Remove groups that contain certain strings

Question

I have an issue about removing the groups that contain certain strings in its rows for example if includes .. I would like to achive this without breaking the pipeline. I mean without using any join function.

The example data

vals <- c("good","bad",'ugly',"good","bad.","ugly")

    gr <- gl(2,3)

vals gr
1 good  1
2  bad  1
3 ugly  1
4 good  2
5 bad.  2
6 ugly  2

df <- data.frame(vals,gr)

I tried

library(dplyr)
        df%>%
          filter(!grepl("\\.",vals))

which removes only the row that match the condition. But I want to remove entire gr 2.

 vals gr
1 good  1
2  bad  1
3 ugly  1
4 good  2
5 ugly  2

This `df <- data.frame(vals,gr)` should be above the first data frame output you show, otherwise it doesn't make sense (not in chronological order). I suggested this edit but reviewers mistakenly thought it changed anything to your post. — Nakx, Apr 27 '20 at 09:18

score 8 · Accepted Answer · answered Feb 21 '18 at 22:04

8

Maybe something like this:

df %>% group_by(gr) %>% filter(all(!grepl("\\.",vals)))

answered Feb 21 '18 at 22:04

joran

169,992
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1

@Tjebo The `filter` is operating only on one group at a time, and it enforcing the requirement that all `vals` within the group not contain a period. – joran Feb 21 '18 at 22:16
@joran Thanks joran for your elegant solution. By the way regarding the my another post is there any solution yet ?. I tried your last comment but still no luck. all group numbers are different. [special-grouping-number-for-each-pairs](https://stackoverflow.com/questions/48913362/special-grouping-number-for-each-pairs) – Alexander Feb 21 '18 at 22:21
@joran I understand the !grepl filter, but I do not understand why 'gr 2' is filtered out. Is this because it evaluates group 1 as TRUE and then group 2 is FALSE?? – tjebo Feb 21 '18 at 22:21
@Tjebo No, remember it's operating on each group in isolation. So when the filter is _only_ acting on the observations within each group. – joran Feb 21 '18 at 22:23
@joran Ah, the bell rings. Sorry, this took a while to get into my head... oops.. Cheers!! – tjebo Feb 21 '18 at 22:27

MKR · Answer 2 · 2018-02-22T04:52:39.967

2

Another option could be using %in% operator.

df %>% 
 filter(!(gr %in% unique(ifelse(grepl("\\.",vals),gr,NA) )))

#  vals gr
#1 good  1
#2  bad  1
#3 ugly  1

edited Feb 22 '18 at 04:52

answered Feb 21 '18 at 22:45

MKR

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IIUC, the OP has requested to remove the entire group in which the certain string occurs. Your solution does only remove the specific row, the other members of `gr` 2 have not been removed. – Uwe Feb 21 '18 at 23:44
@Uwe Thanks to point it out. I have corrected my mistake. Actually I was working on solution using `mutate` and `mapply` something like `df %>% mutate(InValidGroup = ifelse(mapply(grepl, "\\.",vals),gr,NA) ) %>% filter(!(gr %in% unique(InValidGroup))) %>% select(-InValidGroup)` and messed up while adding answer. Your ans looks good too. – MKR Feb 22 '18 at 04:54

score 2 · Answer 3 · answered Feb 21 '18 at 23:41

The OP has requested to remove the entire group when one of the group members contains a certain string in vals - without breaking the pipe.

The OP explicitely has stated: I mean without using any join function.

However, I believe using an anti-join does not break the pipe:

library(dplyr)
data.frame(vals, gr) %>% 
  anti_join(., filter(., grepl("\\.",vals)), by = "gr")

  vals gr
1 good  1
2  bad  1
3 ugly  1

score 1 · Answer 4 · answered Feb 22 '18 at 03:33

1

Here is one option in base R with subset and table

subset(df, gr %in% names(which(!table(gr, grepl("\\.", vals))[,2])))
#  vals gr
#1 good  1
#2  bad  1
#3 ugly  1

answered Feb 22 '18 at 03:33

akrun

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Remove groups that contain certain strings

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