Suppose you have 4 related vectors:
a.id<-c(1:25, 1:25, 1:25)
a.vals <- c(101:175) # same length as a.id (the values for those IDs)
a.id.levels <- c(1:25)
a.id.ratings <- rep(letters[1:5], times=5) # same length as a.id.levels
What I would like to do is specify a rating from a.ratings (e.g. "e"),
get the vector of corresponding IDs from a.id.levels (via
a.id.levels[a.id.ratings=='e']) and then somehow use those IDs in a.id
to get the corresponding values from a.vals.
I think I can probably write a loop to construct of a vector of
ratings of the same length as a.id so that the ratings match the ID,
and then go from there. Is there a better way? Perhaps using factors
or levels or something?
Thanks,
--Paul
--
Paul Lynch
Aquilent, Inc.
National Library of Medicine (Contractor)
Sounds like you have two different tables and are trying to mine one
based on the other. Try
ref <- data.frame( levels = 1:25,
ratings = rep(letters[1:5], times=5) )
db <- data.frame( vals=101:175, levels=c(1:25, 1:25, 1:25) )
levels.of.interest <- ref$levels[ ref$rating=="a" ]
db$vals[ which(db$levels %in% levels.of.interest) ]
[1] 101 106 111 116 121 126 131 136 141 146 151 156 161 166 171
OR a much more intuitive way is to merge both tables and proceeding as
out <- merge( db, ref, by="levels", all.x=TRUE )
out <- out[ order(out$val), ] # little cleanup
subset( out, ratings=="a" ) # ignore the rownames
levels vals ratings
1 1 101 a
16 6 106 a
31 11 111 a
46 16 116 a
61 21 121 a
3 1 126 a
17 6 131 a
32 11 136 a
47 16 141 a
62 21 146 a
2 1 151 a
18 6 156 a
33 11 161 a
48 16 166 a
63 21 171 a
Then you can do cool things using the apply() family like
tapply( out$vals, out$ratings, mean )
a b c d e
136 137 138 139 140
Check out %in%, merge and apply.
Regards, Adai
Paul Lynch wrote:> Suppose you have 4 related vectors:
>
> a.id<-c(1:25, 1:25, 1:25)
> a.vals <- c(101:175) # same length as a.id (the values for those
IDs)
> a.id.levels <- c(1:25)
> a.id.ratings <- rep(letters[1:5], times=5) # same length as
a.id.levels
>
> What I would like to do is specify a rating from a.ratings (e.g.
"e"),
> get the vector of corresponding IDs from a.id.levels (via
> a.id.levels[a.id.ratings=='e']) and then somehow use those IDs in
a.id
> to get the corresponding values from a.vals.
>
> I think I can probably write a loop to construct of a vector of
> ratings of the same length as a.id so that the ratings match the ID,
> and then go from there. Is there a better way? Perhaps using factors
> or levels or something?
>
> Thanks,
> --Paul
>
On Thu, 2007-03-29 at 19:55 -0400, Paul Lynch wrote:> Suppose you have 4 related vectors: > > a.id<-c(1:25, 1:25, 1:25) > a.vals <- c(101:175) # same length as a.id (the values for those IDs) > a.id.levels <- c(1:25) > a.id.ratings <- rep(letters[1:5], times=5) # same length as a.id.levels > > What I would like to do is specify a rating from a.ratings (e.g. "e"), > get the vector of corresponding IDs from a.id.levels (via > a.id.levels[a.id.ratings=='e']) and then somehow use those IDs in a.id > to get the corresponding values from a.vals. > > I think I can probably write a loop to construct of a vector of > ratings of the same length as a.id so that the ratings match the ID, > and then go from there. Is there a better way? Perhaps using factors > or levels or something? > > Thanks, > --PaulIs this what you want? DF <- data.frame(a.id, a.vals, a.id.levels, a.id.ratings)> DF[DF$a.id.ratings == "e", "a.vals"][1] 105 110 115 120 125 130 135 140 145 150 155 160 165 170 175 or> subset(DF, a.id.ratings == "e", select = a.vals)a.vals 5 105 10 110 15 115 20 120 25 125 30 130 35 135 40 140 45 145 50 150 55 155 60 160 65 165 70 170 75 175 See ?subset HTH, Marc Schwartz
On Thu, 29 Mar 2007, Paul Lynch wrote:> Suppose you have 4 related vectors: > > a.id<-c(1:25, 1:25, 1:25) > a.vals <- c(101:175) # same length as a.id (the values for those IDs) > a.id.levels <- c(1:25) > a.id.ratings <- rep(letters[1:5], times=5) # same length as a.id.levels > > What I would like to do is specify a rating from a.ratings (e.g. "e"), > get the vector of corresponding IDs from a.id.levels (via > a.id.levels[a.id.ratings=='e']) and then somehow use those IDs in a.id > to get the corresponding values from a.vals.see ?factor ?match ( in case a.id.levels does not actually index a.id.ratings) ?split> a.ratings.factor <- factor( a.id.ratings[ match(a.id, a.id.levels) ]) > a.vals[ a.ratings.factor == 'e' ][1] 105 110 115 120 125 130 135 140 145 150 155 160 165 170 175> > split( a.vals, a.ratings.factor ) # more generally$a [1] 101 106 111 116 121 126 131 136 141 146 151 156 161 166 171 $b [1] 102 107 112 117 122 127 132 137 142 147 152 157 162 167 172 [output truncated]> lm( a.vals ~ a.ratings.factor - 1 ) # means of a.valsCall: lm(formula = a.vals ~ a.ratings.factor - 1) Coefficients: a.ratings.factora a.ratings.factorb a.ratings.factorc a.ratings.factord a.ratings.factore 136 137 138 139 140> > I think I can probably write a loop to construct of a vector of > ratings of the same length as a.id so that the ratings match the ID, > and then go from there. Is there a better way? Perhaps using factors > or levels or something?A warning: using factor() in this way a.ratings.factor <- factor( a.id, levels=a.id.levels, labels=a.id.ratings ) will work in this case: a.vals[ a.ratings.factor == 'e' ] but generally will get you into trouble as its creates a factor with 25 non-unique levels. So, split( a.vals, a.ratings.factor ) ends up giving a list of 25 (non-uniquely labelled) components HTH, Chuck> > Thanks, > --Paul > > -- > Paul Lynch > Aquilent, Inc. > National Library of Medicine (Contractor) > > ______________________________________________ > R-help at stat.math.ethz.ch 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. >Charles C. Berry (858) 534-2098 Dept of Family/Preventive Medicine E mailto:cberry at tajo.ucsd.edu UC San Diego http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0901