Comments on: Using MATLAB to Grade

By: Peter Perkins

Peter Perkins — Thu, 01 Dec 2011 18:59:29 +0000

Bob, the Statistics Toolbox User Guide does have a section on dataset arrays if you haven’t looked there:

http://www.mathworks.com/help/toolbox/stats/bqziht7-1.html#bqzihxq

To answer your specific question, you can add two variables in a dataset array by just adding them to create a new variable:

d = dataset(randn(10,1),randn(10,1));
d.Var3 = d.Var1 + d.Var2;

Hope this helps.

By: Loren Shure

Loren Shure — Thu, 01 Dec 2011 18:23:36 +0000

Bob-

I am not aware of other documentation. There are some other posts in this overall blog that use dataset arrays as well. Perhaps those would be useful to you.

–Loren

By: Bob

Bob — Thu, 01 Dec 2011 16:19:32 +0000

Where can I find good reference for using dataset arrays beyond the Matlab help?

Specifically, how can I add two variables in a dataset?

By: Loren

Loren — Tue, 05 Jul 2011 12:55:37 +0000

Patrick-

Since RT-DA is not a valid identifier name in MATLAB, so the name gets replaced with one that can be used in MATLAB.

–Loren

By: Patrick de Man

Patrick de Man — Thu, 23 Jun 2011 12:14:51 +0000

What if I want to use as a variable name ‘RT-DA’?
The dataset heading becomes: RT0x2FDA, as the – seems to be an invalid character for the dataset.

By: Peter Perkins

Peter Perkins — Fri, 18 Mar 2011 13:15:17 +0000

Roman, it looks like you've got the hang of using nominal/ordinal variables in a dataset for creating subsets of your data. The only problem I see in your code is this:

~data.MLIndustryLvl2 == ‘Financial’

Consider the precedence of the operators "~" and "==". data.MLIndustryLvl2 is not a logical, so you want either

~(data.MLIndustryLvl2 == ‘Financial’)

data.MLIndustryLvl2 ~= ‘Financial’

By: Roman

Roman — Fri, 18 Mar 2011 09:49:27 +0000

May I further ask:

% Extract all bonds with currency EUR and rating between BBB3 and A1 and not being Financials:
data( data.Rating >= ‘BBB3′ & data.Rating <=’A1′ & ~data.MLIndustryLvl2 == ‘Financial’ …
& data.ISOCurrency ==’EUR’, {‘Description’,'Ticker’,'MLIndustryLvl3′,’YldToWorst’,'OAS’, ‘ModDurToWorst’,'Rating’})
Matlab error message:
??? Undefined function or method ‘not’ for input arguments of type ‘nominal’.

How can I efficiently code and then screen my database for all bonds but the ‘financial’ ones (in MLIndustryLvl2 I have 7 different bond types, in ML IndustryLvl3 it’s 31 and Lvl4 is 88)

By: Roman

Roman — Fri, 18 Mar 2011 08:23:03 +0000

Thanks, Peter

Yes, I do use and need dataset, because my data is 14514 rows and 36 columns with different types of bond variables:
- many currencies
- Price, Yield, Duration, ISIN number information of different bonds
- Ratings, Sectors, Names, Maturities
I try to replicate the webinar about dataset arrays. Goal is: extract out of large universe subsamples of bonds with predefined characteristics such as certain ratings, sectors, within certain time-to-maturity brackets. See following code:

data( data.Rating >= ‘AA3′, {‘Ticker’, ‘YldToWorst’, ‘Rating’})

data( data.Rating >= ‘BBB3′ & data.Rating <=’A1′ & data.MLIndustryLvl2 == ‘Industrials’ …
& data.MLIndustryLvl3 == ‘Healthcare’ & data.ISOCurrency ==’USD’ & data.MaturityDate <= datenum(‘Dec-01-2014′), …
{‘Ticker’, ‘YldToWorst’, ‘ModDurToWorst’,'Rating’})

Kind regards Roman

By: Peter Perkins

Peter Perkins — Fri, 18 Mar 2011 01:32:53 +0000

Roman, I'm not exactly sure what's what in your code snippets, so I'll have to take a guess. data looks like it might be a dataset array, with MaturityDate as one of its variables, and MaturityDate looks like it might be a cell array of datestrs where some are like '01.02.2003' and some are like '02-Mar-2004'. It appears that you've applied arrayfun to that cell array to get all "/"'s, and reassign in place. But I'm not sure what you're up to with datasetfun. It is intended to apply the same function to several variables in a dataset array, and since you only seem to be working on one variable, there'd be no point in using it. If all you want to do is to convert MaturityDate from datestr to datenum, you could use the same arrayfun strategy. But the following is a vectorized version:

MaturityDate = {'01.02.2003'; '02-Mar-2004'; '03.04.2005'; '04-May-2006'; '05.06.2007'};
data = dataset(MaturityDate)

MaturityDateDN = zeros(length(data.MaturityDate),1);
fmt10 = cellfun(@(x)length(x)==10,data.MaturityDate);
MaturityDateDN(fmt10) = datenum(data.MaturityDate(fmt10),'dd.mm.yyyy');
fmt11 = ~fmt10;
MaturityDateDN(fmt11) = datenum(data.MaturityDate(fmt11),'dd-mmm-yyyy');

data.MaturityDate = MaturityDateDN

When I run this, data starts out as data = MaturityDate '01.02.2003' '02-Mar-2004' '03.04.2005' '04-May-2006' '05.06.2007' and ends up as data = MaturityDate 7.3161e+05 7.3201e+05 7.324e+05 7.328e+05 7.332e+05

By: Roman

Roman — Thu, 17 Mar 2011 12:26:36 +0000

I use dataset with a large mixed matrix of values and strings
I have one variable with 14’500 maturity dates of bonds. Unfortunately the maturity dates come in 2 varieties:
1.) dd.mm.yyyy
2.) dd-mmm-yyyy
changing to more appropriate date formats works fine:
data.MaturityDate=arrayfun(@(x)strrep(x,’.',’/'),data.MaturityDate);
data.MaturityDate=arrayfun(@(x)strrep(x,’-',’/'),data.MaturityDate);
However, trying to be efficient with dates by using datenum does not work. I tried many variants, even the following loop does not work:
for i=1,size(data,1);
if length(data.MaturityDate{i})==10;
data.MaturityDate{i}=datenum(data.MaturityDate{i},’dd/mm/yyyy’);
elseif length(data.MaturityDate{i})==11;
data.MaturityDate{i}=datenum(data.MaturityDate{i},’dd/mmm/yyyy’);
else
t=’X’
end
end

Reading the blog I tried to use datasetfun

However I cannot figure out how to define (if possible) a datasetfun like:

f=@(x) length(x)==10, x=datenum(x,’dd/mm/yyyy’),…
length(x)==11, x=datenum(x,’dd/mmm/yyyy’)

data.MaturityDate=datasetfun(f,dataMaturityDate);

Is there an efficient solution?

Regards Roman