n
th number after the m
th occurrence of a pattern.R/num-after.R
str_nth_number_after_mth.Rd
Given a string, a pattern and natural numbers n
and m
, find the n
th
number after the m
th occurrence of the pattern.
str_nth_number_after_mth(
string,
pattern,
n,
m,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
str_nth_number_after_first(
string,
pattern,
n,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
str_nth_number_after_last(
string,
pattern,
n,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
str_first_number_after_mth(
string,
pattern,
m,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
str_last_number_after_mth(
string,
pattern,
m,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
str_first_number_after_first(
string,
pattern,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
str_first_number_after_last(
string,
pattern,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
str_last_number_after_first(
string,
pattern,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
str_last_number_after_last(
string,
pattern,
decimals = FALSE,
leading_decimals = decimals,
negs = FALSE,
sci = FALSE,
big_mark = "",
leave_as_string = FALSE,
commas = FALSE
)
A character vector.
The pattern to look for.
The default interpretation is a regular expression, as described in stringi::about_search_regex.
To match a without regular expression (i.e. as a human would), use
coll(). For details see stringr::regex()
.
Vectors of integerish values. Must be either length 1 or have
length equal to the length of string
. Negative indices count from the
back: while 1
and 2
correspond to first and second, -1
and -2
correspond to last and second-last. 0
will return NA
.
Do you want to include the possibility of decimal numbers
(TRUE
) or not (FALSE
, the default).
Do you want to allow a leading decimal point to be the start of a number?
Do you want to allow negative numbers? Note that double negatives are not handled here (see the examples).
Make the search aware of scientific notation e.g. 2e3 is the same as 2000.
A character. Allow this character to be used as a thousands
separator. This character will be removed from between digits before they
are converted to numeric. You may specify many at once by pasting them
together e.g. big_mark = ",_"
will allow both commas and underscores.
Internally, this will be used inside a []
regex block so e.g. "a-z"
will behave differently to "az-"
. Most common separators (commas, spaces,
underscores) should work fine.
Do you want to return the number as a string (TRUE
)
or as numeric (FALSE
, the default)?
Deprecated. Use big_mark
instead.
A numeric or character vector.
Other numeric extractors:
str_extract_numbers()
,
str_nth_number()
,
str_nth_number_before_mth()
string <- c(
"abc1abc2abc3abc4abc5abc6abc7abc8abc9",
"abc1def2ghi3abc4def5ghi6abc7def8ghi9"
)
str_nth_number_after_mth(string, "abc", 1, 3)
#> [1] 3 7
str_nth_number_after_mth(string, "abc", 2, 3)
#> [1] 4 8
str_nth_number_after_first(string, "abc", 2)
#> [1] 2 2
str_nth_number_after_last(string, "abc", -1)
#> [1] 9 9
str_first_number_after_mth(string, "abc", 2)
#> [1] 2 4
str_last_number_after_mth(string, "abc", 1)
#> [1] 9 9
str_first_number_after_first(string, "abc")
#> [1] 1 1
str_first_number_after_last(string, "abc")
#> [1] 9 7
str_last_number_after_first(string, "abc")
#> [1] 9 9
str_last_number_after_last(string, "abc")
#> [1] 9 9