allocate values avoiding loop
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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?
Respuesta aceptada
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
3 comentarios
Rahel Braun
el 12 de Nov. de 2018
>> 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
el 12 de Nov. de 2018
Más respuestas (2)
Bruno Luong
el 8 de Nov. de 2018
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
5 comentarios
Bruno Luong
el 8 de Nov. de 2018
Editada: Bruno Luong
el 8 de Nov. de 2018
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
el 12 de Nov. de 2018
Bruno Luong
el 12 de Nov. de 2018
Editada: Bruno Luong
el 12 de Nov. de 2018
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
el 12 de Nov. de 2018
Bruno Luong
el 12 de Nov. de 2018
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.
Cris LaPierre
el 8 de Nov. de 2018
grpAvg = splitapply(@mean,p,t);
pAvg = grpAvg(t);
[t k p pAvg]
2 comentarios
Cris LaPierre
el 8 de Nov. de 2018
Editada: Cris LaPierre
el 8 de Nov. de 2018
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
el 12 de Nov. de 2018
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
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