    2.0000    3.2289
    3.0000    2.0000    1.0000    0.3800
    3.0000    2.0000    2.0000  -14.5021

and I want

V =

    1.0000    1.0000    1.0000    0.1435   -2.5851
    1.0000    1.0000    2.0000   -5.3137   -2.5851
    1.0000    2.0000    1.0000   -6.7921   -7.6780
    1.0000    2.0000    2.0000   -8.5640   -7.6780
    2.0000    1.0000    1.0000   -2.3356   -9.6810
    2.0000    1.0000    2.0000  -17.0264   -9.6810
    2.0000    2.0000    1.0000   12.6423   10.4214
    2.0000    2.0000    2.0000    8.2006   10.4214
    3.0000    1.0000    1.0000    2.7997    2.7260
    3.0000    1.0000    2.0000    2.6523    2.7260
    3.0000    2.0000    1.0000   -4.9816    4.1026
    3.0000    2.0000    2.0000   13.1869    4.1026

[Stephen23]

>> M = [1,1,1,0.1435;1,1,2,-5.3137;1,2,1,-6.7921;1,2,2,-8.5640;2,1,1,-2.3356;2,1,2,-17.0264;2,2,1,12.6423;2,2,2,8.2006;3,1,1,2.7997;3,1,2,2.6523;3,2,1,-4.9816;3,2,2,13.1869]
M =
    1.00000    1.00000    1.00000    0.14350
    1.00000    1.00000    2.00000   -5.31370
    1.00000    2.00000    1.00000   -6.79210
    1.00000    2.00000    2.00000   -8.56400
    2.00000    1.00000    1.00000   -2.33560
    2.00000    1.00000    2.00000  -17.02640
    2.00000    2.00000    1.00000   12.64230
    2.00000    2.00000    2.00000    8.20060
    3.00000    1.00000    1.00000    2.79970
    3.00000    1.00000    2.00000    2.65230
    3.00000    2.00000    1.00000   -4.98160
    3.00000    2.00000    2.00000   13.18690
>> [~,~,idx] = unique(M(:,1:end-2),'rows'); % indices of row groups.
>> V = accumarray(idx,M(:,end),[],@mean); % mean of each group.
>> M(:,5) = V(idx)
M =
    1.00000    1.00000    1.00000    0.14350   -2.58510
    1.00000    1.00000    2.00000   -5.31370   -2.58510
    1.00000    2.00000    1.00000   -6.79210   -7.67805
    1.00000    2.00000    2.00000   -8.56400   -7.67805
    2.00000    1.00000    1.00000   -2.33560   -9.68100
    2.00000    1.00000    2.00000  -17.02640   -9.68100
    2.00000    2.00000    1.00000   12.64230   10.42145
    2.00000    2.00000    2.00000    8.20060   10.42145
    3.00000    1.00000    1.00000    2.79970    2.72600
    3.00000    1.00000    2.00000    2.65230    2.72600
    3.00000    2.00000    1.00000   -4.98160    4.10265
    3.00000    2.00000    2.00000   13.18690    4.10265

[Rahel Braun]

Thank you!

Answer 2

[Bruno Luong]

    A=[...
        1.0000    1.0000   -1.1471
        1.0000    2.0000   -1.0689
        2.0000    1.0000   -0.8095
        2.0000    2.0000   -2.9443
        3.0000    1.0000    1.4384
        3.0000    2.0000    0.3252 ]

    [~,~,J] = unique(A(:,1));
    p_t= accumarray(J, A(:,3), [], @(x) mean(x,'omitnan'), NaN);
    [A p_t(J)]

Result

ans =

      1.0000    1.0000   -1.1471   -1.1080
      1.0000    2.0000   -1.0689   -1.1080
      2.0000    1.0000   -0.8095   -1.8769
      2.0000    2.0000   -2.9443   -1.8769
      3.0000    1.0000    1.4384    0.8818
      3.0000    2.0000    0.3252    0.8818

[Bruno Luong]

