I've compared the solutions.
*Solution 1:*
myf <- function( df1, df2 ){
cond <- df2$a > min(df1$a)
if( cond )
{
idx <- which( df1$a == min(df1$a) )
df1[idx, ] <- df2[1, ]
}
df1
}
# On a larger example,
set.seed(4530)
tst <- data.frame(item = 1:1000,a = rnorm(1000),b = rnorm(1000)) # large
data frame
u<-tst
system.time(
for(i in 1:100000){
y<-data.frame(item=(1000+i),a=rnorm(1),b=rnorm(1)) # small data frame,
every time new
u <- myf(u, y)
})
Took me about 31.90 sec
*Solution 2:*
set.seed(4530)
x <- data.frame(item = 1:1000,a = rnorm(1000),b = rnorm(1000)) # large data
frame
system.time(
for(i in 1:100000){
y<-data.frame(item=(1000+i),a=rnorm(1),b=rnorm(1)) # small data frame,
every time new
u[intersect(which(u$a < y$a),which.min(u$a)),] <- y
})
The solution is correct (despite warnings) but took longer - about 48.84
sec.
Dimitri
On Wed, Jan 30, 2013 at 3:27 PM, Dimitri Liakhovitski <
dimitri.liakhovit...@gmail.com> wrote:
> In realy, values in a will be not integers, but numeric. They will never
> be identical, but it could be that they are pretty close - I don't know
> after how many points after the comma matter.
> Dimitri
>
> On Wed, Jan 30, 2013 at 2:06 PM, arun <smartpink...@yahoo.com> wrote:
>
>> Hi,
>> Any chance x$a to have the same number repeated?
>>
>> If `Item` and `a` are unique, I guess both the solutions should work.
>>
>> set.seed(1851)
>> x<-
>> data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:45,20,replace=F),b=sample(20:50,20,replace=F),stringsAsFactors=F)
>> y<- data.frame(item="z",a=3,b=10,stringsAsFactors=F)
>>
>> x[intersect(which(x$a < y$a),which.min(x$a)),]
>> # item a b
>> #17 c 1 48
>> x[x$a==which.min(x$a[x$a<y$a]),]
>> # item a b
>> #17 c 1 48
>> #or
>>
>> x[x$a%in%which.min(x$a[x$a<y$a]),]
>> # item a b
>> #17 c 1 48
>>
>> x[x$a%in%which.min(x$a[x$a<y$a]),]<-y
>>
>> tail(x)
>> # item a b
>> #15 q 45 30
>> #16 g 10 23
>> #17 z 3 10
>> #18 r 15 39
>> #19 l 18 45
>> #20 t 35 33
>>
>> #However, if `item` column is unique, but `a` is not, then the one I
>> mentioned previously arise.
>> set.seed(1851)
>> x1<-
>> data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:10,20,replace=T),b=sample(20:50,20,replace=F),stringsAsFactors=F)
>> y1<- data.frame(item="z",a=3,b=10,stringsAsFactors=F)
>>
>>
>> x1[intersect(which(x1$a < y1$a),which.min(x1$a)),]
>> # item a b
>> #3 s 1 41
>> x1[x1$a==which.min(x1$a[x1$a<y1$a]),]
>> # item a b
>> #3 s 1 41
>> #11 h 1 46
>> #17 c 1 48
>> x1[x1$a==which.min(x1$a[x1$a<y1$a]),]<- y1
>> A.K.
>>
>>
>> ________________________________
>> From: Dimitri Liakhovitski <dimitri.liakhovit...@gmail.com>
>> To: arun <smartpink...@yahoo.com>
>> Cc: R help <r-help@r-project.org>; Jessica Streicher <
>> j.streic...@micromata.de>
>> Sent: Wednesday, January 30, 2013 1:49 PM
>> Subject: Re: [R] Fastest way to compare a single value with all values in
>> one column of a data frame
>>
>>
>> Sorry - I should have clarified:
>> My identifiers (in column "item") will always be unique. In other words,
>> one entry in column "item" will never be repeated - neither in x nor in y.
>> Dimitri
>>
>>
>> On Wed, Jan 30, 2013 at 1:27 PM, Dimitri Liakhovitski <
>> dimitri.liakhovit...@gmail.com> wrote:
>>
>> Thank you, everyone! I'll try to test those different approaches. Really
>> appreciate your help!
>> >Dimitri
>> >
>> >
>> >On Wed, Jan 30, 2013 at 11:03 AM, arun <smartpink...@yahoo.com> wrote:
>> >
>> >HI,
>> >>
>> >>Sorry, my previous solution doesn't work.
