How do I separate date and time of a raw date?

>> a = untitled.startDate(1:10)

a = 

    10×1 categorical array

       2013-07-10 00:01:00.0 
       2013-07-17 00:01:00.0 
       2013-06-29 00:01:00.0 
       2013-07-17 00:01:00.0 
       2013-05-11 00:01:00.0 
       2013-05-31 00:01:00.0 
       2013-06-06 00:01:00.0 
       2013-08-31 00:01:00.0 
       2013-05-21 00:01:00.0 
       2013-08-06 00:01:00.0 

>>  a.Format = 'dd-MMM-yyyy'
Error using categorical/subsasgn (line 87)
Attempt to assign field of non-structure array.

[Walter Roberson]

What data was used to create the categorical array? Was categorical() applied to datetime() values, or was it applied to a cell array of character vectors?

[aishah]

What I did was, I imported the data into a 'table'. And I convert the data to 'Categorical Array'.

[aishah]

The imported data type is 'datetime'.

Answer 1

[Ahmet Cecen]

arrayfun(@(x) strsplit(x,' '),string(a),'UniformOutput',false)

string(a) - this makes the data easily process-able, converts to a string array.

@(x) strsplit(x,' ') - this is a function that splits each string into parts whenever a space(' ') is located(which works for your case)

arrayfun - applies the above function to each "row" in the string array.

[aishah]

>> arrayfun(@(x) strsplit(x,' '),string(a),'UniformOutput',false)

ans =

10×1 cell array

    [1×2 string]
    [1×2 string]
    [1×2 string]
    [1×2 string]
    [1×2 string]
    [1×2 string]
    [1×2 string]
    [1×2 string]
    [1×2 string]
    [1×2 string]

[Ahmet Cecen]

Yep, dates will be the first, time will be the second element.

ans{1}{1} %dont use ans though, assign something to it.

[aishah]

>> b = ans{1}{1}

b =

    '2013-07-10'

[aishah]

sorry, how do I list out all 10 row with the separated elements?

[Ahmet Cecen]

Ah, here:

b = reshape([k{:}],2,length(a))'

Now:

b(:,1)
b(:,2)

[aishah]

b = reshape([k{:}],2,length(a))'

Undefined variable "k" or class "k".

[aishah]

celldisp(b)

b{1} =

    "2013-07-10"    "00:01:00.0"

b{2} =

    "2013-07-17"    "00:01:00.0"

b{3} =

    "2013-06-29"    "00:01:00.0"

b{4} =

    "2013-07-17"    "00:01:00.0"

b{5} =

    "2013-05-11"    "00:01:00.0"

b{6} =

    "2013-05-31"    "00:01:00.0"

b{7} =

    "2013-06-06"    "00:01:00.0"

b{8} =

    "2013-08-31"    "00:01:00.0"

b{9} =

    "2013-05-21"    "00:01:00.0"

b{10} =

    "2013-08-06"    "00:01:00.0"

Answer 2

[Peter Perkins]

Don't do any of that.

You've read the timestamps in as datetimes. Don't fight that. Assuming you have a table T:

T.StartTime = timeofday(T.StartDate);
T.StartDate = dateshift(T.StartDate,'start','day);

[Steph]

Hi, I am excatly trying to seperate datetimes into day and time columns seperately using matlab 2017b. Peter, unfortunately your code says unrecognized property:StartDate

[Walter Roberson]

The original poster of the question had a table or struct field named StartDate .

If you have a table which has a variable which is in datetime format then in the code that Peter showed, replace the T.StartTime with the name of your table variables

T.StartTime = timeofday(T.NameOfYourDatetimeColumnGoesHere);
T.NameOfYourDatetimeColumnGoesHere = dateshift(T.NameOfYourDatetimeColumnGoesHere, 'start', 'day');

[Steph]

Dear Walter' Thank you very much for taking time to write to me.

This works perfectly.

I am trying to change this table to a double array so I can perform some regressions on it. Hence I used table2 array, now it says it can not concatenate the specified table variables, all inputs must be date/time character vectors or date/time strings.

I think after this conversion, matlab still treats these date and time columns as datetime formatted.

How can I get rid of this problem and convert the table to a double array without losing the date and time information? Do you have any idea?

I need to keep date and time information as they will be valuable for further analysis.

[Walter Roberson]

Does it make sense to perform regression on a time of day? Do you not worry about differences in results if the time is represented as 100^2 * hour + 100 * minutes + seconds, compared to 60^2 * hour + 60 * minutes + seconds ? Do you not worry about differences in results if you represent date as 100^2 * month_number + 100 * day_number + two_digit_year, compared to 100^2 * two_digit_year + 100 * month_number + day_number, compared to year plus decimal fraction of a fixed-length year, compared to year plus decimal fraction of year taking into account leap year and leap seconds ?

I think you should reconsider doing regression on dates or times of day.

Perhaps it would make sense to instead calculate a duration in seconds relative to the earliest datetime ? Perhaps it would make sense to table2array(TheTable(:,3:end)) so you are not extracting the date or time information ?

[Steph]

I missteated my problem I think, I am doing regressions on the data that is collected on each of the specific days. Not on the dates or days themselves

[Walter Roberson]

Then you do not need to convert the dates + times or datetime objects to numeric. What you need to do is be able to group your data based upon date. You should have a look at splitapply() -- or convert the table to a timetable() object and use retime()