How to remove the year from a datetime table column

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Paul Barrette
Paul Barrette el 17 de Sept. de 2023
Comentada: Star Strider el 20 de Sept. de 2023
I have a datetime table with
'01-Oct-2014'
'02-Oct-2014'
'03-Oct-2014'
'04-Oct-2014'
'05-Oct-2014'
and I would like to get the same table (or an additional column in that table) but with day-month only for each row
  7 comentarios
Paul Barrette
Paul Barrette el 18 de Sept. de 2023
Movida: Dyuman Joshi el 20 de Sept. de 2023
Sorry, I should have provided more info. I am plotting (with the 'scatter' command) five traces on the same graph, one per year (e.g., 2014 to 2018 inclusive). The horizontal axis for each year ranges from 01-Oct to 01-June (values for the vertical axis are from another matrix). But because the year is part of the object, these traces plot one after the other along that axis, not one on top of each other, which is what I'm looking for. To answer your question, I would store them as 'day-month' if only I could extract this from the current 'day-month-year' data.
ymd, as suggested earlier by @Star Strider, turns the datetime values into numeric arrays, but I'd like the 'day-month' label to appear on the horizontal axis of the graph.
Dyuman Joshi
Dyuman Joshi el 18 de Sept. de 2023
Movida: Dyuman Joshi el 20 de Sept. de 2023
How about this?
%Data
y = datetime(["01-Oct-2014";
"02-Oct-2014"
"03-Oct-2014"
"04-Oct-2014"
"05-Oct-2014"])
y = 5×1 datetime array
01-Oct-2014 02-Oct-2014 03-Oct-2014 04-Oct-2014 05-Oct-2014
%Get the day
d = day(y)
d = 5×1
1 2 3 4 5
%Get the month, with the name in short form
m = month(y,'shortname')
m = 5×1 cell array
{'Oct'} {'Oct'} {'Oct'} {'Oct'} {'Oct'}
%Single Quotation marks gives cell array
%Double Quotation marks gives string array
z = compose("%d-%s", d, string(m))
z = 5×1 string array
"1-Oct" "2-Oct" "3-Oct" "4-Oct" "5-Oct"

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Star Strider
Star Strider el 18 de Sept. de 2023
Editada: Star Strider el 18 de Sept. de 2023
The ymd function may be worth exploring.
EDIT — (18 Sep 2023 at 23:08)
Here is an example that also includes a scatter plot —
DT = datetime(['01-Oct-2014'
'02-Oct-2014'
'03-Oct-2014'
'04-Oct-2014'
'05-Oct-2018'
'01-Oct-2018'
'02-Oct-2018'
'03-Oct-2018'
'04-Oct-2018'
'05-Oct-2018']);
T1 = table(DT,randn(size(DT))+[zeros(5,1);4*ones(5,1)])
T1 = 10×2 table
DT Var2 ___________ _________ 01-Oct-2014 -0.082131 02-Oct-2014 0.47899 03-Oct-2014 0.25127 04-Oct-2014 0.99845 05-Oct-2018 0.39241 01-Oct-2018 4.2423 02-Oct-2018 5.3109 03-Oct-2018 4.1244 04-Oct-2018 3.8779 05-Oct-2018 4.609
[y,m,d] = ymd(DT); % Date Components
ix = findgroups(m, d); % Indices Of Months & Days
scattery = accumarray(ix, T1{:,2}, [], @(x){x}); % Accumulate 'Var2' Values By Day & Month
mc = month(DT,'shortname');
% compose('%s-%02d',string(mc),day(DT))
x = unique(compose('%s-%02d',string(mc),day(DT)),'stable');
dtx = datetime(x,'InputFormat','MMM-dd');
y = cell2mat(scattery.'); % Convert Cell Array To Numeric Array & Transpose
figure
scatter(dtx, y(1,:), 's', 'filled', 'DisplayName','2014')
hold on
scatter(dtx, y(2,:), 's', 'filled', 'DisplayName','2018')
hold off
grid
legend('Location','best')
I am not certain what you want to do.
.
  2 comentarios
Paul Barrette
Paul Barrette el 20 de Sept. de 2023
This works - thanks @Star Strider. Though this is intricate coding for me (an opportunity the learn!). I did try the solution offered by @Dyuman Joshi, much simpler, but I could not make it work.
Star Strider
Star Strider el 20 de Sept. de 2023
As always, my pleasure!

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