[R] Is there an equivalent * operator?

René Capell rene.ca at web.de
Thu Jul 24 14:57:13 CEST 2008


Hi Robin,

to answer your question: I think you are looking for a regular expression, have a look at regex().

In your particular case, where all the strings show the same structure, you can also use substr() to cut out the substring of interest.
More generally, you could consider to convert your your string-dates to real date information with strptime().
After doing so, you can easily calculate functions on various date combinations with format() and tapply().

> date<-strptime(data$date,format="%d-%B-%y")
> tapply(data$temperature,format(date,format="%m"),FUN=mean,na.rm=F)
> # or, if you only want the results for september, use which()
> mean(data$temperature[which(format(date,format="%m")==9)])

Read the help files on the details. :)

Cheers, René


> -----Ursprüngliche Nachricht-----
> Von: "Williams, Robin" <robin.williams at metoffice.gov.uk>
> Gesendet: 24.07.08 14:18:13
> An: <r-help at r-project.org>
> Betreff: [R] Is there an equivalent * operator?


> Hi all, 
> Is there an equivalent to the general * operator in R, where * can
> represent any character? 
>   I have a dataset, one column being date, date ranging between
> April-September, 97-06. I would like to be able to acquire the data for
> a specific month, say September, so that I can take average temperatures
> for the month, etc. I thought I would be able to do something like 
> data.s1 <- subset(data,date=="**-Sep-**") 
> data.s1 
> but when I do 
> data.s1 
> I get "0 rows or 0-length row.names". 
>   Any idea of a simple way I can perform this task? 
> All dates are in the form xx-xxx-xx, EG 30-Sep-06. 
> Thanks for any assistance.  
>  
> 
> Robin Williams
> Met Office summer intern - Health Forecasting
> robin.williams at metoffice.gov.uk 
> 
>  
> 
> 	[[alternative HTML version deleted]]
> 
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> 


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