string <- '    some text on line one; 
and then some text on line two     '

Trimming Whitespace

"Trimming" whitespace typically refers to removing both leading and trailing whitespace from a string. This may be done using a combination of the previous examples. gsub is used to force the replacement over both the leading and trailing matches.

Prior to R 3.2.0

gsub(pattern = "(^ +| +$)",
     replacement = "",
     x = string)

[1] "some text on line one; \nand then some text on line two"

R 3.2.0 and higher

trimws(x = string)

[1] "some text on line one; \nand then some text on line two"

Removing Leading Whitespace

Prior to R 3.2.0

sub(pattern = "^ +", 
    replacement = "",
    x = string)

[1] "some text on line one; \nand then some text on line two     "

R 3.2.0 and higher

trimws(x = string,
       which = "left")

[1] "some text on line one; \nand then some text on line two     "

Removing Trailing Whitespace

Prior to R 3.2.0

sub(pattern = " +$",
    replacement = "",
    x = string)

[1] "    some text on line one; \nand then some text on line two"

R 3.2.0 and higher

trimws(x = string,
       which = "right")

[1] "    some text on line one; \nand then some text on line two"

Removing All Whitespace

gsub(pattern = "\\s",   
     replacement = "",
     x = string)

[1] "sometextonlineone;andthensometextonlinetwo"

Note that this will also remove white space characterse such as tabs (\t), newlines (\r and \n), and spaces.

It is a common practice to name files using the date as prefix in the following format: YYYYMMDD, for example: 20170101_results.csv. A date in such string format can be verified using the following regular expression:

\\d{4}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])

The above expression considers dates from year: 0000-9999, months between: 01-12 and days 01-31.

For example:

> grepl("\\d{4}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])", "20170101")
[1] TRUE
> grepl("\\d{4}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])", "20171206")
[1] TRUE
> grepl("\\d{4}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])", "29991231")
[1] TRUE

Note: It validates the date syntax, but we can have a wrong date with a valid syntax, for example: 20170229 (2017 it is not a leap year).

> grepl("\\d{4}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])", "20170229")
[1] TRUE

If you want to validate a date, it can be done via this user defined function:

is.Date <- function(x) {return(!is.na(as.Date(as.character(x), format = '%Y%m%d')))}

Then

> is.Date(c("20170229", "20170101", 20170101))
[1] FALSE  TRUE  TRUE

The following regex includes 50 states and also Commonwealth/Territory (see www.50states.com):

regex <- "(A[LKSZR])|(C[AOT])|(D[EC])|(F[ML])|(G[AU])|(HI)|(I[DLNA])|(K[SY])|(LA)|(M[EHDAINSOT])|(N[EVHJMYCD])|(MP)|(O[HKR])|(P[WAR])|(RI)|(S[CD])|(T[NX])|(UT)|(V[TIA])|(W[AVIY])"

For example:

> test <- c("AL", "AZ", "AR", "AJ", "AS", "DC", "FM", "GU","PW", "FL", "AJ", "AP")
> grepl(us.states.pattern, test)
 [1]  TRUE  TRUE  TRUE FALSE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE FALSE
> 

Note:

If you want to verify only the 50 States, then we recommend to use the R-dataset: state.abb from state, for example:

> data(state)
> test %in% state.abb
[1]  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE    

We get TRUE only for 50-States abbreviations: AL, AZ, AR, FL.

The following regular expression:

us.phones.regex <- "^\\s*(\\+\\s*1(-?|\\s+))*[0-9]{3}\\s*-?\\s*[0-9]{3}\\s*-?\\s*[0-9]{4}$" 

Validates a phone number in the form of: +1-xxx-xxx-xxxx, including optional leading/trailing blanks at the beginning/end of each group of numbers, but not in the middle, for example: +1-xxx-xxx-xx xx is not valid. The - delimiter can be replaced by blanks: xxx xxx xxx or without delimiter: xxxxxxxxxx. The +1 prefix is optional.

Let's check it:

us.phones.regex <- "^\\s*(\\+\\s*1(-?|\\s+))*[0-9]{3}\\s*-?\\s*[0-9]{3}\\s*-?\\s*[0-9]{4}$"

phones.OK <- c("305-123-4567", "305 123 4567", "+1-786-123-4567", 
    "+1 786 123 4567", "7861234567", "786 - 123   4567", "+ 1 786 - 123   4567")

phones.NOK <- c("124-456-78901", "124-456-789", "124-456-78 90", 
    "124-45 6-7890", "12 4-456-7890")

Valid cases:

> grepl(us.phones.regex, phones.OK)
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> 

Invalid cases:

 > grepl(us.phones.regex, phones.NOK)
[1] FALSE FALSE FALSE FALSE FALSE
> 

Note:

• \\s Matches any space, tab or newline character

Since both R and regex share the escape character ,"\", building correct patterns for grep, sub, gsub or any other function that accepts a pattern argument will often need pairing of backslashes. If you build a three item character vector in which one items has a linefeed, another a tab character and one neither, and hte desire is to turn either the linefeed or the tab into 4-spaces then a single backslash is need for the construction, but tpaired backslashes for matching:

x <- c( "a\nb", "c\td", "e    f")
x  # how it's stored
   #  [1] "a\nb"   "c\td"   "e    f"
cat(x)   # how it will be seen with cat
#a
#b c    d e    f

gsub(patt="\\n|\\t", repl="    ", x)
#[1] "a    b" "c    d" "e    f"

Note that the pattern argument (which is optional if it appears first and only needs partial spelling) is the only argument to require this doubling or pairing. The replacement argument does not require the doubling of characters needing to be escaped. If you wanted all the linefeeds and 4-space occurrences replaces with tabs it would be:

gsub("\\n|    ", "\t", x)
#[1] "a\tb" "c\td" "e\tf"

There are two ever-so-slightly different engines of regular expressions implemented in R. The default is called POSIX-consistent; all regex functions in R are also equipped with an option to turn on the latter type: perl = TRUE.

Look-ahead/look-behind

perl = TRUE enables look-ahead and look-behind in regular expressions.

• "(?<=A)B" matches an appearance of the letter B only if it's preceded by A, i.e. "ABACADABRA" would be matched, but "abacadabra" and "aBacadabra" would not.