Subscribing into a timetable using only the time values

Hello,

I have created a timetable using the following code:

sz = [120 4]; %Dimensionen of the Table
varTypes = {'string','string' 'string','double'}; %Types of Variables
varNames = {'Radlader','Bagger','Gabelstapler','Lesitung Gesamt'}; %Headings of the table

AL = timetable('Size',sz,'VariableTypes',varTypes,'RowTimes',dt,'VariableNames',varNames); %The table with Aggregatslasten
S = timerange('08:30:00','16:00:00')
AL(S,1:3) = {"Ja"}

I would like the word "Ja" to appear in the first to third column, when the time is between 8:30 and 16:00.

With the code at hand, I get the error message: "A timetable with datetime row times cannot be subscripted using duration values."

How can I solve this?

Note that my timetable is not just for a single day, but a whole week. So I have to keep the dates as well! Specifying the dates does not really work, as the dates are chosen by the user, so they can change, I need a way for the table to only look at the time values.

Thank you for your help!

Answer 1

[Star Strider]

Try this —

sz = [120 4]; %Dimensionen of the Table
varTypes = {'string','string' 'string','double'}; %Types of Variables
varNames = {'Radlader','Bagger','Gabelstapler','Lesitung Gesamt'}; %Headings of the table
dt = datetime('now')+hours(0:sz(1)-1)*4;                                                    % Create 'datetime' Array
dt = duration(hour(dt),minute(dt),second(dt));
AL = timetable('Size',sz,'VariableTypes',varTypes,'RowTimes',dt,'VariableNames',varNames); %The table with Aggregatslasten
S = timerange('08:30:00','16:00:00')

S = 
	timetable timerange subscript:

		Select timetable rows with times in the half-open interval:
		[08:30:00, 16:00:00)

	See Select Timetable Data by Row Time and Variable Type.

AL(S,1:3) = {"Ja"}

AL = 120×4 timetable
      Time      Radlader      Bagger      Gabelstapler    Lesitung Gesamt
    ________    _________    _________    ____________    _______________

    14:41:31    "Ja"         "Ja"          "Ja"                  0       
    18:41:31    <missing>    <missing>     <missing>             0       
    22:41:31    <missing>    <missing>     <missing>             0       
    02:41:31    <missing>    <missing>     <missing>             0       
    06:41:31    <missing>    <missing>     <missing>             0       
    10:41:31    "Ja"         "Ja"          "Ja"                  0       
    14:41:31    "Ja"         "Ja"          "Ja"                  0       
    18:41:31    <missing>    <missing>     <missing>             0       
    22:41:31    <missing>    <missing>     <missing>             0       
    02:41:31    <missing>    <missing>     <missing>             0       
    06:41:31    <missing>    <missing>     <missing>             0       
    10:41:31    "Ja"         "Ja"          "Ja"                  0       
    14:41:31    "Ja"         "Ja"          "Ja"                  0       
    18:41:31    <missing>    <missing>     <missing>             0       
    22:41:31    <missing>    <missing>     <missing>             0       
    02:41:31    <missing>    <missing>     <missing>             0       

Experiment to get different results.

.

[Noush]

Thank you, this parshly works!

However in your solution, the dates disappear from the time column. I still need the dates to show up. I tried adding another column by using a datetime I have defined before as "Datum" like so: AL(:,1)= {Datum}

however that somehow doesnt work. It gives me the error:

"Right hand side of an assignment must be a datetime array or text representing dates and times."

I don't get that? Datum is a datetime, so what is that error supposed to mean?

[Star Strider]

My pleasure.

‘I still need the dates to show up.’

That was not obvious. It also may not be possible with the original approach, since apparently calendarDuration (that will display those times) does not work with the rest of it. The error is that it is not the same as a duration vector in this instance, so a duration array approach will fail.

