Hi Phillip,
The following seems to do what you want:
phdf<-read.table(text="v1 v2 v3 v4 v5 code
0 0 0 0 0 1
1 4 0 0 0 1
1 1 0 0 0 1
1 0 1 0 0 1
2 0 1 0 0 1
0 1 0 0 0 1
0 1 2 0 0 1
0 1 2 3 0 1
0 2 3 4 4 1
0 0 0 2 3 1",
header=TRUE,
stringsAsFactors=FALSE)
rules<-list("x[1]==0&&x[2]==0&&x[3]==0&&x[4]==0&&x[5]==0",
"x[1]==1&&x[2]==1","x[1]==0&&x[2]==1&&x[3]==2")
outcomes<-c(1,5,10)
apply_rule<-function(x,rule) return(eval(parse(text=rule)))
for(ri in 1:length(rules))
phdf[apply(phdf,1,apply_rule,rules[[ri]]),"code"] <- outcomes[ri]
and can be expanded to the number of rules that you want. BUT, you
have not specified a non-match value, so your initial values for
"code" will persist.
Jim
On Wed, Oct 2, 2019 at 12:32 PM Phillip Heinrich <herd_...@cox.net> wrote:
>
> With the snippet of data below I’m trying to do an if/then type of thing:
> row 1 – if all five variables equal 0 then code equals 1;
> row 3 – if v1 = 1 and v2 = 1 then code = 5;
> row 7 – if v1 = 0 and v2 = 1 and v3 = 2 then code = 10
>
> There are 24 codes in the complete database.
>
>
> v1 v2 v3 v4 v5 code
> 1 0 0 0 0 0 1
> 2 1 4 0 0 0 1
> 3 1 1 0 0 0 1
> 4 1 0 1 0 0 1
> 5 2 0 1 0 0 1
> 6 0 1 0 0 0 1
> 7 0 1 2 0 0 1
> 8 0 1 2 3 0 1
> 9 0 2 3 4 4 1
> 10 0 0 0 2 3 1 I understand that the mapply function can do things
> like this but I have been reading documentation and poking around with Google
> but am getting nowhere. Any advise whould be greatly appreciated.
> [[alternative HTML version deleted]]
>
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