How can I calculate the mean and std with the retime function?

My data is in a "timetable" format of one year of measurements and its frequency is 1 minute. I want to calculate the daily average for a year accompanied by the std. For this I use the function retime with the periodicity that I want and indicating that I want the average. In the help it says that you can add a function to calculate this std. Can you help me?

days_cloud_indices = retime(yaer_cloud_indices,"daily","mean")

Thanks in advance

[Star Strider]

The name-value pair agruments likely need to be in single quotes. You would need to do separate calculations for the mean and std:

days_cloud_indices_mean = retime(yaer_cloud_indices,'daily','mean')
days_cloud_indices_std = retime(yaer_cloud_indices,'daily',@std)

I did not test this.

[Peter Perkins]

Yes, that's the way to get arbitrary daily summaries.

>> tt = timetable(rand(10,1),'StartTime',datetime(2020,4,14),'TimeStep',hours(9))
tt =
  10×1 timetable
            Time             Var1  
    ____________________    _______
    14-Apr-2020 00:00:00    0.25108
    14-Apr-2020 09:00:00    0.61604
    14-Apr-2020 18:00:00    0.47329
    15-Apr-2020 03:00:00    0.35166
    15-Apr-2020 12:00:00    0.83083
    15-Apr-2020 21:00:00    0.58526
    16-Apr-2020 06:00:00    0.54972
    16-Apr-2020 15:00:00    0.91719
    17-Apr-2020 00:00:00    0.28584
    17-Apr-2020 09:00:00     0.7572
>> ttMeans = retime(tt,'daily','mean')
ttMeans =
  4×1 timetable
       Time         Var1  
    ___________    _______
    14-Apr-2020    0.44681
    15-Apr-2020    0.58925
    16-Apr-2020    0.73346
    17-Apr-2020    0.52152
>> ttStds = retime(tt,'daily',@std)
ttStds =
  4×1 timetable
       Time         Var1  
    ___________    _______
    14-Apr-2020    0.18392
    15-Apr-2020    0.23961
    16-Apr-2020    0.25984
    17-Apr-2020     0.3333
    

Double quotes are fine for retime's parameters, post R2018b-ish.

[Aries Dwi Siswanto]

I have tried

ttStds = retime(tt,'daily',@std)

but some cases did not work.

Here the error messages:

Error using timetable/retime (line 140)

Unable to concatenate incompatible values returned by the function 'std' when applied to groups in the variable

'Var1':

Dimensions of arrays being concatenated are not consistent.

How to solve this problem?

Thank you for everyone who help to solve this problem.

[Walter Roberson]

Experiment with

ttStds = retime(tt,'daily',@(v)std(v,[],1))

When there are multiple variables being processed but a group turns out to only have one entry, then mean() or std() can end up taking the mean or std of the variables, because by default they work along the first non-singular dimension.

[Peter Perkins]

Yes, what Walter says is mostl likely what's happening. retime allows you to say 'mean' instead of @mean, along with a few other canned aggregations, and those automatically do what Walter suggests, but std is not on that list.

[Aries Dwi Siswanto]

Hi Walter Roberson,

Thank you very much for your script and also the explanation related to the problem and how to solve.

I have tried and well work. Great.