Replacing two characters I timed all the methods in the current answers along with one extra. With an input string of abc&def#ghi and replacing & -> \& and # -> \#, the fastest way was to chain together the replacements like this: text.replace(‘&’, ‘\&’).replace(‘#’, ‘\#’). Timings for each function: a) 1000000 loops, best of 3: … Read more
As in 2.x, use str.replace(). Example: >>> ‘Hello world’.replace(‘world’, ‘Guido’) ‘Hello Guido’
Try using a list comprehension and a conditional expression. >>> a=[1,2,3,1,3,2,1,1] >>> [4 if x==1 else x for x in a] [4, 2, 3, 4, 3, 2, 4, 4]
The third parameter of String.prototype.replace() function was never defined as a standard, so most browsers simply do not implement it. The best way is to use regular expression with g (global) flag. var myStr=”this,is,a,test”; var newStr = myStr.replace(/,/g, ‘-‘); console.log( newStr ); // “this-is-a-test” Still have issues? It is important to note, that regular expressions … Read more
Or maybe you could wrap ‘Facts’ round a <span> as follows: .pvw-title span { display: none; } .pvw-title:after { content: ‘whatever it is you want to add’; } <div class=”pvw-title”><span>Facts</span></div>
Here is a short example that should do the trick with regular expressions: import re rep = {“condition1”: “”, “condition2”: “text”} # define desired replacements here # use these three lines to do the replacement rep = dict((re.escape(k), v) for k, v in rep.iteritems()) #Python 3 renamed dict.iteritems to dict.items so use rep.items() for latest … Read more
String are immutable in Java. You can’t change them. You need to create a new string with the character replaced. String myName = “domanokz”; String newName = myName.substring(0,4)+’x’+myName.substring(5); Or you can use a StringBuilder: StringBuilder myName = new StringBuilder(“domanokz”); myName.setCharAt(4, ‘x’); System.out.println(myName);
This tested snippet should do it: import re line = re.sub(r”</?\[\d+>”, “”, line) Edit: Here’s a commented version explaining how it works: line = re.sub(r””” (?x) # Use free-spacing mode. < # Match a literal ‘<‘ /? # Optionally match a “https://stackoverflow.com/” \[ # Match a literal ‘[‘ \d+ # Match one or more digits … Read more
Not sure what would perform faster but this would be more reliable when it comes to extension like .jpeg or .html x.replace(/\.[^/.]+$/, “”)
You need to escape the . because it has the meaning of “an arbitrary character” in a regular expression. mystring = mystring.replace(/\./g,’ ‘)