Subsetting rows of data frame by charater patterns (grepl) in a for loop

Question

I am attempting to subset a data frame by removing rows containing certain charater patterns, which are stored in a vector. My issue is that only the last pattern of the vector is removed from my data frame. How can I make my loop work iteratively, so that all patterns stored in the vector are removed from my data frame?

Mock input:

df<-data.frame(organism=c("human_longname","cat_longname","bird_longname","virus_longname","bat_longname","pangolian_longname"),size=c(6,4,2,1,3,5))
df
   organism            size
1     human_longname     6
2       cat_longname     4
3      bird_longname     2
4     virus_longname     1
5       bat_longname     3
6 pangolian_longname     5

used code and output:

vectors<-c("bat","virus","pangolian")
for(i in vectors){df_1<-df[!grepl(i,df$organism),]}
df_1
  organism             size
1    human_longname      6
2      cat_longname      4
3     bird_longname      2
4    virus_longname      1
5      bat_longname      3

Expected output

df_1
  organism             size
1    human_longname      6
2      cat_longname      4
3     bird_longname      2

Try this: `df[!df$organism %in% c("bat","virus","pangolian"),]` — Duck, Aug 03 '20 at 15:36
Thanks a lot guys. I prefer these elegant solutions. However, I am still curious to learn how this could have been achieved with the for loop :) — Rikki Franklin Frederiksen, Aug 03 '20 at 15:51
Sorry guys. I forgot to mention that I HAVE to subset by pattern, as my strings in the data frame are very long. I will change to mock data to better show this. But thanks a lot for your answers. — Rikki Franklin Frederiksen, Aug 03 '20 at 16:16
@RikkiFranklinFrederiksen I have updated the solution to work with the new data. Please check if you like. — Duck, Aug 03 '20 at 16:45

Duck · Accepted Answer · 2020-08-03T17:14:45.497

1

You can try this:

df[!df$organism %in% c("bat","virus","pangolian"),]

  organism size
1    human    6
2      cat    4
3     bird    2

Update: Based on new data, here an approach using grepl(). These functions can be used to avoid loops:

#Vectors
vectors<-c("bat","virus","pangolian")
#Format
vectors2 <- paste0(vectors,collapse = '|')
#Avoid loop
df[!grepl(pattern = vectors2,df$organism),]

        organism size
1 human_longname    6
2   cat_longname    4
3  bird_longname    2

Also just for curious, here maybe a not optimal loop to do the same task creating a new dataframe and an index:

#Create index
index <- c()
#Loop
for(i in 1:dim(df)[1])
{
  if(grepl(vectors2,df$organism[i])==F) 
  {
    index <- c(index,i)
  }
  ndf <- df[index,]
}

ndf

        organism size
1 human_longname    6
2   cat_longname    4
3  bird_longname    2

edited Aug 03 '20 at 17:14

answered Aug 03 '20 at 15:45

Duck

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It works! Thank you so much Duck. It seems like such an easy task, but I have been struggling with it for hours trying with for loops and apply() functions to no avail. – Rikki Franklin Frederiksen Aug 03 '20 at 17:08
@RikkiFranklinFrederiksen Great!! Most of functions in `R` are vectorized like `grepl` so you can avoid loops :) – Duck Aug 03 '20 at 17:09
@RikkiFranklinFrederiksen Just for curious, I have added the way to do the task with loop, maybe not the most optimal compared to the rest of solutions but you can check. – Duck Aug 03 '20 at 17:15
Thanks for showing me Duck. Your first solution is definitely more easy for me to grasp :). – Rikki Franklin Frederiksen Aug 03 '20 at 17:36
@RikkiFranklinFrederiksen Totally agree with you :) – Duck Aug 03 '20 at 17:39

Subsetting rows of data frame by charater patterns (grepl) in a for loop

1 Answers1