The old loyal ACCUMARRAY is still the king of the speed

n = 1e6;
ntest = 10;
time = zeros(2,ntest);
for i = 1:ntest
    g = ceil(100*rand(n,1));
    t = rand(n,1);

      tic
      grpAvg1 = splitapply(@mean,t,g);
      time(1,i) = toc;

      tic
      grpAvg2 = accumarray(g,t)./accumarray(g,1);
      time(2,i) = toc;
  end
  time = mean(time,2);

fprintf('splitapply time = %f s\n', time(1)); % splitapply time = 0.158369 s
fprintf('accumarray time = %f s\n', time(2)); % accumarray time = 0.029645 s

[Rahel Braun]

Thank you for your answer, and yes accumarray is great! with a loop it took me almost a whole afternoon to calculate means and now it takes only a minute! However, I abbreviated my question, because I thought it would work the same way. Do you know how I can solve the problem if I have 3 or more groups and want to calculate the mean for each group and reassign it? Like in the comment above https://ch.mathworks.com/matlabcentral/answers/428818-allocate-values-avoiding-loop#comment_636183

[Bruno Luong]

Unless I misunderstood something, assignment with the mean already answered, look at the last statement

[A p_t(J)]

where p_t is accumarray output (average) and J is the third output of UNIQUE.

Assign it to A if you like.

[Rahel Braun]

My problem was that I didn't know how to use unique() properly with 3 groups, and then the accumarray gave me a 3*2 matrix and I couldn't assign that. But thank you anyway.

[Bruno Luong]

My problem was that I didn't know how to use unique() properly with 3 groups,

Stephen already answered by just add 'ROWS' argument, to have one identification (third output) by for each 1x3 row (your "groups").

BTW, you might not noticed by using

accumarray(...,data) ./ accumarray(...,1)

is always fater than

 accumarray(...,data, ..., @mean)

if speed is matter for you.

Answer 3

[Cris LaPierre]

Consider using the splitapply function. Assuming you have variable t, k and p already defined:

grpAvg = splitapply(@mean,p,t);
pAvg = grpAvg(t);
[t k p pAvg]

[Cris LaPierre]

If your grouping variable is not as clean as it is in this example, you can use the findgroups function to create an index of the unique values in your grouping variable.

[Cris LaPierre]

Using your updated matrix from a comment, here is a robust way to achieve what you want using findgroups and splitapply (assuming variable t,k,l, and p exist and represent the columns of M):

M = [t k l p]
G = findgroups(t,k);
grpAvg = splitapply(@mean,p,G);
pAvg = grpAvg(G);
V = [t k l p pAvg]

The original matrix M is

M =

      1.0000    1.0000    1.0000    0.1435
      1.0000    1.0000    2.0000   -5.3137
      1.0000    2.0000    1.0000   -6.7921
      1.0000    2.0000    2.0000   -8.5640
      2.0000    1.0000    1.0000   -2.3356
      2.0000    1.0000    2.0000  -17.0264
      2.0000    2.0000    1.0000   12.6423
      2.0000    2.0000    2.0000    8.2006
      3.0000    1.0000    1.0000    2.7997
      3.0000    1.0000    2.0000    2.6523
      3.0000    2.0000    1.0000   -4.9816
      3.0000    2.0000    2.0000   13.1869

And resulting matrix V is

V =

    1.0000    1.0000    1.0000    0.1435   -2.5851
    1.0000    1.0000    2.0000   -5.3137   -2.5851
    1.0000    2.0000    1.0000   -6.7921   -7.6780
    1.0000    2.0000    2.0000   -8.5640   -7.6780
    2.0000    1.0000    1.0000   -2.3356   -9.6810
    2.0000    1.0000    2.0000  -17.0264   -9.6810
    2.0000    2.0000    1.0000   12.6423   10.4215
    2.0000    2.0000    2.0000    8.2006   10.4215
    3.0000    1.0000    1.0000    2.7997    2.7260
    3.0000    1.0000    2.0000    2.6523    2.7260
    3.0000    2.0000    1.0000   -4.9816    4.1026
    3.0000    2.0000    2.0000   13.1869    4.1026