>> >>This should work for your dataset:
>> >>set.seed(1851)
>> >>x<-
>> data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F)
>> >>y<- data.frame(item="f",a=3,b=10,stringsAsFactors=F)
>> >> x[x$a%in%which.min(x[x$a<y$a,]$a),]<- y #if there are multiple minimum
>> values
>> >>
>> >>set.seed(1241)
>> >>x1<-
>> data.frame(item=sample(letters[1:10],1e4,replace=TRUE),a=sample(1:30,1e4,replace=TRUE),b=sample(1:100,1e4,replace=TRUE),stringsAsFactors=F)
>> >>y1<- data.frame(item="f",a=3,b=10,stringsAsFactors=F)
>> >>length(x1$a[x1$a==1])
>> >>#[1] 330
>> >> system.time({x1[x1$a%in%which.min(x1[x1$a<y1$a,]$a),]<- y1})
>> >># user system elapsed
>> >> # 0.000 0.000 0.001
>> >>length(x1$a[x1$a==1])
>> >>#[1] 0
>> >>
>> >>
>> >>#For some reason, it is not working when the multiple number of minimum
>> values > some value
>> >>
>> >>set.seed(1241)
>> >>x1<-
>> data.frame(item=sample(letters[1:10],1e5,replace=TRUE),a=sample(1:30,1e5,replace=TRUE),b=sample(1:100,1e5,replace=TRUE),stringsAsFactors=F)
>> >>y1<- data.frame(item="f",a=3,b=10,stringsAsFactors=F)
>> >>length(x1$a[x1$a==1])
>> >>#[1] 3404
>> >>x1[x1$a%in%which.min(x1[x1$a<y1$a,]$a),]<- y1
>> >> length(x1$a[x1$a==1])
>> >>#[1] 3404 #not getting replaced
>> >>
>> >>#However, if I try:
>> >>set.seed(1241)
>> >> x1<-
>> data.frame(item=sample(letters[1:10],1e6,replace=TRUE),a=sample(1:5000,1e6,replace=TRUE),b=sample(1:100,1e6,replace=TRUE),stringsAsFactors=F)
>> >> y1<- data.frame(item="f",a=3,b=10,stringsAsFactors=F)
>> >> length(x1$a[x1$a==1])
>> >>#[1] 208
>> >> system.time(x1[x1$a%in%which.min(x1[x1$a<y1$a,]$a),]<- y1)
>> >>#user system elapsed
>> >> # 0.124 0.016 0.138
>> >> length(x1$a[x1$a==1])
>> >>#[1] 0
>> >>
>> >>
>> >>#Tried Jessica's solution:
>> >>set.seed(1851)
>> >> x<-
>> data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F)
>> >> y<- data.frame(item="f",a=3,b=10,stringsAsFactors=F)
>> >> x[intersect(which(x$a < y$a),which.min(x$a)),] <- y
>> >>
>> >> x
>> >># item a b
>> >>#1 a 8 25
>> >>#2 a 10 26
>> >>#3 f 3 10 #replaced
>> >>#4 e 15 26
>> >>#5 b 13 20
>> >>#6 a 5 23
>> >>#7 d 4 29
>> >>#8 e 2 24
>> >>#9 c 7 30
>> >>#10 e 14 24
>> >>#11 d 2 20
>> >>#12 e 10 21
>> >>#13 c 13 27
>> >>#14 d 12 23
>> >>#15 b 11 26
>> >>#16 e 5 22
>> >>#17 c 1 26 #it is not replaced
>> >>#18 a 8 21
>> >>#19 e 10 26
>> >>#20 c 2 22
>> >>
>> >>
>> >>
>> >>
>> >>A.K.
>> >>
>> >>
>> >>
>> >>
>> >>
>> >>----- Original Message -----
>> >>From: Dimitri Liakhovitski <dimitri.liakhovit...@gmail.com>
>> >>To: r-help <r-help@r-project.org>
>> >>Cc:
>> >>Sent: Tuesday, January 29, 2013 4:11 PM
>> >>Subject: [R] Fastest way to compare a single value with all values in
>> one column of a data frame
>> >>
>> >>
>> >>Hello!
>> >>
>> >>I have a large data frame x:
>> >>x<-data.frame(item=letters[1:5],a=1:5,b=11:15) # in actuality, x has
>> 1000
>> >>rows
>> >>x$item<-as.character(x$item)
>> >>I also have a small data frame y with just 1 row:
>> >>y<-data.frame(item="f",a=3,b=10)
>> >>y$item<-as.character(y$item)
>> >>
>> >>I have to decide if y$a is larger than the smallest of all the values in
>> >>x$a. If it is, I want y to replace the whole row in x that has the
>> lowest
>> >>value in column a.
>> >>This is how I'd do it.
>> >>
>> >>if(y$a>min(x$a)){
>> >> whichmin<-which(x$a==min(x$a))
>> >> x[whichmin,]<-y[1,]
>> >>}
>> >>
>> >>
>> >>I am wondering if there is a faster way of doing it. What would be the
>> >>fastest possible way? I'd have to do it, unfortunately, many-many times.
>> >>
>> >>Thank you very much!
>> >>
>> >>--
>> >>Dimitri Liakhovitski
>> >>
>> >>gfk.com <http://marketfusionanalytics.com/>
>> >>
>> >> [[alternative HTML version deleted]]
>> >>
>> >>______________________________________________
>> >>R-help@r-project.org mailing list
>> >>https://stat.ethz.ch/mailman/listinfo/r-help
>> >>PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
>> >>and provide commented, minimal, self-contained, reproducible code.
>> >>
>> >>
>> >
>> >
>> >--
>> >
>> >Dimitri Liakhovitski
>> >gfk.com
>>
>>
>> --
>>
>> Dimitri Liakhovitski
>> gfk.com
>>
>
>
>
> --
> Dimitri Liakhovitski
> gfk.com <http://marketfusionanalytics.com/>
>
--
Dimitri Liakhovitski
gfk.com <http://marketfusionanalytics.com/>
[[alternative HTML version deleted]]
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