Finding the difference in timeseries values

If I have the below timetable, how do I find the difference between values? Diff() doesn't work. I've tried diff(TableName.Variables), however that removes the Time column and VariableNames.

            Time            Humidity    TemperatureF    PressureHg
    ____________________    ________    ____________    __________

    15-Nov-2015 00:00:00       48.9         51.45          29.61  
    15-Nov-2015 06:00:00       48.9         51.45           29.6  
    15-Nov-2015 12:00:00     49.025         51.45          29.61  
    15-Nov-2015 18:00:00       48.9        51.225         29.607  
    16-Nov-2015 00:00:00       48.5          51.4          29.61  

Answer 1

[Star Strider]

This is kludgy, however I can find no other way to do what you want:

C = {'15-Nov-2015 00:00:00'       48.9         51.45          29.61  
     '15-Nov-2015 06:00:00'       48.9         51.45           29.6  
     '15-Nov-2015 12:00:00'     49.025         51.45          29.61  
     '15-Nov-2015 18:00:00'       48.9        51.225         29.607  
     '16-Nov-2015 00:00:00'       48.5          51.4          29.61}; 

T1 = cell2table(C, 'VariableNames',{'Time','Humidity','TemperatureF','PressureHg'})                                     % Create Table From Data
T1.Time = datetime(T1{:,1}, 'InputFormat','dd-MMM-yyyy HH:mm:ss');                                                      % Create Table From Data

DiffArray = diff(T1{:,2:4})                                                                                             % Column Differences
DT = table(datetime(diff(datenum(T1{:,1})),'ConvertFrom','datenum','Format','HH:mm'),'VariableNames',{'DiffTime'});     % Time Differences
T2 = [DT array2table(DiffArray, 'VariableNames',{'DiffHumidity','DiffTemperatureF','DiffPressureHg'})]  

producing:

T2 =

  4×4 table

    DiffTime    DiffHumidity    DiffTemperatureF    DiffPressureHg
    ________    ____________    ________________    ______________

     06:00              0                 0              -0.01    
     06:00          0.125                 0               0.01    
     06:00         -0.125            -0.225             -0.003    
     06:00           -0.4             0.175              0.003   

.

Answer 2

[Peter Perkins]

David, you are correct that diff(TableName.Variables) removes the times and the variable names, in fact, it returns a numeric matrix not a timetable!

Applying diff gives you one fewer rows, so one way or another you need to create a new timetable. If you are only diff'ing one variable, then

ttDiff = timetable(diff(tt.X),'RowTimes',tt.Time(2:end),'VariableNames',{'X_diff'})

is one way to go. For three variables (maybe even for just one in this case), varfun is probably the best:

>> tt
tt =
  5×3 timetable
            Time            Humidity    TemperatureF    PressureHg
    ____________________    ________    ____________    __________
    15-Nov-2015 00:00:00       48.9         51.45          29.61  
    15-Nov-2015 06:00:00       48.9         51.45           29.6  
    15-Nov-2015 12:00:00     49.025         51.45          29.61  
    15-Nov-2015 18:00:00       48.9        51.225         29.607  
    16-Nov-2015 00:00:00       48.5          51.4          29.61  
>> ttDiff = varfun(@diff,tt)
ttDiff =
  4×3 timetable
            Time            diff_Humidity    diff_TemperatureF    diff_PressureHg
    ____________________    _____________    _________________    _______________
    15-Nov-2015 00:00:00            0                  0               -0.01     
    15-Nov-2015 06:00:00        0.125                  0                0.01     
    15-Nov-2015 12:00:00       -0.125             -0.225              -0.003     
    15-Nov-2015 18:00:00         -0.4              0.175               0.003     

There's also a long example in the doc that talks about "how to perform calculations by using the numeric and categorical data that the table contains" at length:

https://www.mathworks.com/help/matlab/matlab_prog/computations-with-numeric-data-in-table-or-timetable.html

Answer 3

[Siddharth Bhutiya]

If you are using R2023a, then you can now directly call diff on a timetable. Here is an example,

>> tt

tt =

  5x3 timetable

            Time            Humidity    TemperatureF    PressureHg
    ____________________    ________    ____________    __________

    15-Nov-2015 00:00:00       48.9         51.45          29.61  
    15-Nov-2015 06:00:00       48.9         51.45           29.6  
    15-Nov-2015 12:00:00     49.025         51.45          29.61  
    15-Nov-2015 18:00:00       48.9        51.225         29.607  
    16-Nov-2015 00:00:00       48.5          51.4          29.61  

>> diff(tt)

ans =

  4x3 timetable

            Time            Humidity    TemperatureF    PressureHg
    ____________________    ________    ____________    __________

    15-Nov-2015 00:00:00          0             0          -0.01  
    15-Nov-2015 06:00:00      0.125             0           0.01  
    15-Nov-2015 12:00:00     -0.125        -0.225         -0.003  
    15-Nov-2015 18:00:00       -0.4         0.175          0.003  

See this documentation page for other arithmetic operations that are now allowed on tables and timetables.