The reason can be inferred from the difference between NUL : No character. It is used for filling in time or filling space on the surface (such as surface of platter) of storage device where there are no data. We’ll use this character when we’ll be doing programming for data wipers (destructive and non-destructive both) … Read more
Use the following regex replacement: Find: myObject\[(.*?)\] Replace: myObject.get($1) If the index is an integer, you may replace (.*?) with (\d+). The pair of unescaped parentheses creates a capturing group that we may reference from the replacement pattern using $ + Group ID. $1 will insert the index into the replacement result.
This functionality is already implemented in Commons Lang’s StringUtils class. StringUtils.replaceEach(String text, String[] searchList, String[] replacementList)
Rinku is a drop-in replacement for Rails 3.1 auto_link. Auto-linking functionality has been removed from Rails 3.1, and is instead offered as a standalone gem, rails_autolink. You can choose to use Rinku instead. require ‘rails_rinku’ The rails_rinku package monkeypatches Rails with an auto_link method that mimics 100% the original one, parameter per parameter. It’s just … Read more
Give this a shot: @echo off setlocal call :FindReplace “findstr” “replacestr” input.txt exit /b :FindReplace <findstr> <replstr> <file> set tmp=”%temp%\tmp.txt” If not exist %temp%\_.vbs call :MakeReplace for /f “tokens=*” %%a in (‘dir “%3” /s /b /a-d /on’) do ( for /f “usebackq” %%b in (`Findstr /mic:”%~1″ “%%a”`) do ( echo(&Echo Replacing “%~1” with “%~2” in … Read more
A substitution would work nicely. :%s/}/\0\r/g Replace } with the whole match \0 and a new line character \r. or :%s/}/&\r/g Where & also is an alternative for the whole match, looks a bit funny though in my opinion. Vim golfers like it because it saves them a keystroke 🙂 \0 or & in the … Read more
You can try this sed: sed -i.bak ‘s/^\(VAR5=\).*/\1VALUE10/’ file It gives: VAR1=VALUE1 VAR2=VALUE2 VAR3=VALUE3 VAR4=VALUE4 VAR5=VALUE10 VAR6=VALUE6
Numeric columns have no ,, so converting to strings is not necessary, only use DataFrame.replace with regex=True for substrings replacement: df = df.replace(‘,’,”, regex=True) Or: df.replace(‘,’,”, regex=True, inplace=True) And last convert strings columns to numeric, thank you @anki_91: c = df.select_dtypes(object).columns df[c] = df[c].apply(pd.to_numeric,errors=”coerce”)
You can use regex myString.replace(/[^\w\s!?]/g,”); This will replace everything but a word character, space, exclamation mark, or question. Character Class: \w stands for “word character”, usually [A-Za-z0-9_]. Notice the inclusion of the underscore and digits. \s stands for “whitespace character”. It includes [ \t\r\n]. If you don’t want the underscore, you can use just [A-Za-z0-9]. … Read more
Use this: str.replace(/\s+/g, ”); Instead of this: str.trim()