Although there are more columns and observations, my dataframe looks like the following:
dt <- data.frame(hid = c(1, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4),
syear = c(2000, 2001, 2003, 2003, 2003, 2000, 2000, 2001, 2001, 2002, 2002),
employlvl = c("Full-time", "Part-time", "Part-time", "Unemployed", "Unemployed",
"Full-time", "Full-time", "Full-time", "Unemployed", "Part-time",
"Full-time"),
relhead = c("Head", "Head", "Head", "Partner", "other", "Head",
"Partner", "Head", "Partner", "Head", "Partner"))
| hid | syear | employlvl | relhead |
|-----|-------|-------------|-----------------------|
| 1 | 2000 | Full-time | Head |
| 2 | 2001 | Part-time | Head |
| 2 | 2003 | Part-time | Head |
| 2 | 2003 | Unemployed | Partner |
| 2 | 2003 | Unemployed | other |
| 4 | 2000 | Full-time | Head |
| 4 | 2000 | Full-time | Partner |
| 4 | 2001 | Full-time | Head |
| 4 | 2001 | Unemployed | Partner |
| 4 | 2002 | Part-time | Head |
| 4 | 2002 | Full-time | Partner |
I would like to create another column which indicates the employmentlevel of the Partner and hope to get the following output:
| hid | syear | employlvl | relhead | Partner |
|-----|-------|-------------|-----------------------|-------------------|
| 1 | 2000 | Part-time | Head | NA |
| 2 | 2001 | Part-time | Head | NA |
| 2 | 2003 | Part-time | Head | Unemployed |
| 2 | 2003 | Unemployed | Partner | NA |
| 2 | 2003 | Unemployed | other | NA |
| 4 | 2000 | Full-time | Head | Full-time |
| 4 | 2000 | Full-time | Partner | NA |
| 4 | 2001 | Full-time | Head | Unemployed |
| 4 | 2001 | Unemployed | Partner | NA |
| 4 | 2002 | Part-time | Head | Full-time |
| 4 | 2002 | Full-time | Partner | NA |
Currently, I am using the following code:
library(dplyr)
library(tidyr)
dt2 <- dt %>%
group_by(hid, syear) %>%
filter(n() > 1) %>%
filter(`relhead` != "Child") %>%
spread(relhead, employlvl) %>%
mutate(Relation = "Head") %>%
rename(`Employment Partner` = Partner) %>%
select(-Head)
dt3 <- dt %>%
left_join(dt2, by = c("hid", "syear", "relhead" = "Relation"))
The code works absolutely fine for this small data set. But as soon as I try for my whole data, I get the following:
Error: Data source must be a dictionary
As stated in other answers this is caused by non unique names. I was able to reproduce error by modifying your example (third element of relhead)
dt <- data.frame(
hid = c(1, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4),
syear = c(2000, 2001, 2003, 2003, 2003, 2000, 2000, 2001, 2001, 2002, 2002),
employlvl = c("Full-time", "Part-time", "Part-time", "Unemployed", "Unemployed",
"Full-time", "Full-time", "Full-time", "Unemployed", "Part-time",
"Full-time"),
relhead = c("Head", "Head", "Employment Partner", "Partner", "other", "Head",
"Partner", "Head", "Partner", "Head", "Partner")
)
In that case spread creates first "Employment Partner" column and rename creates second. You should check if any of "Employment Partner", "Relation" (and maybe hid, syear) is in dt$relhead (first one gives you error, second one is overwrite by mutate(Relation=...)).
Minimal reproducible example:
data_frame(g = c("a1","a2","a3"), i=1) %>%
spread(g, i) %>%
rename(a1 = a3) %>%
select(-a1)
Important update: now we have more informative error message:
Error in `rename()`:
! Names must be unique.
ā These names are duplicated:
* "Employment Partner" at locations 3 and 6.
Run `rlang::last_trace()` to see where the error occurred.