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
>>and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
>
>--
>
>Dimitri Liakhovitski
>gfk.com
--
Dimitri Liakhovitski
gfk.com
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