Select rows from data.frame ending with a specific character string in R

24,195

Solution 1

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"))

Solution 2

I modified the data a bit so that there were words that ended in te or de.

> PV
     LogFreq   Word PhonCV  FreqDev
1593     140 blahte    CVC 5.480774
482      139    had    CVC 5.438114
1681     138 aaaade   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 kwamde   CCVC 5.182154
483      132 hadden CVC-CV 5.139494
436      131   ging    CVC 5.096834
734      130 moeste  CVVCC 5.054174
1171     129  stond  CCVCC 5.011514
1654     128  zagde    CVC 4.968854
1620     127 werden CVC-CV 4.926194
1683     126 zouden CVV-CV 4.883534

# Add a column to PV that you can visually check the regular expression matches.
PV$Match <- grepl(pattern = "(de|te)$", PV$Word)

# Subset PV data frame to show only TRUE matches
PV <- PV[PV$Match == FALSE, ]

The result is shown below

     LogFreq   Word PhonCV  FreqDev Match
482      139    had    CVC 5.438114 FALSE
1662     137    zei    CVV 5.352794 FALSE
1619     136   werd   CVCC 5.310134 FALSE
1592     135  waren CVV-CV 5.267474 FALSE
620      134    kon    CVC 5.224814 FALSE
483      132 hadden CVC-CV 5.139494 FALSE
436      131   ging    CVC 5.096834 FALSE
1171     129  stond  CCVCC 5.011514 FALSE
1620     127 werden CVC-CV 4.926194 FALSE
1683     126 zouden CVV-CV 4.883534 FALSE

Solution 3

Using grep

grep -xvE '.{17}(de|te).*' file.txt
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HernanLG
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HernanLG

Updated on July 05, 2022

Comments

  • HernanLG
    HernanLG almost 2 years

    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.