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Extract a range of data acording to the date range from a table

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Mark Rodger
Mark Rodger 2018 年 2 月 1 日
コメント済み: Guillaume 2018 年 2 月 7 日
I want to separate months of the year from a table into another table for each season.
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Mark Rodger
Mark Rodger 2018 年 2 月 1 日
It's part of my project where I need to show the wind speed for each season. The table has 2 years of wind speed with two variables: Time and WindSpeed. Could you help me with this?
Mark Rodger
Mark Rodger 2018 年 2 月 7 日
I forgot to mention that the table has 2 years worth of data (2013-2014). Is there a way to extract one season of one year

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Peter Perkins
Peter Perkins 2018 年 2 月 1 日
There's only four seasons, best to not overthink this.
spring = t(ismember(month(t.Time),1:3),:);
etc. But Guillaume's advice is worth thinking about.
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Fraser McMurray
Fraser McMurray 2018 年 2 月 7 日
編集済み: Fraser McMurray 2018 年 2 月 7 日
If you want to view a specific year you can try this:
april = x(ismember(month(x.time),4),:);
april2013 = april(ismember(year(april.time),2013),:);
Guillaume
Guillaume 2018 年 2 月 7 日
Again, it is very likely that you don't need to separate the table at all and that it may be more efficient to work on the whole table at once.
For example, if you wanted to calculate the mean wind speed per season per year:
[group, season, year] = findgroups(discretize(month(yourtable.time), [1 4 10 12]), year(yourtable.time));
meanwindspeed = splitapply(@mean, yourtable.windspeed, group);
result = table(season, year, meanwindspeed)

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