What about s<C-r>0, this can be repeated with . s deletes the character under the curser and puts you into insert mode. <C-r>0 inserts register 0, which holds the yanked text. You can also use s and type foo manually, which is also repeatable with .
Your answer is in the Matcher#appendReplacement documentation. Just put your function call in the while loop. [The appendReplacement method] is intended to be used in a loop together with the appendTail and find methods. The following code, for example, writes one dog two dogs in the yard to the standard-output stream: Pattern p = Pattern.compile(“cat”); … Read more
You can use REGEXP_REPLACE since Oracle 10: SELECT REGEXP_REPLACE(‘+34 (947) 123 456 ext. 2013’, ‘[^0-9]+’, ”) FROM DUAL This example returns 349471234562013. Alternative syntaxes include: POSIX character classes: ‘[^[:digit:]]+’ Perl-influenced extensions (since Oracle 11): ‘\D+’
Using Pandas str.replace and regex: df[“new_column”] = df[‘review’].str.replace(‘[^\w\s]’,”)
If all values are unique then you should reverse the dict first to get an efficient solution: >>> subs = { … “Houston”: “HOU”, … “L.A. Clippers”: “LAC”, … … } >>> rev_subs = { v:k for k,v in subs.iteritems()} >>> [rev_subs.get(item,item) for item in my_lst] [‘L.A. Clippers’, ‘Houston’, ’03/03 06:11 PM’, ‘2.13’, ‘1.80’, ’03/03 … Read more
Try this: text.replace(/\n\s*\n/g, ‘\n’); This basically looks for two line breaks with only whitespace in between. And then it replaces those by a single line break. Due to the global flag g, this is repeated for every possible match. edit: is it possibile to leave a double line break instead of a single Sure, simplest … Read more
You can use capturing groups with alternations matching either string boundaries or a character not _ (still using a word boundary): var re = regexp.MustCompile(`(^|[^_])\bproducts\b([^_]|$)`) s := re.ReplaceAllString(sample, `$1.$2`) Here is the Go demo and a regex demo. Notes on the pattern: (^|[^_]) – match string start (^) or a character other than _ \bproducts\b … Read more
In JavaScript, strings are immutable, which means the best you can do is to create a new string with the changed content and assign the variable to point to it. You’ll need to define the replaceAt() function yourself: String.prototype.replaceAt = function(index, replacement) { return this.substring(0, index) + replacement + this.substring(index + replacement.length); } And use … Read more
Finding text in a PDF can be inherently hard because of the graphical nature of the document format — the letters you are searching for may not be contiguous in the file. That said, CAM::PDF has some search-replace capabilities and heuristics. Give changepagestring.pl a try and see if it works on your PDFs. To install: … Read more
Check the help for \%u: /\%d /\%x /\%o /\%u /\%U E678 \%d123 Matches the character specified with a decimal number. Must be followed by a non-digit. \%o40 Matches the character specified with an octal number up to 0377. Numbers below 040 must be followed by a non-octal digit or a non-digit. \%x2a Matches the character … Read more