Equivalent function to excel's replace

Question

I would like to add quotation marks to hundreds of words (surrounded by blank spaces) and separate them by a comma. Preferably, by using only one or a few functions to avoid a manual replacement due to temporal expenditure.

Exemplary Problem:

words <- c("Paris Milan Berlin")

Output should be: "Paris","Milan","Berlin"

I've already tried gsub() and str_extract(), however I did not get the desirable outcome.

Hi, now that I read the answers, I think it is not clear whether you would like to have a character **vector** (with three elements in this case) or a single scalar string with the text `"Paris","Milan","Berlin"` — Valeri Voev, Dec 02 '19 at 12:38

GKi · Answer 1 · 2019-12-02T13:50:15.557

3

You can use gsub to place " around words and place , between them.

x <- gsub("[[:blank:]]+", ",", gsub('(\\b[[:alnum:]]+\\b)', '"\\1"', words))
x
#[1] "\"Paris\",\"Milan\",\"Berlin\""

noquote(x)
#"Paris","Milan","Berlin"

Or event shorter as suggested in the comments by @zx8754

paste0('"', gsub(' ', '","',  words), '"')

edited Dec 02 '19 at 13:50

answered Dec 02 '19 at 12:27

GKi

37,245
2
26
48

2

Maybe? `gsub(' ','","', words, fixed = TRUE)` – zx8754 Dec 02 '19 at 12:34
@zx8754 Oh yes, that's much shorter! Thanks. But the leading and trailing `"` is missing. I add them with `paste`. – GKi Dec 02 '19 at 13:06
Thank you a lot. That works but how can I avoid these back slashes in the outcome? – Maria26 Dec 02 '19 at 13:09
@Maria26 The backslashes are just how R displays a quote. You can use `noquote`, `cat` or `writeLines` to to display the string without the backslash escapes. See https://stackoverflow.com/q/45362944/10488504 – GKi Dec 02 '19 at 13:52

score 2 · Answer 2 · answered Dec 02 '19 at 12:36

2

With base R (given that you always want to split on a space) you can do as simple as

unlist(strsplit(words, split = " "))

answered Dec 02 '19 at 12:36

Valeri Voev

1,982
9
25

score 1 · Answer 3 · answered Dec 02 '19 at 12:27

1

There might be several ways. This is one of them:

library(tokenizers)
words <- c("Paris Milan Berlin")
tokenize_words(words, simplify =TRUE)

> tokenize_words(words, simplify =TRUE)
[1] "paris"  "milan"  "berlin"

answered Dec 02 '19 at 12:27

Zhiqiang Wang

6,206
2
13
27

ThomasIsCoding · Answer 4 · 2019-12-02T13:33:05.317

1

In addition to the method of using strsplit (mentioned by @Valeri Voev), another way is to use regmatches() and gregexpr, i.e.,

regmatches(words,gregexpr("[[:alnum:]]+",words))[[1]]

which gives

> regmatches(words,gregexpr("[[:alnum:]]+",words))[[1]]
[1] "Paris"  "Milan"  "Berlin"

To make a string output, the complete code is shown as below:

words <- c("Paris Milan Berlin")

r <- regmatches(words,gregexpr("[[:alnum:]]+",words))[[1]]

res1 <- toString(sapply(r, function(v) paste0('"',v,'"')))
# > res1
# [1] "\"Paris\", \"Milan\", \"Berlin\""

res2 <- toString(sapply(r, function(v) paste0("'",v,"'")))
# > res2
# [1] "'Paris', 'Milan', 'Berlin'"

edited Dec 02 '19 at 13:33

answered Dec 02 '19 at 12:40

ThomasIsCoding

96,636
9
24
81

That looks great! However, there are also words with numbers such as "Paris01" or "Milan04". If I use your functions, the numbers disappear. I just get "Paris","Milan" instead of "Paris01", "Milan04". Do you have an idea? :) Thanks a lot!! – Maria26 Dec 02 '19 at 13:02
@Maria26 Then you can try `[[:alnum:]]`, please see my updates – ThomasIsCoding Dec 02 '19 at 13:06
Great! It works but the commas are missing between the words please see the following: [1] "Paris","Milan","Berlin" – Maria26 Dec 02 '19 at 13:17
@Maria26 do you want `'"Paris", "Milan" ,"Berlin"'` or `'"Paris, Milan, Berlin"'` with quotations in the string? – ThomasIsCoding Dec 02 '19 at 13:21
The first one :) – Maria26 Dec 02 '19 at 13:27
@Maria26 I added outputs with single and double quotations, you can choose either one according to your request – ThomasIsCoding Dec 02 '19 at 13:33

Equivalent function to excel's replace

4 Answers4