Redundant variable naming issue with purrr::map and dplyr - r

Question

I would guess that this is a basic issue that isn't specific to purrr at all, but it caught me off guard in this context. The general answer would be great if this isn't about how purrr and dplyr play together.
I tried to name a variable I was "mapping" over the same as the variable in the d.f. I wanted to match, and it led to problems. Can someone explain why my first attempt to generate pairwise differences fails?
It seems like a variable scoping issue or something with redundant names but I don't know exactly what is wrong. Obviously, I found a workaround.
Imagine I have data like mydf below and there are a lot of variables, and I want to compute the difference in the values of those variables between each pair of sites:
#four sites
site<-rep(c("j", "k", "l", "m"), 3)
#some measurment
val<-1:12
#some variable
vari<-c(rep(1,4), rep(2, 4), rep(3,4))
mydf<-data.frame(site, val, vari)
#compute pairwise differences between values at each site for each variable
outp<-map_dfr(1:3, function(vari){
dists<-as.numeric(dist(mydf %>% filter(vari==vari) %>% select(val), method="manhattan"))
names(dists)<-c("jk","jl", "jm", "kl", "km", "lm" )
dists
return(data.frame(t(dists), vari=vari))
})
# looks like there was an issue with using "vari"
outp
#but use a different name for the same variable and it works fine
outp2<-map_dfr(1:3, function(a){
dists<-as.numeric(dist(mydf %>% filter(vari==a) %>% select(val), method="manhattan"))
names(dists)<-c("jk","jl", "jm", "kl", "km", "lm" )
dists
return(data.frame(t(dists), vari=vari))
})
outp2
edit as noted in comments and answer below, the issue here is in variable usage in dplyr::filter and not with purrr

Answer 1

If you run some simplified code it might make sense. For example:
# Remove the vector `vari` to avoid confusion.
rm(vari)
# Run using `map` and a simplified function.
map(1:3, function(vari) filter(mydf, vari==vari))
The above call to map returns a list of three dataframes, each identical to mydf:
[[1]]
site val vari
1 j 1 1
2 k 2 1
3 l 3 1
4 m 4 1
5 j 5 2
6 k 6 2
7 l 7 2
8 m 8 2
9 j 9 3
10 k 10 3
11 l 11 3
12 m 12 3
[[2]]
site val vari
1 j 1 1
2 k 2 1
3 l 3 1
4 m 4 1
5 j 5 2
6 k 6 2
7 l 7 2
8 m 8 2
9 j 9 3
10 k 10 3
11 l 11 3
12 m 12 3
[[3]]
site val vari
1 j 1 1
2 k 2 1
3 l 3 1
4 m 4 1
5 j 5 2
6 k 6 2
7 l 7 2
8 m 8 2
9 j 9 3
10 k 10 3
11 l 11 3
12 m 12 3
It's obvious that filter(vari == vari) is comparing mydf$vari with itself, which will simply return an exact copy of mydf. This is a good behavior because we always know what filter will compare. Try the same thing with a temporary variable x:
map(1:3, function(x) filter(mydf, vari==x))
Which returns the expected subsets:
[[1]]
site val vari
1 j 1 1
2 k 2 1
3 l 3 1
4 m 4 1
[[2]]
site val vari
1 j 5 2
2 k 6 2
3 l 7 2
4 m 8 2
[[3]]
site val vari
1 j 9 3
2 k 10 3
3 l 11 3
4 m 12 3
This is basically what you did in your "workaround" – which I would describe as valid code using proper conventions, i.e. not a workaround at all.
aosmith already pointed out that you can use tidy evaluation. This is pretty neat, and it definitely has it's use cases, but I think it would reflect bad practice in this particular context. Using a temporary variable would make your code less ambiguous and thus more readable. It also makes sense because we are really dealing with two different things: vari is a vector containing the (repeated) values 1, 2, and 3 while x is in essence a temporary loop variable that is either 1 or 2 or 3, depending on the iteration.