Arrayfun with a timetable

1 visualización (últimos 30 días)
Allie
Allie el 10 de Nov. de 2020
Respondida: Peter Perkins el 19 de Nov. de 2020
I need to extract items from a timetable (timetable1) matching specific dates (time2).
timetable1=[
'01-Jan-1983' -2.374
'01-Feb-1983' -2.290
'01-Mar-1983' -1.723
'01-Apr-1983' -2.117
'01-May-1983' -2.775
'01-Jun-1983' -2.913
'01-Jul-1983' -2.492
'01-Aug-1983' -2.256
'01-Sep-1983' -1.739
'01-Oct-1983' -1.108
'01-Nov-1983' -0.675
'01-Dec-1983' -0.602]
time2=[
'01-Jan-1983'
'01-Feb-1983'
'01-Mar-1983'
'01-Apr-1983'
'01-May-1983'
'01-Jun-1983'
'01-Jul-1983']
for j=1:12
timetable2(j,:)=arrayfun(@(x) timetable1.Var1(timetable1.time==x,j),time2,'UniformOutput','false')
end
Normally I would use the arrayfun code above for this type of function but when I try it with timetable data, it produces this error:
Error using arrayfun
All of the input arguments must be of the same size and shape.
Previous inputs had size 81 in dimension 1. Input #3 has size 1
Is there a way to extract timetable data in this manner?

Respuesta aceptada

Peter Perkins
Peter Perkins el 19 de Nov. de 2020
timetable1 is not a timetable, and probably not even legal MATLAB code. I'm going to assume you mean that you have an actual timetable with datetime row times and one numeric variable. And a list of datetimes at which you want to select rows of your timetable.
Just subscript the timetable with the datetimes. That's it.
>> tt = timetable([1;2;3;4;5],'RowTimes',datetime(2020,11,18:22))
tt =
5×1 timetable
Time Var1
___________ ____
18-Nov-2020 1
19-Nov-2020 2
20-Nov-2020 3
21-Nov-2020 4
22-Nov-2020 5
>> tt(datetime(2020,11,[18 20 22]),:)
ans =
3×1 timetable
Time Var1
___________ ____
18-Nov-2020 1
20-Nov-2020 3
22-Nov-2020 5
You should also take a look at timerange, which is a way to create a subscript that will select rows in a range rather tha exact matches.

Más respuestas (0)

Categorías

Más información sobre Tables en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by