Get the characters after a certain pattern in R - regex










1















I have a data frame with one column:



df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
df$cat


How can I extract the characters that appear after c(", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.



So far I have tried:



substr(x = df$cat, start = 4, stop = 6)


But this results in:



 ""BP" "BP2" "BPT" "CN""


And I only want the output to show 



"BPT" "BP2" "BPT" "CN"





regex string rstudio text-extraction







share|improve this question



















  • 2





    Try lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))

    – Wiktor Stribiżew
    Nov 13 '18 at 13:55











  • Or gsub('^c\(|\)$|\\(")', '\1', df$cat)

    – Wiktor Stribiżew
    Nov 13 '18 at 14:00












  • Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one

    – Syed
    Nov 13 '18 at 14:08






  • 1





    Maybe lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Or unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP

    – Wiktor Stribiżew
    Nov 13 '18 at 14:13












  • Just let know if the first BP in your expected output is a typo or not, I understand you need "BPT" "BP2" "BPT" "CN" and not "BP" "BP2" "BPT" "CN"

    – Wiktor Stribiżew
    Nov 13 '18 at 14:21















1















I have a data frame with one column:



df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
df$cat


How can I extract the characters that appear after c(", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.



So far I have tried:



substr(x = df$cat, start = 4, stop = 6)


But this results in:



 ""BP" "BP2" "BPT" "CN""


And I only want the output to show 



"BPT" "BP2" "BPT" "CN"





regex string rstudio text-extraction







share|improve this question



















  • 2





    Try lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))

    – Wiktor Stribiżew
    Nov 13 '18 at 13:55











  • Or gsub('^c\(|\)$|\\(")', '\1', df$cat)

    – Wiktor Stribiżew
    Nov 13 '18 at 14:00












  • Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one

    – Syed
    Nov 13 '18 at 14:08






  • 1





    Maybe lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Or unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP

    – Wiktor Stribiżew
    Nov 13 '18 at 14:13












  • Just let know if the first BP in your expected output is a typo or not, I understand you need "BPT" "BP2" "BPT" "CN" and not "BP" "BP2" "BPT" "CN"

    – Wiktor Stribiżew
    Nov 13 '18 at 14:21













1












1








1








I have a data frame with one column:



df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
df$cat


How can I extract the characters that appear after c(", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.



So far I have tried:



substr(x = df$cat, start = 4, stop = 6)


But this results in:



 ""BP" "BP2" "BPT" "CN""


And I only want the output to show 



"BPT" "BP2" "BPT" "CN"





regex string rstudio text-extraction







share|improve this question
















I have a data frame with one column:



df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
df$cat


How can I extract the characters that appear after c(", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.



So far I have tried:



substr(x = df$cat, start = 4, stop = 6)


But this results in:



 ""BP" "BP2" "BPT" "CN""


And I only want the output to show 



"BPT" "BP2" "BPT" "CN"





regex string rstudio text-extraction




regex string rstudio text-extraction






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 13 '18 at 15:35







Syed

















asked Nov 13 '18 at 13:47









SyedSyed

8310




8310







  • 2





    Try lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))

    – Wiktor Stribiżew
    Nov 13 '18 at 13:55











  • Or gsub('^c\(|\)$|\\(")', '\1', df$cat)

    – Wiktor Stribiżew
    Nov 13 '18 at 14:00












  • Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one

    – Syed
    Nov 13 '18 at 14:08






  • 1





    Maybe lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Or unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP

    – Wiktor Stribiżew
    Nov 13 '18 at 14:13












  • Just let know if the first BP in your expected output is a typo or not, I understand you need "BPT" "BP2" "BPT" "CN" and not "BP" "BP2" "BPT" "CN"

    – Wiktor Stribiżew
    Nov 13 '18 at 14:21












  • 2





    Try lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))

    – Wiktor Stribiżew
    Nov 13 '18 at 13:55











  • Or gsub('^c\(|\)$|\\(")', '\1', df$cat)

    – Wiktor Stribiżew
    Nov 13 '18 at 14:00












  • Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one

    – Syed
    Nov 13 '18 at 14:08






  • 1





    Maybe lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Or unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP

    – Wiktor Stribiżew
    Nov 13 '18 at 14:13












  • Just let know if the first BP in your expected output is a typo or not, I understand you need "BPT" "BP2" "BPT" "CN" and not "BP" "BP2" "BPT" "CN"

