Highly appreciate for all the help.

I have one more thing to resolve..

Suppose 3 additional records are binded to the previous arbitrary data set.
i.e
> a <-
data.frame(id=c(c("A1","A2","A3","A4","A5"),c("A3","A2","A3","A4","A5")),loc=c("B1","B2","B3","B4","B5"),clm=c(rep(("General"),6),rep("Life",4)))
> b <-
data.frame(id=c("A3","A3","A4"),loc=c("B3","B3","B4"),clm=rep("General",3))
> dat <- rbind(a,b)
> dat
   id loc     clm
1  A1  B1 General
2  A2  B2 General
3  A3  B3 General
4  A4  B4 General
5  A5  B5 General
6  A3  B1 General
7  A2  B2    Life
8  A3  B3    Life
9  A4  B4    Life
10 A5  B5    Life
11 A3  B3 General
12 A3  B3 General
13 A4  B4 General

The records with row number 3, 11 & 12 and records with row number 4 & 13
are identical.

     id  loc    clm                id  loc   clm
3  A3  B3 General          4 A4 B4 General
11 A3  B3 General        13 A4 B4 General
12 A3  B3 General

The provided solutions does not perform 1 to 1 matching. (i.e all the
matching duplicated records are removed..)

The desired output is:

     id   loc  clm
1   A1  B1 General
6   A3  B1 General
11 A3  B3 General
12 A3  B3 General
13 A4  B4 General

Are there solution to this problem with 'merging' function or other
alternative method?

Thanks



Steven







On Thu, Oct 29, 2009 at 10:30 PM, Adaikalavan Ramasamy <
a.ramasamy@imperial.ac.uk> wrote:

> Here is another way based on pasting ids as hinted below:
>
>
> a <- data.frame(id=c(c("A1","A2","A3","A4","A5"),
>                   c("A3","A2","A3","A4","A5")),
>                   loc=c("B1","B2","B3","B4","B5"),
>                   clm=c(rep(("General"),6),rep("Life",4)))
>
> a$uid <- paste(a$id, ".", a$loc, sep="")
>
> out <- tapply( a$clm, a$uid, paste ) # can also add collapse=","
> $A1.B1
> [1] "General"
>
> $A2.B2
> [1] "General" "Life"
>
> $A3.B1
> [1] "General"
>
> $A3.B3
> [1] "General" "Life"
>
> $A4.B4
> [1] "General" "Life"
>
> $A5.B5
> [1] "General" "Life"
>
>
> Then here are those with single policies.
>
> > out[ which( sapply(out, length) == 1 ) ]
> $A1.B1
> [1] "General"
>
> $A3.B1
> [1] "General"
>
>
>
>
> David Winsemius wrote:
>
>> On Oct 28, 2009, at 9:30 PM, Steven Kang wrote:
>>
>> Dear R users,
>>>
>>>
>>> Basically, from the following arbitrary data set:
>>>
>>> a <-
>>> data
>>> .frame
>>> (id
>>> =
>>> c
>>> (c
>>> ("A1
>>> ","A2
>>> ","A3
>>> ","A4
>>> ","A5
>>> "),c
>>> ("A3
>>> ","A2
>>> ","A3
>>> ","A4","A5")),loc=c("B1","B2","B3","B4","B5"),clm=c(rep(("General"),
>>> 6),rep("Life",4)))
>>>
>>> a
>>>>
>>>   id   loc  clm
>>> 1  A1  B1 General
>>> 2  A2  B2 General
>>> 3  A3  B3 General
>>> 4  A4  B4 General
>>> 5  A5  B5 General
>>> 6  A3  B1 General
>>> 7  A2  B2    Life
>>> 8  A3  B3    Life
>>> 9  A4  B4    Life
>>> 10 A5  B5    Life
>>>
>>> I desire removing records (highlighted records above) with identical
>>>  values
>>> in each fields ("id" & "loc") but with different value of "clm" (i.e
>>> according to category)
>>>
>>
>> Take a look at this merge operation on separate rows of "a".
>>
>>  > merge( a[a$clm=="Life", ], a[a$clm=="General", ] , by=c("id",  "loc"),
>> all=T)
>>   id loc clm.x   clm.y
>> 1 A1  B1  <NA> General
>> 2 A2  B2  Life General
>> 3 A3  B1  <NA> General
>> 4 A3  B3  Life General
>> 5 A4  B4  Life General
>> 6 A5  B5  Life General
>>
>> Assignment of that object and selection with is.na should complete the
>>  process.
>>
>>  > a2m <- merge( a[a$clm=="Life", ], a[a$clm=="General", ] ,  by=c("id",
>> "loc"), all=T)
>>
>>  > a2m[ is.na(a2m$clm.x) | is.na(a2m$clm.y), ]
>>   id loc clm.x   clm.y
>> 1 A1  B1  <NA> General
>> 3 A3  B1  <NA> General
>>
>> Alternate methods might include paste-ing id to loc and removing
>>  duplicates.
>>
>>
>> i.e
>>>
>>>> categ <- table(a$id,a$clm)
>>>> categ
>>>>
>>>    General Life
>>>  A1       1    0
>>>  A2       1    1
>>>  A3       2    1
>>>  A4       1    1
>>>  A5       1    1
>>>
>>> The desired output is
>>>
>>>   id   loc  clm
>>> 1  A1  B1 General
>>> 6  A3  B1 General
>>>
>>> Because the data set I am working on is quite big (~ 800,000 x 20)
>>> with majority of the fields values being long strings, looping  turned
>>> out to
>>> be very inefficient in comapring individual rows..
>>>
>>> Are there any alternative efficient methods in implementing this
>>>  problem?
>>> Steven
>>>
>>
>

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