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PORegular Expression to split on specific character ONLY if that character is not in a pair
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copied!<p>After finding the fastest string replace algorithm in <a href="https://stackoverflow.com/questions/1919096/mass-string-replace-in-python">this thread</a>, I've been trying to modify one of them to suit my needs, particularly <a href="https://stackoverflow.com/questions/1919096/mass-string-replace-in-python/1919221#1919221">this one</a> by gnibbler.</p> <p>I will explain the problem again here, and what issue I am having.</p> <p>Say I have a string that looks like this:</p> <pre><code>str = "The &yquick &cbrown &bfox &Yjumps over the &ulazy dog" </code></pre> <p>You'll notice a lot of locations in the string where there is an ampersand, followed by a character (such as "&y" and "&c"). I need to replace these characters with an appropriate value that I have in a dictionary, like so:</p> <pre><code>dict = {"y":"\033[0;30m", "c":"\033[0;31m", "b":"\033[0;32m", "Y":"\033[0;33m", "u":"\033[0;34m"} </code></pre> <p>Using gnibblers solution provided in my previous thread, I have this as my current solution:</p> <pre><code>myparts = tmp.split('&') myparts[1:]=[dict.get(x[0],"&"+x[0])+x[1:] for x in myparts[1:]] result = "".join(myparts) </code></pre> <p>This works for replacing the characters properly, and does not fail on characters that are not found. The only problem with this is that there is no simple way to <strong>actually</strong> keep an ampersand in the output. The easiest way I could think of would be to change my dictionary to contain:</p> <pre><code>dict = {"y":"\033[0;30m", "c":"\033[0;31m", "b":"\033[0;32m", "Y":"\033[0;33m", "u":"\033[0;34m", "&":"&"} </code></pre> <p>And change my "split" call to do a regex split on ampersands that are NOT followed by other ampersands.</p> <pre><code>>>> import re >>> tmp = "&yI &creally &blove A && W &uRootbeer." >>> tmp.split('&') ['', 'yI ', 'creally ', 'blove A ', '', ' W ', 'uRootbeer.'] >>> re.split('MyRegex', tmp) ['', 'yI ', 'creally ', 'blove A ', '&W ', 'uRootbeer.'] </code></pre> <p>Basically, I need a Regex that will split on the <strong>first ampersand of a pair</strong>, and every <strong>single</strong> ampersand, to allow me to escape it via my dictionary.</p> <p>If anyone has any better solutions please feel free to let me know.</p>

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