    – Wiktor Stribiżew
    Nov 13 '18 at 14:21







2




2





Try lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))

– Wiktor Stribiżew
Nov 13 '18 at 13:55





Try lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))

– Wiktor Stribiżew
Nov 13 '18 at 13:55













Or gsub('^c\(|\)$|\\(")', '\1', df$cat)

– Wiktor Stribiżew
Nov 13 '18 at 14:00






Or gsub('^c\(|\)$|\\(")', '\1', df$cat)

– Wiktor Stribiżew
Nov 13 '18 at 14:00














Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one

– Syed
Nov 13 '18 at 14:08





Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one

– Syed
Nov 13 '18 at 14:08




1




1





Maybe lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Or unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP

– Wiktor Stribiżew
Nov 13 '18 at 14:13






Maybe lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Or unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP

– Wiktor Stribiżew
Nov 13 '18 at 14:13














Just let know if the first BP in your expected output is a typo or not, I understand you need "BPT" "BP2" "BPT" "CN" and not "BP" "BP2" "BPT" "CN"

– Wiktor Stribiżew
Nov 13 '18 at 14:21





Just let know if the first BP in your expected output is a typo or not, I understand you need "BPT" "BP2" "BPT" "CN" and not "BP" "BP2" "BPT" "CN"

– Wiktor Stribiżew
Nov 13 '18 at 14:21












1 Answer
1






active

oldest

votes


















1














You may use



df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
## => [1] "BPT" "BP2" "BPT" "CN"


See the R demo online.



Notes




  • gsub('\', '', df$cat, fixed=TRUE) removes all backslashes. You may use gsub('\"', '"', df$cat, fixed=TRUE) if you only plan to remove backslashes before ".


  • eval(parse(text=x))[[1]] parses the vector and returns the first item


  • lapply helps traverse the whole data you have. See Using sapply and lapply.





share|improve this answer






















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    1 Answer
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    1 Answer
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    active

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    You may use



    df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
    df$cat <- as.character(df$cat)
    unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
    ## => [1] "BPT" "BP2" "BPT" "CN"


    See the R demo online.



    Notes




    • gsub('\', '', df$cat, fixed=TRUE) removes all backslashes. You may use gsub('\"', '"', df$cat, fixed=TRUE) if you only plan to remove backslashes before ".


    • eval(parse(text=x))[[1]] parses the vector and returns the first item


    • lapply helps traverse the whole data you have. See Using sapply and lapply.





    share|improve this answer



























      1














      You may use



      df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
      df$cat <- as.character(df$cat)
      unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
      ## => [1] "BPT" "BP2" "BPT" "CN"


      See the R demo online.



      Notes




      • gsub('\', '', df$cat, fixed=TRUE) removes all backslashes. You may use gsub('\"', '"', df$cat, fixed=TRUE) if you only plan to remove backslashes before ".


      • eval(parse(text=x))[[1]] parses the vector and returns the first item


      • lapply helps traverse the whole data you have. See Using sapply and lapply.





      share|improve this answer

























        1












        1








        1







        You may use



        df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
        df$cat <- as.character(df$cat)
        unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
        ## => [1] "BPT" "BP2" "BPT" "CN"


        See the R demo online.



        Notes




        • gsub('\', '', df$cat, fixed=TRUE) removes all backslashes. You may use gsub('\"', '"', df$cat, fixed=TRUE) if you only plan to remove backslashes before ".


        • eval(parse(text=x))[[1]] parses the vector and returns the first item


        • lapply helps traverse the whole data you have. See Using sapply and lapply.





        share|improve this answer













        You may use



        df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
        df$cat <- as.character(df$cat)
        unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
        ## => [1] "BPT" "BP2" "BPT" "CN"


        See the R demo online.



        Notes




        • gsub('\', '', df$cat, fixed=TRUE) removes all backslashes. You may use gsub('\"', '"', df$cat, fixed=TRUE) if you only plan to remove backslashes before ".


        • eval(parse(text=x))[[1]] parses the vector and returns the first item


        • lapply helps traverse the whole data you have. See Using sapply and lapply.






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 13 '18 at 15:22









        Wiktor StribiżewWiktor Stribiżew

        316k16134215




        316k16134215





























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