Going back in time 1 week or a year

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ektor
ektor 2023-7-14
编辑： Voss 2023-7-17
Dear all,
I attach a data set of weekly observations on an index.
Suppose that I am in week 2023/6/16 (yellow cell). For this cell I want to obtain the corresponding index value
Then I want to go back one week (7 days) ; that is to 2023/6/9 (yellow cell) and calculate its corresponding index value and then calculate the percentage change between the two values.
I used something like
t = datetime(2023,6,16);
t2 = dateshift(t,'dayofweek', 7,'previous')
but it does not work. I am sure that there is a mistake in my code. Is there a way to do that.
Also, assuming that we are in week 2023/6/16, I want to calculate the mean, min, max of the index values for the last 12 months. Is there a way to do that?
I would very much appreciate some guidance.
Thank you in advance!

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采纳的回答

Voss
Voss 2023-7-14
编辑：Voss 2023-7-17
Ran in:
T = readtable('data.xlsx')
T = 1390×2 table
Date Index ___________ _____ 02-Jan-2018 2281 03-Jan-2018 2426 04-Jan-2018 2715 05-Jan-2018 2791 08-Jan-2018 2826 09-Jan-2018 2820 10-Jan-2018 2677 11-Jan-2018 2384 12-Jan-2018 2296 15-Jan-2018 2231 16-Jan-2018 2010 17-Jan-2018 1743 18-Jan-2018 1611 19-Jan-2018 1493 22-Jan-2018 1445 23-Jan-2018 1525
"Suppose that I am in week 2023/6/16 (yellow cell). For this cell I want to obtain the corresponding index value
Then I want to go back one week (7 days) ; that is to 2023/6/9 (yellow cell) and calculate its corresponding index value and then calculate the percentage change between the two values."
dt = datetime(2023,6,9); % June 9th, 2023
row_dt = find(T.Date == dt);
row_a_week_before_dt = find(T.Date == dt - caldays(7));
Index_dt = T{row_dt,'Index'}
Index_dt = 1514
Index_a_week_before_dt = T{row_a_week_before_dt,'Index'}
Index_a_week_before_dt = 1116
percent_change = 100*(Index_dt-Index_a_week_before_dt)/Index_a_week_before_dt
percent_change = 35.6631
"Also, assuming that we are in week 2023/6/16, I want to calculate the mean, min, max of the index values for the last 12 months. Is there a way to do that?"
dt = datetime(2023,6,16); % June 16th, 2023
row_dt = find(T.Date == dt);
row_a_year_before_dt = find(T.Date > dt - calyears(1), 1);
Index_last_year = T{row_a_year_before_dt:row_dt,'Index'};
mean_last_year = mean(Index_last_year)
mean_last_year = 1.6455e+03
min_last_year = min(Index_last_year)
min_last_year = 271
max_last_year = max(Index_last_year)
max_last_year = 3031
  2 个评论
ektor
ektor 2023-7-14
编辑：ektor 2023-7-14
Hi Voss,
Thank you a lot.
Peter Perkins
Peter Perkins 2023-7-17
This answer is incorrect. Do NOT subtract days or years for this purpose.
As the doumentation says, these are "exact-length time" units equal to exactly 24hrs and 365.2425*24hrs. You may be using unzoned datetimes today, but eventually, you will use time zones, in which case days will give you the wrong answer.
years will already give you the wrong answer.
Use caldays and calyears, as Steve Lord says.
I also recommend that you use timetables, and look into things like groupsummary or retime, to compute data summaries over each week, or year or whatever. Your life will be much easier.

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更多回答（1 个）

Steven Lord
Steven Lord 2023-7-15
Ran in:
Subtract the appropriate calendar duration.
t = datetime('today')
t = datetime
15-Jul-2023
oneWeekAgo = t - calweeks(1)
oneWeekAgo = datetime
08-Jul-2023
oneYearAgo = t - calyears(1)
oneYearAgo = datetime
15-Jul-2022
Don't try to subtract 1 year as a duration. That's 365.24 days, not exactly 1 year.
oneYearAgoAlmost = t - years(1)
oneYearAgoAlmost = datetime
14-Jul-2022 18:10:48

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