How to sort and rearrange time series data into random groups?

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Parthu P
Parthu P 2019-11-15
评论： Parthu P 2019-11-15
Hi, I have 100 years time series data.
Here is one month data as example:
= = = = = = = = = = = = = = = = = = = = = = = = = = = = == = = = = = = = == = = = = = = = = = = = = = = = = = = = = = = =
Date temperature differrnce
01-Mar-2011 00:00:00 1.5764
02-Mar-2011 00:00:00 2.0028
03-Mar-2011 00:00:00 1.9728
04-Mar-2011 00:00:00 1.7604
05-Mar-2011 00:00:00 1.6378
06-Mar-2011 00:00:00 1.3025
07-Mar-2011 00:00:00 1.9193
08-Mar-2011 00:00:00 2.0134
09-Mar-2011 00:00:00 2.1462
10-Mar-2011 00:00:00 2.1556
11-Mar-2011 00:00:00 2.14
12-Mar-2011 00:00:00 1.9263
13-Mar-2011 00:00:00 2.9042
14-Mar-2011 00:00:00 2.8062
15-Mar-2011 00:00:00 1.6993
16-Mar-2011 00:00:00 2.3935
17-Mar-2011 00:00:00 1.9432
18-Mar-2011 00:00:00 1.9997
19-Mar-2011 00:00:00 2.0371
20-Mar-2011 00:00:00 1.632
21-Mar-2011 00:00:00 2.2297
22-Mar-2011 00:00:00 1.8848
23-Mar-2011 00:00:00 1.129
24-Mar-2011 00:00:00 1.5175
25-Mar-2011 00:00:00 1.7745
26-Mar-2011 00:00:00 1.5799
27-Mar-2011 00:00:00 1.8735
28-Mar-2011 00:00:00 2.3939
29-Mar-2011 00:00:00 1.9723
30-Mar-2011 00:00:00 1.3674
31-Mar-2011 00:00:00 1.1814
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
How to rearrange this data into random groups based on date and time?
For example:
Group1: 01-Mar-2011 00:00:00 to 20-Mar-2011 00:00:00
Group2: 06-Mar-2011 00:00:00 to 14-Mar-2011 00:00:00
Thanks in advance.
  2 个评论
Walter Roberson
Walter Roberson 2019-11-15
Is the implication that there can be overlaps?
The groups are not fixed length? Is there a maximum and mininum length? But they are also continuous, no gaps right?
Parthu P
Parthu P 2019-11-15
Thanks Walter,
Yes, the group will overlap and there is no fixed length and no gaps.
I have to sort and rearrange these data randomly (only based on date/time) from continuous 100 years daily time-series.

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Walter Roberson
Walter Roberson 2019-11-15
readtable(). If necessary, add the duration variable corresponding to the time, to the datetime variable for the date -- though in what you show us, the times are all 0:0:0 so you probably do not need to do that.
Now sort() the table based upon the time.
After that it is just a matter of using
sort(randi(height(TheTable),2,number_of_sequences))
to have the first row correspond to the row index in the table for the start, and the second row correspond to the row index in the table to stop at.
Note: this does not distribute sizes equally. Smaller sequences are favoured; the descrease looks to be pretty much linear as size of the sequence increases.
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Walter Roberson
Walter Roberson 2019-11-15
Where I wrote number_of_sequences, that was the number of different random sequences you want to create, not an indication of which part of the table you want to select from.
timetable() objects can be indexed by date, with there being a helper function for time ranges: https://www.mathworks.com/help/matlab/ref/timerange.html
For you, instead,
number_of_sequences = 83; %set as appropriate
mask = isbetween(A.Date, datetime(2011,1,3), datetime(2011,3,20) );
subtable = B(mask,:);
rand_subsets = sort(randi(height(subtable), 2, number_of_sequences));
rand_subsets = cell(number_of_sequences,1);
for K = 1 : number_of_sequences
rand_subsets{K} = subtable(rand_subsets(1,K):rand_subsets(2,K), :);
end

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