R help - Mar 2012 - Extracting numbers from a character variable of different types

Hello,

I have a file which contains a column with age, which is represented in the
two following patterns

1. "007/A" or ''007/a" or ''7 /a" ..... In
this case A or a means year and I
would like to extract only the numeric values eg 7 in the above case if this
pattern exits in a line of file.

2. "004/M" or "004/m" where M or m means month ...... for
these lines I
would like to first extract the numeric value of Month eg. 4  and then
convert it into a value of years, which would be 0.33 eg 4 divided by 12.

Can anyone help?

Thank you 

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On Mar 18, 2012, at 10:44 AM, irene wrote:
> Hello,
>
> I have a file which contains a column with age, which is represented  
> in the
> two following patterns
>
> 1. "007/A" or ''007/a" or ''7 /a" .....
In this case A or a means
> year and I
> would like to extract only the numeric values eg 7 in the above case  
> if this
> pattern exits in a line of file.
>
> 2. "004/M" or "004/m" where M or m means month ......
for these
> lines I
> would like to first extract the numeric value of Month eg. 4  and then
> convert it into a value of years, which would be 0.33 eg 4 divided  
> by 12.
I thought it easier to get to months as an initial step:

 > dfrm <- read.table(text="'007/A'\n'007/a' \n
'7 /a '\n '004/
M'\n'004/m'")

 > dfrm$agenew <- sub("(^\\d+\\s*)(/)([aA])","\\1 *
12", dfrm$V1)
 > dfrm$agenew2 <- sub("(^\\d+\\s*)(/)([mM])","\\1",
dfrm$agenew)
 > dfrm$agenew2
[1] "007 * 12" "007 * 12" "7  * 12 "
"004"      "004"
 > eval(parse(text=dfrm$agenew2))
[1] 4
 > sapply(dfrm$agenew2, function(x) eval(parse(text=x)) )
007 * 12 007 * 12 7  * 12       004      004
       84       84       84        4        4

>
> Can anyone help?
>
> Thank you
>
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> View this message in context:
http://r.789695.n4.nabble.com/Extracting-numbers-from-a-character-variable-of-different-types-tp4482248p4482248.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT

Assume your year value is 

x<-007/A

You want to replace all non-numeric characters (i.e. letters and
punctuation) and all zeros with nothing.

gsub('[[:alpha:]]|[[:punct:]]|0','',x)

Let's say you have a vector with both month and year values (you can
separate them). Now we need to identify the cells that have a month or year
indicator

x<-c("007/A","007/a","003/M","003/m")

grep("/A|/a",x) #cells in x with year information
grep("/M|/m",x) #cells in x with month information

To remove all characters, punctuation, and 0s from x, do:

gsub('[[:alpha:]]|[[:punct:]]|0','',x)

which you can also do specifically for the cells that identify months and
years, respectively:

years<-gsub('[[:alpha:]]|[[:punct:]]|0','',x[grep("/A|/a",x)])
#years
years
months<-gsub('[[:alpha:]]|[[:punct:]]|0','',x[grep("/M|/m",x)])
#months
months

Convert the resulting character vectors into numeric vectors by
as.numeric(as.character(years)) , for example.

HTH,
Daniel





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It worked perfectly!

Thank you

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