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…

Accepted Answer

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"))
```

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

Solution

Problem