My lack of understanding is, using the example that follows, I thought that:
df2 <- df1 %>% left_join(code_label, by = c("team"="code"))
and, in a function,
df2 <- df1 %>% left_join(code_label, by = c(x='code'))
would be evaluated identically but they aren't, is there a way of wrapping 'x' so they are?
df1_id <- c(1, 2, 3, 4, 5, 6)
team <- c(1, 2, 1, 6, 4, 1)
year <- c(2014, 2014, 2009, 2020, 2015, 2017)
df1 <- data.frame(df1_id, team, year)
code_id <- c(1,2,3,4,5,6)
code <- c(1,2,3,4,5,6)
label <- c("team_A", "team_B","team_C","team_D","team_E","team_F")
code_label <- data.frame(code_id, code,label)
df2 <- df1 %>% left_join(code_label, by = c("team"="code"))
gives a result.But I want to pass the column name "team" of df1 as an un-named object (I think is the right description?) to a function as follows:
f_show_labels_with_codes(x = "team")
f_show_labels_with_codes <- function(x)
{
print(x)
df2 <- df1 %>% left_join(code_label, by = c(x='code'))
}
but it generates an error:
Error: Join columns must be present in data.
x Problem with `x`
I had a look at the documentation around enquo etc, it's new to me. But print(x) within the function it already returns [1] "team". This question looks similar but is not quite the same I think.
CodePudding user response:
We may use a named vector with setNames
f_show_labels_with_codes <- function(x) {
print(x)
df1 %>%
left_join(code_label, by = setNames('code', x))
}
-testing
> f_show_labels_with_codes(x = "team")
[1] "team"
df1_id team year code_id label
1 1 1 2014 1 team_A
2 2 2 2014 2 team_B
3 3 1 2009 1 team_A
4 4 6 2020 6 team_F
5 5 4 2015 4 team_D
6 6 1 2017 1 team_A