Home/ Questions/Q 8708419
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-13T04:08:02+00:00

I’m using R and I have a data.frame with nearly 2,000 entries that looks

  • 0

I’m using R and I have a data.frame with nearly 2,000 entries that looks as follows:

> head(PVs,15)
     LogFreq   Word PhonCV  FreqDev
1593     140    was    CVC 5.480774
482      139    had    CVC 5.438114
1681     138    zou   CVVC 5.395454
1662     137    zei    CVV 5.352794
1619     136   werd   CVCC 5.310134
1592     135  waren CVV-CV 5.267474
620      134    kon    CVC 5.224814
646      133   kwam   CCVC 5.182154
483      132 hadden CVC-CV 5.139494
436      131   ging    CVC 5.096834
734      130  moest  CVVCC 5.054174
1171     129  stond  CCVCC 5.011514
1654     128    zag    CVC 4.968854
1620     127 werden CVC-CV 4.926194
1683     126 zouden CVV-CV 4.883534

What I want to do is to create a new data.frame that is equal to PVs, except that all entries having as a member of the “Word” column a string of character that does NOT end in either “te” or “de” removed. i.e. All words not ending in either “de” or “te” should be removed from the data.frame.

I know how to slectively remove entries from data.frames using logical operators, but those work when you’re setting numeric criteria. I think to do this I need to use regular expressions, but sadly R is the only programming language I “know”, so I’m far from knowing what type of code to use here.

I appreciate your help.
Thanks in advance.

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    Method 1

    You can use grepl with an appropraite regular expression. Consider the following:

    x <- c("blank","wade","waste","rubbish","dedekind","bated")
grepl("^.+(de|te)$",x)
[1] FALSE  TRUE  TRUE FALSE FALSE FALSE

    The regular expression says begin (^) with anything any number of times (.+) and then find either de or te ((de|te)) then end ($).

    So for your data.frame try,

    subset(PVs,grepl("^.+(de|te)$",Word))

    Method 2

    To avoid the regexp method you can use a substr method instead.

    # substr the last two characters and test
substr(x,nchar(x)-1,nchar(x)) %in% c("de","te")
[1] FALSE  TRUE  TRUE FALSE FALSE FALSE

    So try:

    subset(PVs,substr(Word,nchar(Word)-1,nchar(Word)) %in% c("de","te"))
    • 0