Finding a maximum period

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Kate
Kate il 18 Gen 2017
Commentato: Kate il 18 Gen 2017
Hey Guys,
I'm trying to wrap my brain around how to find n maximum or minimum values in a row. I've used grpstats to generate mean values, but need to find the three highest or lowest in a row, so I can't use a sort function:
[m, g]=grpstats(data(:, 2), data(:, 1), {'mean', 'gname'})
m =
16.4605
19.6645
19.0127
20.0246
21.0003
20.0367
15.7379
22.7125
19.9485
20.4098
21.6320
17.6872
g =
'1'
'2'
'3'
'4'
'5'
'6'
'7'
'8'
'9'
'10'
'11'
'12'
Any ideas or functions to suggest? Thanks and all the best.
  2 Commenti
Walter Roberson
Walter Roberson il 18 Gen 2017
Are you looking for a moving maximum, where every point is replaced by the maximum of the point and the point before and the next point? Are you looking for a single group of three values in a row that has the highest average out of all of the sliding possibilities?
Kate
Kate il 18 Gen 2017
The later, I'm looking for the 3 highest (or lowest) consecutive values. This is a metric called Mean Temperature of the Warmest Quarter. Thanks Walter.

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Risposta accettata

Image Analyst
Image Analyst il 18 Gen 2017
If you're looking for the group of 3 elements that have the highest sum (and mean) then you can use conv() or movmean()
threeElementSum = conv(data, [1,1,1], 'same');
[maxSum, indexOfMax] = max(threeElementSum);
This will tell you where the center of the 3 element window is where the values inside that 3 element window have the highest sum of any window location in the data. Is that what you're looking for?
  1 Commento
Kate
Kate il 18 Gen 2017
Perfect, convolution is exactly what I was looking for. I appreciate it.

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Più risposte (1)

Alexandra Harkai
Alexandra Harkai il 18 Gen 2017
If you're looking for the n highest and lowest entry for each row of input data, you can still use the sort function:
[sorted, row_idx] = sort(data, 2);
sorted will contain the sorted data, row_idx will contain the row-wise indices, so:
sorted(:,1:n) % n smallest elements in each row
sorted(:,(end+1-n):end) % n largest elements in each row
row_idx(1,1:n) % indices in rows for the n smallest elements in each row
row_idx(1,(end+1-n):end) % indices in rows for the n largest elements in each row
  1 Commento
Kate
Kate il 18 Gen 2017
Modificato: Kate il 18 Gen 2017
It's a great crack, but essentially I can't sort the values, because then the months are split up. Here's an example result of your code:
row_idx((end+1-n):end, 1)
ans =
5
11
8
Indicating May, November and August, which aren't proximal in time. I appreciate the help though.

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