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

Question

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.

James · Accepted Answer · 2012-10-22 15:55:11Z

24

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"))

edited Oct 22, 2012 at 15:55

answered Oct 22, 2012 at 13:24

James

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5 Comments

HernanLG Over a year ago

Thanks @James I tried this but I don't seem to get the results I want. I applied your command just as suggested for my data but the words that were left were in no way those that ended only in "de" and "te". Indeed almost 100 entries were removed, but I don't know which, or according to which criteria. I then tried creating a new column that had TRUE of FALSE if the entry adapted to the regular expression, but the results don't look good, as, for example, the word "aandrong" has TRUE, but clearly doesn't end in my desired endings. Any idea what might be going wrong?

James Over a year ago

@Hernan_L I've added an alternative method to the answer.

James Over a year ago

@Hernan_L And now I've fixed the regexp, so that should work too.

HernanLG Over a year ago

Many thanks James. I used your new regular expression and now it works perfectly. What is the additional $ symbol doing? Anyways, the subset function also worked, but I'm sticking with the regexp, as I will be using it to select more types of words. Again, many thanks.

James Over a year ago

@Hernan_L The $ symbol matches the empty string at the end of the line. It ensures that the match is at the end of the work. Try it without it with the example vector I provided.

RossB · Accepted Answer · 2012-10-22 15:57:34Z

3

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

edited Oct 22, 2012 at 15:57

answered Oct 22, 2012 at 15:32

RossB

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1 Comment

HernanLG Over a year ago

Thanks @RossB , the idea to add an additional column with the logical values is a good idea, as it allows me to see the effects of the function on the original data. Also, the regexp is working fine. Thanks!

Ωmega · Accepted Answer · 2012-10-22 13:35:46Z

1

Using grep

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

answered Oct 22, 2012 at 13:35

Ωmega

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1 Comment

HernanLG Over a year ago

Thx @Ωmega for your answer. Could you be a bit more specific as to what this command does and how may I apply it to my data?

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