This RegEx would do it I think: \s(?=.*\s\d*\.)
Looks for space - \s
Before any strings followed by space, numbers, period
text <- "STR ING 01. Remainder of the string"
stringr::str_replace_all(text, "\\s(?=.*\\s\\d*\\.)", "")
Should do it I think!
On 2020-07-28 21:34, Dennis Fisher wrote:
It is possible that there will be > 1 space. But, most likely only
one (i.e., a solution for one space will suffice; a solution for > 1
space would be even better)
Dennis Fisher MD
P < (The "P Less Than" Company)
Phone / Fax: 1-866-PLessThan (1-866-753-7784)
www.PLessThan.com [1]
On Jul 28, 2020, at 1:29 PM, cpolw...@chemo.org.uk wrote:
On 2020-07-28 21:20, Dennis Fisher wrote:
R 4.0.2
OS X
Colleagues
I have strings that contain a space in an unexpected location.
The
intended string is:
“STRING 01. Remainder of the string"
However, variants are:
“STR ING 01. Remainder of the string"
“STRIN G 01. Remainder of the string"
I would like a general approach to deleting a space, but only if
it
appears before the period. Any suggestions on a regular
expression
for this?
You aren't deleting the space before 0? Is that in the requirement?
Links:
------
[1] http://www.plessthan.com/
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