Time vector in curve fitting toolbox

Question

I want to use the curve fitting toolbox to model my day which consists of two vectors, time and measurement. Here is a screenshot of my data file (I will use time and DOWN only).

How do I feed the time vector as my x vector in the toolbox? It's currently datetime type vector and I can't find it in my list when I select my variables, what is the right format to use?

Tommaso Belluzzo Tommaso Belluzzo · Accepted Answer · 2018-02-17T12:48:14

Here is my excel sample (I reduced the amount of observations a little bit, but this won't affect the final outcome):

DATE        TIME        DOWN
---------   ----------  -----
1/27/2018   8:00:03 PM  5,375
1/27/2018   8:00:06 PM  5,375
1/27/2018   8:00:10 PM  5,375
1/27/2018   8:00:13 PM  5,375
1/27/2018   8:00:17 PM  5,375
1/27/2018   8:00:20 PM  4,855
1/27/2018   8:00:27 PM  4,855
1/27/2018   8:00:31 PM  4,855
1/27/2018   8:00:35 PM  4,855
1/27/2018   8:00:38 PM  4,855

I usually prefer to use the readtable function together with tweaked import options as follows:

opts = detectImportOptions('data.xlsx');
opts.VariableTypes = {'datetime' 'datetime' 'double'};
data = readtable('data.xlsx',opts);

The data variable will then be parsed as follows:

DATE (datetime)         TIME (datetime)         DOWN (double)
----------------------  ----------------------  ----------------
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:03'  5,37500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:06'  5,37500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:10'  5,37500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:13'  5,37500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:17'  5,37500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:20'  4,85500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:27'  4,85500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:31'  4,85500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:35'  4,85500000000000
'27-Jan-2018 00:00:00'  '31-Dec-1899 20:00:38'  4,85500000000000

As you can see, when the time is not defined it is defaultized to 00:00:00; the same goes for the date, which is defaultized to 31-Dec-1899. Since the reference date is static and never changes, in my opinion the best thing to do is to merge the first two columns of your table in order to produce coherent datenum values to use as X Data (datetime variables cannot be used in the Curve Fitting Toolbox, you have to convert them to serial format):

[y,m,d] = ymd(data.DATE);
[h,mn,s] = hms(data.TIME);
x = datenum(y,m,d,h,mn,s)

x =
    737087.833368056
    737087.833402778
    737087.833449074
    737087.833483796
    737087.833530093
    737087.833564815
    737087.833645833
    737087.83369213
    737087.833738426
    737087.833773148

The values to be used as Y Data in the cftool can instead be extracted directly from the data column DOWN:

y = data.DOWN;

Time vector in curve fitting toolbox

1 Answers