rerror-handlingdplyr

Error: Data source must be a dictionary (dplyr)


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


Solution

  • 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.
    āœ– These names are duplicated:
      * "Employment Partner" at locations 3 and 6.
    Run `rlang::last_trace()` to see where the error occurred.