QStringList Sorting

[JohannesMunk]

Ok, I see, too quick
But with "abc-2.00234" and "abc-2.12352" n1 == n2 == 2?

No! "abc-2." would fall into the common prefix (while s1.at(i) == s2.at(i)). Thus: n1="00234", n2="12352"

Anyway, may it not be better/quicker first to determinate the number on the end of the string, and then chop the x characters (x = length of the determinated number)? Of course it depends of the string list you compare. Many different prefixes exterminating from the front would be faster...

Well the last number could be equal. I think the approach to find the biggest common prefix and then look for the difference is the way to go. Thats how a natural/human sort does work anyway.

Also interesting would be a speed comparison with a member reg expression...

There is no doubt in my mind, that the explicit while loop will be faster. Even when you use a member RegExp, so that its mask-parsing is only done once.. the reg-exp-fitting code is much more general and thus carries a lot of overhead, compared to this specialized solution..

The code would perhaps be more readable.. But I'm not so sure :->

Greetings!

Joh

[JohannesMunk]

Thinking about it again.. my natural sort doesn't work for "a2002, a20001". Here n1 would be ="2" and n2="01". So n2.toInt() < n1.toInt().. which is wrong, when you look at the whole number..

So from the first different character we also need to look backward and add all adjacent numbers:

bool compareNames(const QString& s1,const QString& s2)
{
   // ignore common prefix..
   int i = 0;
   while ((i < s1.length()) && (i < s2.length()) && (s1.at(i).toLower() == s2.at(i).toLower())) ++i;
   ++i;
   // something left to compare?
   if ((i < s1.length()) && (i < s2.length())) 
   {
      // get number prefix from position i - doesnt matter from which string
      int k = i-1;QString n = "";
      while ((k >= 0) && (s1.at(k).isNumber())) {n = s1.at(k)+n;--k;}
 
      // get relevant/signficant number string for s1
      k = i;QString n1 = "";
      while ((k < s1.length()) && (s1.at(k).isNumber()) {n1 += s1.at(k);++k;}
 
      // get relevant/signficant number string for s2
      k = i;QString n2 = "";
      while ((k < s2.length()) && (s2.at(k).isNumber()) {n2 += s2.at(k);++k;}
 
      // got two numbers to compare?
      if (!n1.isEmpty() && !n2.isEmpty())
      {
          return (n+n1).toInt() < (n+n2).toInt();
      }
      else {
          // not a number has to win over a number.. number could have ended earlier... same prefix..
         if (!n1.isEmpty()) return false;
         if (!n2.isEmpty()) return true;
         return s1.at(i) < s2.at(i);
      }
   }
   else {
      // shortest string wins
      return s1.length() < s2.length();
   }
}

bool compareNames(const QString& s1,const QString& s2)
{
   // ignore common prefix..
   int i = 0;
   while ((i < s1.length()) && (i < s2.length()) && (s1.at(i).toLower() == s2.at(i).toLower())) ++i;
   ++i;
   // something left to compare?
   if ((i < s1.length()) && (i < s2.length())) 
   {
      // get number prefix from position i - doesnt matter from which string
      int k = i-1;QString n = "";
      while ((k >= 0) && (s1.at(k).isNumber())) {n = s1.at(k)+n;--k;}

      // get relevant/signficant number string for s1
      k = i;QString n1 = "";
      while ((k < s1.length()) && (s1.at(k).isNumber()) {n1 += s1.at(k);++k;}
 
      // get relevant/signficant number string for s2
      k = i;QString n2 = "";
      while ((k < s2.length()) && (s2.at(k).isNumber()) {n2 += s2.at(k);++k;}
 
      // got two numbers to compare?
      if (!n1.isEmpty() && !n2.isEmpty())
      {
          return (n+n1).toInt() < (n+n2).toInt();
      }
      else {
          // not a number has to win over a number.. number could have ended earlier... same prefix..
         if (!n1.isEmpty()) return false;
         if (!n2.isEmpty()) return true;
         return s1.at(i) < s2.at(i);
      }
   }
   else {
      // shortest string wins
      return s1.length() < s2.length();
   }
}

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[^Nisok^]

I want to do the same thing sort of a class QStrings as integers. But I don't know where to put the functinon compareNames. As a class function? Or should I extend the qSort class?

[faldzip]

qSort is not a class but function.
make your function compareNames as a global function or static class method or a struct with operator() (and pass object of that struct).

[soxs060389]

I'd suggest first to determine the name sheme before throwing solutions at each other and then just use the one that fits the purpose....

[QwertyManiac]

For an implementation of Natural-Order Sorting (Case sensitive or non) you can have a look at the GPL code by Tobias Koenig <tokoe@kde.org> in these links: [Header - qnatsort.h] [Source - qnatsort.cpp]

An implementation using qSort() is explained in the comments - you simply have to pass the included function.