HI,
By splitting that code into smaller components:
res1<-strsplit(v2,"_") # will split the elements into substrings wherever it
matches substring "_". Result will be a list
head(res1)
#[[1]]
#[1] "sex" "1"
#[[2]]
#[1] "sex" "2"
#
[[3]]
#[1] "sex" "3"
#
#[[4]]
#[1] "age" "height" "1" "1"
#[[5]]
#[1] "age" "height" "2" "2"
#[[6]]
#[1] "age" "height" "3" "3"
#lists can be processed with lapply(). I used paste() to rearrange the split
components so that it matches with v1
lapply(res1,function(x)
paste(x[1],x[3],x[2],x[4],sep="_")) # in this case I change the order
of components, You can try changing the order or the number of
#components and see how it works.
#For example
unlist(lapply(res1,function(x) paste(x[1],x[3],sep="_"))) #with just 1st and
3rd component
# [1] "sex_NA" "sex_NA" "sex_NA" "age_1" "age_2" "age_3" "age_4" "age_5"
#[9] "age_6" "age_1" "age_2"
res2<-unlist(lapply(strsplit(v2,"_"),function(x)
paste(x[1],x[3],x[2],x[4],sep="_")))
#Last part is indexing:
#grep() regular expression
v1[1]
#[1] "age_1"
grep(v1[1],res2)
#[1] 4 10 #this means 4th and 10 element of res2 matches with first
element of v1[1]
res2[4]
#[1] "age_1_height_1"
#The
goal is to get the corresponding element of v2 that matches with v1.
Since the order of res2 and v2 are the same, we can index it to get the
matching v2 #elements.
v2[grep(v1[1],res2)]
#[1] "age_height_1_1" "age_height_1_7"
I hope this helps you.
A.K.
________________________________
From: Fares Said <frespi...@hotmail.com>
To: smartpink...@yahoo.com
Sent: Thursday, November 1, 2012 2:09 PM
Subject: RE: [R] Return Vector Component
Hi,
I think this is better, but can you explain to me what this code do
function(x) paste(x[1],x[3],x[2],x[4],sep="_"),
Thanks
> Date: Thu, 1 Nov 2012 10:17:56 -0700
> From: smartpink...@yahoo.com
> Subject: Re: [R] Return Vector Component
> To: frespi...@hotmail.com
> CC: r-help@r-project.org
>
> HI,
>
> May be this is a bit closer:
> res1<-unlist(lapply(strsplit(v2,"_"),function(x)
> paste(x[1],x[3],x[2],x[4],sep="_")))
> v1[1]
> #[1] "age_1"
>
> v2[grep(v1[1],res1)]
> #[1] "age_height_1_1" "age_height_1_7"
>
> v1[8]
> #[1] "height_2"
> v2[grep(v1[8],res1)]
> #[1] "age_height_2_2"
> A.K.
>
>
>
>
> ----- Original Message -----
> From: frespider <frespi...@hotmail.com>
> To: r-help@r-project.org
> Cc:
> Sent: Thursday, November 1, 2012 10:20 AM
> Subject: Re: [R] Return Vector Component
>
>
>
> Hi A.K.
>
> Thank you so much for replying this could work but the problem you recreate
> the newv1, to match v2 but I don't want to do that. I just want to check if
> v2 has that component from v1
>
> THanks
> Date: Thu, 1 Nov 2012 06:32:03 -0700
> From: ml-node+s789695n4648123...@n4.nabble.com
> To: frespi...@hotmail.com
> Subject: Re: Return Vector Component
>
>
>
> Hi,
>
>
> Not sure whether I understand it correctly.
>
> v1 <-
> c("age_1","age_2","age_3","age_4","age_5","age_6","height_1","height_2","height_3","height_4","height_5","height_6","height_7","height_8")
>
>
> v2 <-
> c("sex_1","sex_2","sex_3","age_height_1_1","age_height_2_2","age_height_3_3","age_height_4_4","age_height_5_5","age_height_6_6",
>
> "age_height_1_7","age_height_2_8")
>
> v3<-gsub("(age{0,1}\\_).*","\\1",v1)
>
> v4<-gsub("age{0,1}\\_(.*)","\\1",v1)
>
> v5<-gsub("(height{0,1}\\_).*","\\1",v1)
>
> v6<-gsub("height{0,1}\\_(.*)","\\1",v1)
>
> newv1<-paste0(v3[grep("age",v3)],v5[grep("height",v5)],v4[!v4%in%v4[grep("height",v4)]],"_",v6[!v6%in%v6[grep("age",v6)]])
>
> newv1[newv1%in%v2]
>
> #[1] "age_height_1_1" "age_height_2_2" "age_height_3_3" "age_height_4_4"
>
> #[5] "age_height_5_5" "age_height_6_6" "age_height_1_7" "age_height_2_8"
>
> A.K.
>
>
>
>
>
>
>
>
>
>
>
>
> If you reply to this email, your message will be added to the
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> NAML
>
>
>
>
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