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Group mean for repeated samples

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Colin Edgar
Colin Edgar 2015-11-3
评论： Colin Edgar 2015-11-4
Have a looping sample with index 1-5 through time, I need averages for each period and index (not simple group-wise mean of id 1-5). Complicated by occasional non-even sample numbers, i.e. usually there are 2 data points for each id, but occasionally 3 and also by the hh and mm time stamp not changing in step with id, i.e. the same minute could exist for several id's. Data like so (actual dataset larger and more complicated):
yyyy mm dd hh mm rec id data1 data2
2015 5 16 12 54 65433 1 402.2262 3.598117
2015 5 16 12 54 65434 1 401.9153 3.596588
2015 5 16 12 54 65435 2 401.9349 3.597093
2015 5 16 12 54 65436 2 402.1344 3.598488
2015 5 16 12 54 65437 3 401.6603 3.599219
2015 5 16 12 55 65438 3 402.3899 3.596839
2015 5 16 13 3 65439 4 401.6629 3.574918
2015 5 16 13 3 65440 4 401.9409 3.575383
2015 5 16 13 3 65441 5 402.3803 3.574625
2015 5 16 13 3 65442 5 402.174 3.571742
2015 5 16 13 24 65443 1 401.9677 3.568859
2015 5 16 3 24 65444 1 401.7614 3.565976
Could use a moving window-type approach:
for i = 1:floor(n/(period)
means1(i) = mean(T1((i*period-(period-1)):(i*period)));
means2(i) = mean(T2((i*period-(period-1)):(i*period)));
...but occasionally there are less than expected numbers in the period.
This is nice solution if I wanted all ID 1 etc means:
[ii,jj] = ndgrid(id,1:size(data,2));
bb = accumarray([ii(:),jj(:)], data(:), [], @mean);
meanout = data - bb(id,:);
So what I want as result is:
id n mean1 mean2
1 2 402.07075 3.5973525
2 2 402.03465 3.5977905
3 2 402.0251 3.598029
4 2 401.8019 3.5751505
5 2 402.27715 3.5731835
1 2 401.86455 3.5674175
Any advice appreciated, could not find info for this specific issue. Some clever use of accumarray will probably work, but I could not conquer it.
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dpb
dpb 2015-11-3
Think it might help to post a sufficient section of the file to illustrate the problem(s) tried to describe. I, at least, got lost trying to figure out what is in the data and isn't and where the time identifiers referred to came from...oh, now I see essentially the same comment in Cyclist's Answer.
Colin Edgar
Colin Edgar 2015-11-3
I have posted better data sorry about that....

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采纳的回答

Guillaume
Guillaume 2015-11-3
It looks like you can't use the date to generate (with unique as the cyclist suggested) a key for accumarray. But since your data is sorted, really, the only thing you want is to find where each period starts. That's easily done, that's when the diff of id is negative. So:
period = cumsum([0; diff(id)<0]); %asuming id is a column vector
[pid, ~, subs] = unique([period id], 'rows');
mean1 = accumarray(subs, data1, [], @mean);
mean2 = accumarray(subs, data2, [], @mean);
count = accumarray(subs, ones(size(subs)));
result = array2table([pid, count, mean1, mean2], 'VariableNames', {'period', 'id', 'count', 'mean1', 'mean2'})
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Guillaume
Guillaume 2015-11-4
If you have used the exact same table format for output as I've shown in my example, it's dead easy to convert the tall table into the wide format you want with unstack:
result = unstack(result, {'mean1', 'mean2'}, 'id')
You may want to use the optional NewDataVariableNames to rename the columns, e.g:
result = unstack(result, {'mean1', 'mean2'}, 'id', 'NewDataVariableNames', ...
arrayfun(@(id) sprintf('id%d', id), unique(result.id), 'UniformOutput', false))
Colin Edgar
Colin Edgar 2015-11-4
Thank you for being so helpful, I appreciate it :)

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更多回答（1 个）

the cyclist
the cyclist 2015-11-3
I don't fully understand how to identify the rows that should be grouped, but it seems that if you can figure that out, then you can use accumarray.
One possibility that comes to mind is to use the unique function (with the "rows" argument) on your original data. You identify the ids, times, or whatever that are unique, and should be aggregated.
The third argument of unique is an index from the unique values back to the original, and can be used as the first argument of accumarray, telling it exactly which rows to aggregate.
Make sense? My answer here is an example of the technique. In that example I only needed to look at one column to find the unique values, but the concept would extend to your case where multiple values are required to identify the "key".
  1 个评论
Colin Edgar
Colin Edgar 2015-11-3
Great thank you for the guidance! I looked at your answer to the similar problem. I can see that:
[uniqueID,~,idx] = unique(mydata(:,7));
gets: uniqueID = [1,2,3,4,5]
idx[1,1,2,2,3,3,4,4,5,5,1,1]
I still can't get the means the way I want with idx. See in my updated example data, I already have a unique sequence number, maybe can use that.

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