How to remove the year from a datetime table column

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 Comments

Simply change the format of the data -
%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
%Change the format of the datetime array
y.Format = 'dd-MMM';
y
y = 5×1 datetime array
01-Oct 02-Oct 03-Oct 04-Oct 05-Oct
Thanks @Dyuman Joshi. However, this code simply changes the formatting, but the underlying year data is still stored in the DATETIME object. What I'm trying to do is extract the day-month from each value.
Well, then you will have to get the day and the month data separately and store in a format you want to, as the datetime data type requires year as an input.
So, the question is how do you want store them?
The day and month functions exist.
Yes, I was going to provide that in my comment above, but thought that it would be better to answer after OP provides details on what the expected outcome is.
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.
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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 Accepted Answer

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 Comments

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.

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