Color me puzzled. Can you express the run more clearly in Boolean logic?
If someone has five policies: 3 Life and 2 General ... is he in or out?
Applying the alternate strategy to that data set I get:
out <- tapply( dat$clm, dat$uid, paste ,collapse=",")
>
> out
A1.B1
A2.B2 A3.B1
"General"
"General,Life" "General"
A3.B3
A4.B4 A5.B5
"General,Life,General,General"
"General,Life,General" "General,Life"
Please explain why you want A3.B3.
--
David.
On Oct 29, 2009, at 8:56 PM, Steven Kang wrote:
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.ramas...@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
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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