0

This seems simple enough but I could not find a solution on this site. I would simply like to remove all columns from a dataframe if they contain a particular partial string match of "gs://". The table has 100's of columns and looks like this:

enter image description here

Wiktor Stribiżew
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Moneeb Irshad Bajwa
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  • Replace image in your question with its text and use its field separator. – Cyrus Jan 31 '23 at 22:03
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    In the future, please post copy-pasteable data using `dput()`, [not an image](https://meta.stackoverflow.com/a/285557/17303805). See [How to make a great R reproducible example](https://stackoverflow.com/q/5963269/17303805) for more details. – zephryl Jan 31 '23 at 22:07

3 Answers3

2

Using this example data:

dat <- data.frame(
  x = c("gs://red", "orange"), 
  y = c("yellow", "gs://green"), 
  z = c("blue", "indigo")
)

dat
#          x          y      z
# 1 gs://red     yellow   blue
# 2   orange gs://green indigo

Index the dataframe using grepl():

dat[!sapply(dat, \(x) any(grepl("gs://", x)))]
#        z
# 1   blue
# 2 indigo
zephryl
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    @RitchieSacramento I see, thank you! Changed to a safer approach using `sapply()`. – zephryl Jan 31 '23 at 22:43
  • I like the other approach! Why wouldn't it find the character "c"? – Moneeb Irshad Bajwa Jan 31 '23 at 22:44
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    The problem is that it would find `"c"`s even if not present in your data. This is because `grepl()` coerces the dataframe using `as.character()`, which (to my surprise) yields `"c(\"gs://red\", \"orange\")" "c(\"yellow\", \"gs://green\")" "c(\"blue\", \"indigo\")"`. – zephryl Jan 31 '23 at 22:46
0

alternatively with dplyr::select_if

library(dplyr)

dat %>% select_if(~!any(str_detect(.,'gs://')))

Created on 2023-01-31 with reprex v2.0.2

       z
1   blue
2 indigo
jkatam
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0

Using select with where

library(dplyr)
library(stringr)
df1 %>%
   select(where(~ all(str_detect(.x, fixed("gs://"), negate = TRUE))))

-output

       z
1   blue
2 indigo
akrun
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