"Adding missing grouping variables" message in dplyr in R

RDplyr

R Problem Overview


I have a portion of my script that was running fine before, but recently has been producing an odd statement after which many of my other functions do not work properly. I am trying to select the 8th and 23rd positions in a ranked list of values for each site to find the 25th and 75th percentile values for each day in a year for each site for 30 years. My approach was as follows (adapted for the four line dataset - slice(3) would be slice(23) for my full 30 year dataset usually):

library(“dplyr”)

mydata

structure(list(station_number = structure(c(1L, 1L, 1L, 1L), .Label = "01AD002", class = "factor"), 
year = 1981:1984, month = c(1L, 1L, 1L, 1L), day = c(1L, 
1L, 1L, 1L), value = c(113, 8.329999924, 15.60000038, 149
)), .Names = c("station_number", "year", "month", "day", "value"), class = "data.frame", row.names = c(NA, -4L))	

  value <- mydata$value
  qu25 <- mydata %>% 
          group_by(month, day, station_number) %>% 
          arrange(desc(value)) %>% 
          slice(3) %>% 
          select(value)

Before, I would be left with a table that had one value per site to describe the 25th percentile (since the arrange function seems to order them highest to lowest). However, now when I run these lines, I get a message:

Adding missing grouping variables: `month`, `day`, `station_number`

This message doesn’t make sense to me, as the grouping variables are clearly present in my table. Also, again, this was working fine until recently. I have tried:

  • detatch(“plyr”) – since I have it loaded before dplyr
  • dplyr:: group_by – placing this directly in the group_by line
  • uninstalling and re-intstalling dplyr, although this was for another issue I was having

Any idea why I might be receiving this message and why it may have stopped working?

Thanks for any help.

Update: Added dput example with one site, but values for January 1st for multiple years. The hope would be that the positional value is returned once grouped, for instance slice(3) would hopefully return the 15.6 value for this smaller subset.

R Solutions


Solution 1 - R

For consistency sake the grouping variables should be always present when defined earlier and thus are added when select(value) is executed. ungroup should resolve it:

qu25 <- mydata %>% 
  group_by(month, day, station_number) %>%
  arrange(desc(value)) %>% 
  slice(2) %>% 
  ungroup() %>%
  select(value)

The requested result is without warnings:

> mydata %>% 
+   group_by(month, day, station_number) %>%
+   arrange(desc(value)) %>% 
+   slice(2) %>% 
+   ungroup() %>%
+   select(value)
# A tibble: 1 x 1
  value
  <dbl>
1   113

Solution 2 - R

Did you update dplyr recently by chance? I wonder if your dplyr::arrange call has been adversely effected by https://blog.rstudio.org/2016/06/27/dplyr-0-5-0/

> Breaking changes > arrange() once again ignores grouping, reverting back > to the behaviour of dplyr 0.3 and earlier. This makes arrange() > inconsistent with other dplyr verbs, but I think this behaviour is > generally more useful. Regardless, it’s not going to change again, as > more changes will just cause more confusion.

Solution 3 - R

You can also convert your tibble to a data frame before your select statement using data.frame(). Then dplyr loses track of your grouping variables and isn't worried about them anymore.

qu25 <- mydata %>% 
      group_by(month, day, station_number) %>% 
      arrange(desc(value)) %>% 
      slice(3) %>% 
      data.frame() %>%
      select(value)

Solution 4 - R

You can convert that tribble to dataframe at the beginning

qu25 <- as.data.frame(mydata) %>% 
  group_by(month, day, station_number) %>% 
  arrange(desc(value)) %>% 
  slice(3) %>% 
  select(value)

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionacersaccharumView Question on Stackoverflow
Solution 1 - RDreyView Answer on Stackoverflow
Solution 2 - RKen LinView Answer on Stackoverflow
Solution 3 - RSpencer EView Answer on Stackoverflow
Solution 4 - RMd. Naimul HasanView Answer on Stackoverflow