From: Gabor Grothendieck <ggrothendieck_at_gmail.com>

Date: Wed 11 Jan 2006 - 07:16:29 EST

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Wed Jan 11 07:24:43 2006

Date: Wed 11 Jan 2006 - 07:16:29 EST

One other thing to try could be soundex. ITs normally used for last names but it might work here too. Google to find the soundex encoding rules. Reviewing the country names might suggest minor modifications to the soundex algorithm to improve it for your case.

On 1/10/06, Gabor Grothendieck <ggrothendieck@gmail.com> wrote:

> You can improve it somewhat by first accepting all the largest

*> matches and removing the rows and columns for those and
**> repeatedly doing that with what is left.
**>
**> On 1/10/06, Werner Wernersen <pensterfuzzer@yahoo.de> wrote:
**> > Thanks for the nice code, Gabor!
**> >
**> > Unfortunately, it seems not to work for my purpose, confuses lots of
**> > countries when I compare two lists of over 150 countries each.
**> > Do you have any other suggestions?
**> >
**> >
**> >
**> > Gabor Grothendieck <ggrothendieck@gmail.com> schrieb:
**> > If they were the same you could use merge. To figure out
**> > the correspondence automatically or semiautomatically, try this:
**> >
**> > x <- c("Canada", "US", "Mexico")
**> > y <- c("Kanada", "United States", "Mehico")
**> > result <- outer(x, y, function(x,y) mapply(lcs2, x, y))
**> > result[] <- sapply(result, nchar)
**> > # try both which.max and which.min and if you are lucky
**> > # one of them will give unique values and that is the one to use
**> > # In this case which.max does.
**> > apply(result, 1, which.max) # 1 2 3
**> >
**> > # calculate longest common subsequence between 2 strings
**> > lcs2 <- function(s1,s2) {
**> > longest <- function(x,y) if (nchar(x) > nchar(y)) x else y
**> > # Make sure args are strings
**> > a <- as.character(s1); an <- nchar(s1)+1
**> > b <- as.character(s2); bn <- nchar(s2)+1
**> >
**> >
**> > # If one arg is an empty string, returns the length of the other
**> > if (nchar(a)==0) return(nchar(b))
**> > if (nchar(b)==0) return(nchar(a))
**> >
**> >
**> > # Initialize matrix for calculations
**> > m <- matrix("", nrow=an, ncol=bn)
**> >
**> > for (i in 2:an)
**> > for (j in 2:bn)
**> > m[i,j] <- if (substr(a,i-1,i-1)==substr(b,j-1,j-1))
**> > paste(m[i-1,j-1], substr(a,i-1,i-1), sep = "")
**> > else
**> > longest(m[i-1,j], m[i,j-1])
**> >
**> > # Returns the distance
**> > m[an,bn]
**> > }
**> >
**> >
**> >
**> > On 1/10/06, Werner Wernersen wrote:
**> > > Hi,
**> > >
**> > > Before I reinvent the wheel I wanted to kindly ask you for your opinion if
**> > there is a simple way to do it.
**> > >
**> > > I want to merge a larger number of tables from different data sources in R
**> > and the matching criterium are country names. The tables are of different
**> > size and sometimes the country names do differ slightly.
**> > >
**> > > Has anyone done this or any recommendation on what commands I should look
**> > at to automize this task as much as possible?
**> > >
**> > > Thanks a lot for your effort in advance.
**> > >
**> > > All the best,
**> > > Werner
**> > >
**> > >
**> > >
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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Wed Jan 11 07:24:43 2006

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