CODEWITHSUNDEEP
r
myfatson
myfatson

Reputation: 549

"Result must have length..." error with 'complete.cases'

This is an extension of the question I asked here where I was looking for a way to automate my labeling of subjects into groups based on if their data matched my filter.

Prior to attempting to the automating labeling, this is what I had. 

library(tidyverse)

df <- structure(list(Subj_ID = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L),
                 Location = c(1, 2, 3, 1, 4, 2, 1, 2, 5)), class = "data.frame", 
            row.names = c(NA, -9L))


df2 <- df %>%
  mutate(group=
       if_else(Subj_ID ==1,
               "Treatment",
       if_else(Subj_ID == 2,
                "Control","Withdrawn")))


complete.df <- df2 %>% filter(complete.cases(.))

In my actual data, there are some rows that have NA's and I need to be able to filter for both complete and incomplete cases so I can review the sub-data sets separately if needed. My new code looks like this which assigns a subject to a group based on if they have a location data point 4 or 5: 

df2 <- df %>%
  mutate(group=
       if_else(Subj_ID ==1,
               "Treatment",
       if_else(Subj_ID == 2,
                "Control","Withdrawn")))

df3 <- df2 %>% ##this chunk breaks filter(complete.cases(.))
  group_by(Subj_ID) %>%
  mutate(group2 = case_when(any(Location == 4) | any(Location == 5) ~ "YES", TRUE ~ "NO"))

complete.df <- df3 %>% filter(complete.cases(.))

Once I generate df3 by mutating df2, my filter(complete.cases(.)) subsequently fails.

Yet, if I were to generate df3 by manual recoding, it works! As so: 

df2 <- df %>%
  mutate(group=
       if_else(Subj_ID ==1,
               "Treatment",
       if_else(Subj_ID == 2,
                "Control","Withdrawn")))

df3 <- df2 %>%
  mutate(group2=
       if_else(Subj_ID ==2 | 
              Subj_ID ==3,
               "TRUE", "FALSE"))

complete.df <- df3 %>% filter(complete.cases(.))

Thoughts?

Upvotes: 2

Views: 667

Answers (1)

akrun
akrun

Reputation: 887058

It would be the group_by attribute which causes the issue and can be solved by ungrouping and then apply the filter. In the OP's last code block (manual coding), it is not creating a grouping attribute and thus it works

library(dplyr)
df3 %>%
       ungroup %>%
       filter(complete.cases(.))

Or instead of complete.cases in filter, we can use !is.na with filter_all without removing the grouping attribute

df3 %>% 
    filter_all(any_vars(!is.na(.)))

OP mentioned about the last code block is working, but it doesn't have any group attribute. If we create one, then it fails too

df3 %>% 
     group_by(group) %>%
     filter(complete.cases(.))

Error: Result must have length 3, not 9

Upvotes: 3

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