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? Dennis Dennis Fisher MD P < (The "P Less Than" Company) Phone / Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com <http://www.plessthan.com/> [[alternative HTML version deleted]]
cpoiw@rt m@iii@g oii chemo@org@uk
2020-Jul-28 20:29 UTC
[R] Removing a space from a string
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?
Only the spaces in STRING. However, if I inadvertently delete the space between STRING and NN, I can add it back in. Dennis Fisher MD P < (The "P Less Than" Company) Phone / Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com <http://www.plessthan.com/>> On Jul 28, 2020, at 1:29 PM, cpolwart at 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?[[alternative HTML version deleted]]
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 <http://www.plessthan.com/>> On Jul 28, 2020, at 1:29 PM, cpolwart at 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?[[alternative HTML version deleted]]
Dear Dennis,
On 2020-07-28 13:20 -0700, Dennis Fisher wrote:
| 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?
Perhaps by using gregexpr to look for
dots, remove spaces from the substring until the first
finding, then pasting it back.
strings <-
c("STRING 01. Remainder of the string.",
"STR ING 01. Remainder of the string.",
"STRIN G 01. Remainder of the string.")
search <- gregexpr("\\.", strings)
lens <- nchar(strings)
FUN <- function(i, strings, search, lens) {
before.dot <- substr(strings[i], 1, search[[i]][1])
before.dot <- gsub(" ", "", before.dot)
after.dot <- substr(strings[i], search[[i]][1]+1, lens[i])
return(paste0(before.dot, after.dot))
}
simplify2array(parallel::mclapply(
X=1:length(strings),
FUN=FUN,
strings=strings,
search=search,
lens=lens))
yields
[1] "STRING01. Remainder of the string."
[2] "STRING01. Remainder of the string."
[3] "STRING01. Remainder of the string."
Yes, I know, the space just before 01
also disappears ...
Best,
Rasmus
On 2020-07-28 23:00 +0200, Rasmus Liland wrote:
|
| Perhaps by using gregexpr to look for
| dots, remove spaces from the substring until the first
| finding, then pasting it back.
|
| strings <-
| c("STRING 01. Remainder of the string.",
| "STR ING 01. Remainder of the string.",
| "STRIN G 01. Remainder of the string.")
|
| search <- gregexpr("\\.", strings)
| lens <- nchar(strings)
| FUN <- function(i, strings, search, lens) {
| before.dot <- substr(strings[i], 1, search[[i]][1])
| before.dot <- gsub(" ", "", before.dot)
| after.dot <- substr(strings[i], search[[i]][1]+1, lens[i])
| return(paste0(before.dot, after.dot))
| }
| simplify2array(parallel::mclapply(
| X=1:length(strings),
| FUN=FUN,
| strings=strings,
| search=search,
| lens=lens))
|
| yields
|
| [1] "STRING01. Remainder of the string."
| [2] "STRING01. Remainder of the string."
| [3] "STRING01. Remainder of the string."
|
| Yes, I know, the space just before 01
| also disappears ...
I forgot about regexpr ... this is
simpler I think:
strings <-
c("STRING 01. Remainder of the string.",
"STR ING 01. Remainder of the string.",
"STRIN G 01. Remainder of the string.")
search <- regexpr("...\\.", strings) # search for the first dot
and three chars in front of it
ml <- attr(search, "match.length")
paste0(
gsub(" ", "", substr(strings, 1, search)),
substr(strings, search, search+ml-1),
substr(strings, search+ml, nchar(strings))
)
/Rasmus
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1. Thanks for the nice reprex. 2. However, I thought there was still a bit of ambiguity. I interpreted your specification to mean: "any number of spaces could occur in the beginning alphabetic part of the strings before one or more digits occur followed by a '.' (a period) and then more stuff after." 3. My strategy was simply to split the strings into the first part consisting of the alphabetic characters and spaces and the second part with the numbers and everything else. Then I just removed the spaces in the first part. You can then concatenate them together again (using paste()) however you like. Thus >x [1] "STRING 01. Remainder of the string" "STR ING 01. Remainder of the string" [3] "STRIN G 01. Remainder of the string"> p1 <-gsub(" ","",gsub("([^[:digit:]]+)[[:digit:]]+\\..*$","\\1",x)) > p2 <- gsub("[^[:digit:]]+([[:digit:]]+\\..*$)","\\1",x) > p1[1] "STRING" "STRING" "STRING"> p2[1] "01. Remainder of the string" "01. Remainder of the string" [3] "01. Remainder of the string" I look forward to better approaches using basic regex's (no additional packages), however. Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Jul 28, 2020 at 1:20 PM Dennis Fisher <fisher at plessthan.com> 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? > > Dennis > > Dennis Fisher MD > P < (The "P Less Than" Company) > Phone / Fax: 1-866-PLessThan (1-866-753-7784) > www.PLessThan.com <http://www.plessthan.com/> > > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >[[alternative HTML version deleted]]
Note that my previous strategy can be expressed slightly more clearly as:
x <- c("STRING 01. Remainder of the string",
"STR ING 01. Remainder of the string",
"STRIN G 01. Remainder of the string",
"STR IN G 01. Remainder of the string") ## more spaces in this last
example entry
rx <- "([^[:digit:]]+)([[:digit:]]+.+)"
> gsub(" ","",gsub(rx,"\\1",x))
[1] "STRING" "STRING" "STRING" "STRING"
> gsub(rx,"\\2",x)
[1] "01. Remainder of the string" "01. Remainder of the
string"
[3] "01. Remainder of the string" "01. Remainder of the
string"
Bert Gunter
On Tue, Jul 28, 2020 at 2:53 PM Bert Gunter <bgunter.4567 at gmail.com>
wrote:
> 1. Thanks for the nice reprex.
