Averaging sections of a vector

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Thishan Dharshana Karandana Gamalathge
评论： Andrei Bobrov 2017-7-28
A=[1 3 -999 -999 -999 -999 4 9 8 -999 -999 -999]; Here I want to get averages of three elements by ignoring -999. But in the final answer, if there were cases with three -999 in a given sector, it needs to be replaced by NaN. For an example, final answer of the above should B=[2 NaN 7 NaN]
Thanks.

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采纳的回答

Steven Lord
Steven Lord 2017-7-28
% Sample data
A=[1 3 -999 -999 -999 -999 4 9 8 -999 -999 -999];
% Replace the 'missing' -999 values with NaN
A(ismissing(A, -999)) = NaN;
% If you have a release that doesn't include ismissing, use ==
A(A == -999) = NaN;
% Take the mean, omitting NaN unless all the elements whose mean
% are being computed are NaN
M = mean(reshape(A, 3, []), 'omitnan')
  1 个评论
Andrei Bobrov
Andrei Bobrov 2017-7-28
% Sample data
>> A=[1 3 -999 3 -999 -999 4 9 8 9 -999 -999];
% Replace the 'missing' -999 values with NaN
A(ismissing(A, -999)) = NaN;
% If you have a release that doesn't include ismissing, use ==
A(A == -999) = NaN;
% Take the mean, omitting NaN unless all the elements whose mean
% are being computed are NaN
M = mean(reshape(A, 3, []), 'omitnan')
M =
2 3 7 9
>>

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更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2017-7-28
z = A;
z(z == -999) = nan;
t = ~isnan(z);
B = accumarray(cumsum(diff([~t(1);t(:)]) ~= 0),z,[],@mean);

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