# Compact way to extract day and hour from a datetime / datestrings format ?

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Sim on 27 Jun 2022
Commented: Voss on 28 Jun 2022
I have a cell array including datetime / datestrings data, i.e. days and hours.
Here following, I show two examples related to the same day, and 24 hours far from each other:
>> time_array(1)
ans =
1×1 cell array
{'17-Jun-2017 0:00:00'}
>> time_array(24)
ans =
1×1 cell array
{'17-Jun-2017 23:00:00'}
Is there any compact way to extract day number (starting from the first day in my dataset) and hour just given the row / index of my dataset, called time_array?
For example, if I give the row / index of my datset, let's say 200:
>> time_array(200)
ans =
1×1 cell array
{'25-Jun-2017 7:00:00'}
is there any compact way to get this output ?
day = 9
hour = 7
Indeed, the 25th June is the day number 9 if we start to count from the day number 1, which is the 17th June.
##### 2 CommentsShowHide 1 older comment
Sim on 28 Jun 2022
Edited: Sim on 28 Jun 2022
Thanks a lot @KSSV, very kind!
I have realised that my question was not clear and I am really sorry!
t.Day
gives me the day that I select from
t
For example:
>> t.Day(200)
ans =
25
gives me "25" because "t(200)" corresponds to '25-Jun-2017 07:00:00',
>> t(200)
ans =
datetime
25-Jun-2017 07:00:00
However, in my question description, I asked something slightly different.. Probably, I should rephrase the question...?
What I wanted to know was the day's number corresponding to "t(200)", starting from the first day in my dataset "t(1)"
>> t(1)
ans =
datetime
17-Jun-2017
In this example, the 25th of June would be the 9th day, if we start counting from the first day of my dataset, i.e. the 17th of June (the 17th of June would be the day number 1, the 18th of June would be the day number 2, and so on...).
In this regard, @Voss provided one possible answer / solution, i.e.
>> floor(days(t(200)-t(1))) + 1
ans =
9

Voss on 27 Jun 2022
Edited: Voss on 27 Jun 2022
time_array = { ...
'17-Jun-2017 0:00:00'; ...
'17-Jun-2017 1:00:00'; ...
'17-Jun-2017 2:00:00'; ...
...
'17-Jun-2017 23:00:00'; ...
'18-Jun-2017 0:00:00'; ...
...
'25-Jun-2017 7:00:00'; ...
...
'31-Jul-2017 13:00:00'; ...
};
t = datetime(time_array)
t = 7×1 datetime array
17-Jun-2017 00:00:00 17-Jun-2017 01:00:00 17-Jun-2017 02:00:00 17-Jun-2017 23:00:00 18-Jun-2017 00:00:00 25-Jun-2017 07:00:00 31-Jul-2017 13:00:00
disp(t(end-1));
25-Jun-2017 07:00:00
floor(days(t(end-1)-t(1))) + 1
ans = 9
t(end-1).Hour
ans = 7
disp(t(end));
31-Jul-2017 13:00:00
floor(days(t(end)-t(1))) + 1
ans = 45
t(end).Hour
ans = 13

Sim on 28 Jun 2022
Edited: Sim on 28 Jun 2022
Thanks a lot @Voss, your code does what I need... In my case, I did in this way and it works:
t = datetime(time_array); % I found out that "time_array" was not in a "datetime" format
index_dataset = 200; % just an example
number_of_days_from_the_first_one = floor(days(t(index_dataset)-t(1))) + 1;
hour_corresponding_to_index_dataset = t(index_dataset).Hour;
Voss on 28 Jun 2022
You're welcome!