> 2. However, I thought there was still a bit of ambiguity. I interpreted
> your specification to mean: "any number of spaces could occur in the
> beginning alphabetic part of the strings before one or more digits occur
> followed by a '.' (a period) and then more stuff after."
> 3. My strategy was simply to split the strings into the first part
> consisting of the alphabetic characters and spaces and the second part with
> the numbers and everything else. Then I just removed the spaces in the
> first part. You can then concatenate them together again (using paste())
> however you like. Thus
>
> >x
> [1] "STRING 01. Remainder of the string" "STR ING 01.
Remainder of the
> string"
> [3] "STRIN G 01. Remainder of the string"
> > p1 <-gsub("
","",gsub("([^[:digit:]]+)[[:digit:]]+\\..*$","\\1",x))
> > p2 <-
gsub("[^[:digit:]]+([[:digit:]]+\\..*$)","\\1",x)
> > p1
> [1] "STRING" "STRING" "STRING"
> > p2
> [1] "01. Remainder of the string" "01. Remainder of the
string"
> [3] "01. Remainder of the string"
>
> I look forward to better approaches using basic regex's (no additional
> packages), however.
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
and
> sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip
)
>
>
> On Tue, Jul 28, 2020 at 1:20 PM Dennis Fisher <fisher at
plessthan.com>
> 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?
>>
>> Dennis
>>
>> Dennis Fisher MD
>> P < (The "P Less Than" Company)
>> Phone / Fax: 1-866-PLessThan (1-866-753-7784)
>> www.PLessThan.com <http://www.plessthan.com/>
>>
>>
>>
>>
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
[[alternative HTML version deleted]]
The first response has to be "how did the spaces get there in the first place?" Can you fix the process that creates the data? If the process sometimes generates one extra space, are you sure it never generates two? But let's treat this purely as a regular expression problem, where if there is a space before a dot you want to delete the first. In vi(1) you would do s/^\([^ .]*\) \([^.]*\)/\1\2/ but apparently there is *supposed* to be a space before the 01, so it is only when there are two or more spaces that one should be deleted, so we'd want s/^\([^ .]*\) \([^ .]* \)/\1\2/ I leave converting that to R as an exercise for the reader. On Wed, 29 Jul 2020 at 08:20, Dennis Fisher <fisher at plessthan.com> 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? > > Dennis > > Dennis Fisher MD > P < (The "P Less Than" Company) > Phone / Fax: 1-866-PLessThan (1-866-753-7784) > www.PLessThan.com <http://www.plessthan.com/> > > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >[[alternative HTML version deleted]]
Richard In reply to your ?first response?, the text was originally in a Word document and it did NOT contain the errant spaces. I used read_docx in the textreadr package to access the text. The spaces were added during that step. I am copying the maintainer of that package to see if he has any idea as to the source. Thanks for your regular expression suggestion. Dennis Dennis Fisher MD P < (The "P Less Than" Company) Phone / Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com <http://www.plessthan.com/>> On Jul 28, 2020, at 5:11 PM, Richard O'Keefe <raoknz at gmail.com> wrote: > > The first response has to be "how did the spaces get there > in the first place?" Can you fix the process that creates > the data? If the process sometimes generates one extra > space, are you sure it never generates two? > > But let's treat this purely as a regular expression > problem, where if there is a space before a dot you want > to delete the first. In vi(1) you would do > > s/^\([^ .]*\) \([^.]*\)/\1\2/ > > but apparently there is *supposed* to be a space before > the 01, so it is only when there are two or more spaces > that one should be deleted, so we'd want > > s/^\([^ .]*\) \([^ .]* \)/\1\2/ > > I leave converting that to R as an exercise for the reader. > > > > > On Wed, 29 Jul 2020 at 08:20, Dennis Fisher <fisher at plessthan.com <mailto:fisher at plessthan.com>> 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? > > Dennis > > Dennis Fisher MD > P < (The "P Less Than" Company) > Phone / Fax: 1-866-PLessThan (1-866-753-7784) > www.PLessThan.com <http://www.plessthan.com/> <http://www.plessthan.com/ <http://www.plessthan.com/>> > > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org <mailto:R-help at r-project.org> mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help <https://stat.ethz.ch/mailman/listinfo/r-help> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html <http://www.r-project.org/posting-guide.html> > and provide commented, minimal, self-contained, reproducible code.[[alternative HTML version deleted]]