[GRASS-user] Re: v.what.rast (large file)

GRASS GRASS user
1

Hi Glynn

thanks for your suggestion,

i do not know the c code...
i must study it...

to walk around the problem
i used a bash script,
its sintax is like :
v.out.ascii | r.what > file_raster
then using awk
i extract the column of raster value from file_raster
and past it in the original vector file (ascii vector points )
then re-import the vector file using v.in.ascii -r (region of my raster)

Il giorno 11/apr/07, alle ore 14:22, Glynn Clements <glynn@gclements.plus.com> ha scritto:

Message: 2
Date: Mon, 9 Apr 2007 09:48:15 +0100
From: Glynn Clements <glynn@gclements.plus.com>
Subject: Re: [GRASS-user] v.what.rast (large file)
To: Massimo Di Stefano <massimodisasha@yahoo.it>
Cc: grassuser@grass.itc.it, grass-dev@grass.itc.it
Message-ID: <17945.64975.520050.289809@cerise.gclements.plus.com>
Content-Type: text/plain; charset=us-ascii

Massimo Di Stefano wrote:

i've a large point vector file (ascii 160 mb)

can i divide my area in many subregion
and then patch the results ?

The categories are modified in-place, so you can just run the command
on each subregion; there's no need to "patch" anything.

AFAICT, the problem is in the code which deletes duplicate categories:

    i = 1;
    while ( i < point_cnt ) {
        if ( cache[i].cat == cache[i-1].cat ) {
      cache[i-1].count++;
      for ( j = i; j < point_cnt - 1; j++ ) {
    cache[j].row = cache[j+1].row;
    cache[j].col = cache[j+1].col;
    cache[j].cat = cache[j+1].cat;
    cache[j].count = cache[j+1].count;
      }
      point_cnt--;
      continue;
  }
        i++;
    }

This is O(n^2), but thiscould be done in O(n). The problem is that
it's shifting the remainder of the array down by one place at each
step, when it could move each entry to its final destination in one
go. E.g. (untested):

    i = j = 1;
    while ( i < point_cnt ) {
        while ( i < point_cnt && cache[i].cat == cache[j-1].cat ) {
            cache[j-1].count++;
            i++;
        }
        cache[j].row = cache[i].row;
        cache[j].col = cache[i].col;
        cache[j].cat = cache[i].cat;
        cache[j].count = cache[i].count;
        i++;
        j++;
    }
    point_cnt = j;