Subscribing into a timetable using only the time values
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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!
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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')
AL(S,1:3) = {"Ja"}
Experiment to get different results.
.
댓글 수: 5
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
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"}
That works, and all the date and time information are preserved!
.
Noush
2021년 12월 4일
Star Strider
2021년 12월 4일
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
2021년 12월 4일
And,
If my Answer helped you solve your problem, please Accept it!
.
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