So, just use regular MATLAB ‘logical indexing’ —

sz = [120 4]; %Dimensionen of the Table
varTypes = {'string','string' 'string','double'}; %Types of Variables
varNames = {'Radlader','Bagger','Gabelstapler','Lesitung Gesamt'}; %Headings of the table
dt = datetime('now')+hours(0:sz(1)-1).'*4;                                                    % Create 'datetime' Array
AL = timetable('Size',sz,'VariableTypes',varTypes,'RowTimes',dt,'VariableNames',varNames) %The table with Aggregatslasten

AL = 120×4 timetable
            Time            Radlader      Bagger      Gabelstapler    Lesitung Gesamt
    ____________________    _________    _________    ____________    _______________

    03-Dec-2021 12:09:14    <missing>    <missing>     <missing>             0       
    03-Dec-2021 16:09:14    <missing>    <missing>     <missing>             0       
    03-Dec-2021 20:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 00:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 04:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 08:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 12:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 16:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 20:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 00:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 04:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 08:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 12:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 16:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 20:09:14    <missing>    <missing>     <missing>             0       
    06-Dec-2021 00:09:14    <missing>    <missing>     <missing>             0       

dtn = rem(datenum(AL.Time),1);                                              % Day Fractions
t1 = rem(datenum('08:30:00', 'HH:mm:ss'),1);                                % Start Time
t2 = rem(datenum('16:00:00', 'HH:mm:ss'),1);                                % End Time
S = dtn>=t1 & dtn<t2;                                                       % Logical Vector
AL(S,1:3) = {"Ja"}

AL = 120×4 timetable
            Time            Radlader      Bagger      Gabelstapler    Lesitung Gesamt
    ____________________    _________    _________    ____________    _______________

    03-Dec-2021 12:09:14    "Ja"         "Ja"          "Ja"                  0       
    03-Dec-2021 16:09:14    <missing>    <missing>     <missing>             0       
    03-Dec-2021 20:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 00:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 04:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 08:09:14    "Ja"         "Ja"          "Ja"                  0       
    04-Dec-2021 12:09:14    "Ja"         "Ja"          "Ja"                  0       
    04-Dec-2021 16:09:14    <missing>    <missing>     <missing>             0       
    04-Dec-2021 20:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 00:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 04:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 08:09:14    "Ja"         "Ja"          "Ja"                  0       
    05-Dec-2021 12:09:14    "Ja"         "Ja"          "Ja"                  0       
    05-Dec-2021 16:09:14    <missing>    <missing>     <missing>             0       
    05-Dec-2021 20:09:14    <missing>    <missing>     <missing>             0       
    06-Dec-2021 00:09:14    <missing>    <missing>     <missing>             0       

That works, and all the date and time information are preserved!

.

[Noush]

Thank you! That worked :)

Can you explain what this does exactly though? I didn't really understand your comments:

dtn = rem(datenum(AL.Time),1);

t1 = rem(datenum('08:30:00', 'HH:mm:ss'),1);

t2 = rem(datenum('16:00:00', 'HH:mm:ss'),1);

[Star Strider]

My pleasure!

Sure!

The datenum function (that has been available for several decades) produces a decimal fraction depicting days to the left of the decimal separator and fractions of days to the right of the decimal separator. Using the rem or mod functions (similar in most respects, although not the same in all) returns the remainder-after-division. So dividing by 1 returns the decimal part (fractions-of-a-day) as the output. (Experiment with that with any decimal fraction to understand how it works.) This makes it easy to compare times-of-day, and using only the day fraction makes the days themselves irrelevant so that it works across all days.

In detail, then:

dtn = rem(datenum(AL.Time),1);                  % Day Fractions Of All Days In The 'Time' Variable
t1 = rem(datenum('08:30:00', 'HH:mm:ss'),1);    % Day Fraction Equivalent For '08:30:00'
t2 = rem(datenum('16:00:00', 'HH:mm:ss'),1);    % Day Fraction Equivalent For '16:00:00'

Then the ‘S’ assignment uses these to create a logical vector to produce the desired end result.

.

[Star Strider]

And,

If my Answer helped you solve your problem, please Accept it!

.