# Error in appliying split apply

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Skydriver on 9 Jun 2019
Commented: Walter Roberson on 10 Jan 2021
I have a problem using splitapply for function mean.
Error using splitapply (line 111)
For N groups, every integer between 1 and N must occur at least once in the vector of group numbers.
Error in try_31032019 (line 65)
LPInew = splitapply(@mean,LPI,bins_index);
G1 = findgroups(amax2);
bins_index = discretize(G1,amaxgrid);
LPInew = splitapply(@mean,LPI,bins_index);
CSR_new = splitapply(@mean,CSR,bins_index);
constN1_60_new = splitapply(@mean,N1_60,bins_index);
constFC_new = splitapply(@mean,Const,bins_index);
Is there any one can help me to solve my problem?
Skydriver on 26 Jun 2019
I want to develop 3d figure with x absis is amaxgrid, y ordinat is Mw grid and to match with the z like state bellow. In further I have to decided that QL should be representing in the z direction with matrix of grid as 36 * 7. Yess I have small mistake to decide 7 not 8.
with this coding:
Mwgrid = linspace(4.75,8.25,7);
amaxgrid = linspace(0.19,0.42,36);
grid = meshgrid(Mwgrid,amaxgrid);
QL_new_1 =zeros(size(grid));
for i=1:length(LPInew)
for j=1:length(Mwgrid)
QL_new_1(i,j) = Coeff(1) + Coeff(2)*constN1_60 + Coeff(3).*constFC + ...
Coeff(4).*log(LPInew(i)) + Coeff(5).*log(Mwgrid(j)) + Coeff(6).*log(constDi);
end
end
QL_zeroes=QL_new_1; % import
QL_zeroes(QL_zeroes<0)=0; % import
PL = normcdf(QL_zeroes); % import
[Xq_cond1,Yq_cond1] = meshgrid(linspace(min(Mwgrid),max(Mwgrid),8),linspace(min(amaxgrid),max(amaxgrid),36)); %import
Vq_cond1 = interp2(Mwgrid, amaxgrid, PL,Xq_cond1,Yq_cond1,'cubic');
Vq_cond2 = interp2(Mwgrid, amaxgrid, PL_new,Xq_cond1,Yq_cond1,'cubic')
figure(4)
surf(Xq_cond1, Yq_cond1, Vq_cond1)
hold on
xlim([4.75 7.6])
xlabel('Magnitude (Mw)');
ylim([0 3])
ylabel('Ground Surface Acceleration (a_{max})');
zlim([0 1])
zlabel('Conditional probability of a_{max} and Mw');

Mil Shastri on 12 Oct 2019
The group numbers provided to splitapply need to be every integer between 1 to N. For example:
This will work:
splitapply(@mean,1:5,[2,2,3,1,1])
and this will not because the group number 3 is missing:
splitapply(@mean,1:5,[2,2,4,1,1])
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Walter Roberson on 10 Jan 2021
No.
You can use findgroups on your current non-consecutive numbers.