Extracting non-capitalized words using Regex - r

Question

I'm trying to extract non-capitalized words using Regex in R. The data contains several columns (e.g. word, word duration, syllable, syllable duration ...), and in the Word column, there are tons of words that are either capitalized (e.g. EAT), non-capitalized (e.g. see), or in curly brackets (e.g. {VAO}). I want to extract all the words that are not capitalized in the word column. The following is a small example data frame with an expected outcome.
file word
1 sp
2 WHAT
3 ISN'
4 'EM
5 O
6 {PPC}
OUTCOME:
"sp", "{PPC}"
> unique(full_dat$word[!grepl("^[A-Z].*[A-Z]|\\d", full_dat$word) & !grepl(" [[:punct:]] ", full_dat$word)]
This result is the following:
[1] "sp" "{OOV}"
[3] "O" "I"
[5] "A" NA
[7] "{XX}" "'S"
[9] "{LG}" "Y"
[11] "B" "'VE"
[13] "N" "{GAP_ANONYMIZATION_NAME}"
[15] "'EM" "W"
[17] "{GAP_ANONYMIZATION}" "K"
This looks good, since I can easily recognize the non-capitalized words, but there are still some capitalized words in this list.... How can I modify the code, so it shows only lower-case words and curly-bracketed words?

Answer 1

With the library stringr you can just simply do that:
library(stringr)
x <- c("HELLO WORLD", "hello world", "Hello World", "hello World", "HeLLo wOrlD")
str_extract(x, "[A-Z]+")
Which results in all the upper cases found in each word:
[1] "HELLO" NA "H" "W" "H"
You can omit NAs by applying the na.omit function, and you will also obtain in which positions there are NAs, that is, in which positions there are not capitalized words:
na.omit(str_extract(x, "[A-Z]+"))
[1] "HELLO" "H" "W" "H"
attr(,"na.action")
[1] 2
attr(,"class")
[1] "omit"
But you can also see in in which positions there are not capitalized words by doing:
is.na(str_extract(x, "[A-Z]+"))
[1] FALSE TRUE FALSE FALSE FALSE
I hope this is helpful 😀