Re: how do I do a LOT of replaces to a string?

Eric Sosman <esosman@ieee-dot-org.invalid>
Wed, 06 May 2009 08:40:18 -0400
Stryder wrote:

I'm trying to do a lot (several hundred) of replacements in a string.
These are just string replacements I'm trying to do, not regular
expressions. Here's the WRONG way to do it, but hopefully it tells
you what it is I'm trying to do...

    static String unescapeString(String string) {
        Iterator i = entitiesHashMap.keySet().iterator();

     What's this for? A left-over from an earlier version?

        for (String key : entitiesHashMap.keySet()) {
            System.out.println(key + ":" + (String) entitiesHashMap.get
            string = string.replaceAll(key, (String)

        return string;

entitiesHashmap is a HashMap with literally hundreds of entries.

     It seems to me there must be some conditions on the
universe of keys and replacements if the transformation is
to be meaningful. For example, if one key is a substring
of another there's an ambiguity depending on which one you
search for first. Or if a key is a substring of some other
key's replacement (or even a partial overlap, in unlucky
situations) you get a similar order dependence.

     What I'm driving at is that the additional conditions
might allow you to simplify the searching and/or to build a
better-tuned data structure. In a really simple case, maybe
all the keys begin with a distinguishing character like "#",
in which case you might proceed by scanning the original
string for a "#" and then looking in your map for successively
longer substrings: "#a", no, "#as", no, "#ask", aha! replace
"#ask" with "query". (If you're *really* lucky there'll be
delimiters at both ends, as in "Hello, [THING]!".)

     Even without such a simple delimiting scheme, suitable
conditions on the keys ought to let you build something like
a modified trie. Or as a really sloppy hack you could glom
all the keys together into one giant regex. For example,
from the transformations

    "aluminium" -> "aluminum"
    "colour" -> "color"
    "parlour" -> "rumpus room"

you might build the regex "(aluminium|colour|parlour)" and
look for a match. Having found a match you'd then have to
consult your map to find the replacement, which seems a bit
of a shame -- but I said "sloppy," did I not? Then you'd
resume searching after the end of the replaced stretch, and
keep on going until there are no more matches.

Eric Sosman

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