Thu, Dec 9 2004
00:16:08
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Request created by jidanni@jidanni.org
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Subject |
v.to.points: better document lfield
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From |
Dan Jacobson <jidanni@jidanni.org>
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Date |
Thu, 09 Dec 2004 04:17:54 +0800
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In the v.to.points documentation, mention what
lfield Line field default: 1
is all about.
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Thu, Dec 9 2004
18:16:27
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Mail sent by neteler@itc.it
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Thu, 9 Dec 2004 18:16:24 +0100
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Markus Neteler <neteler@itc.it>
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Request Tracker <grass-bugs@intevation.de>
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grass5@grass.itc.it
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Subject |
Re: [GRASS5] [bug #2822] (grass) v.to.points: better document lfield
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<20041209171624.GD3487@thuille.itc.it>
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On Thu, Dec 09, 2004 at 12:16:08AM +0100, Request Tracker wrote:
> this bug's URL: http://intevation.de/rt/webrt?serial_num=2822
> -------------------------------------------------------------------------
>
> In the v.to.points documentation, mention what
> lfield Line field default: 1
> is all about.
Was changed a couple of days ago:
v.to.points [-vit] input=name [type=name[,name,...]] output=name
[llayer=value] [dmax=value]
...
llayer Line layer
default: 1
So it refers to line layer one.
Markus
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Mon, Dec 12 2005
12:31:45
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Mail sent by msieczka
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Dan wrote:
>> In the v.to.points documentation, mention what
>> lfield Line field default: 1
>> is all about.
Markus wrote:
>Was changed a couple of days ago:
>
> v.to.points [-vit] input=name [type=name[,name,...]] output=name
> [llayer=value] [dmax=value]
>...
> llayer Line layer
default: 1
> So it refers to line layer one.
Is the "llayer" an output or input option?
I'm asking because it is mentioned after the "output=string" in help, but as
it refers to lines ("Line layer"), seems it is actually about the input.
Maciek |
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Mon, Dec 12 2005
12:31:52
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Area changed to grass6 by msieczka
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Tue, Sep 26 2006
19:21:23
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Mail sent by guest
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I think I've figured this out, after much experimentation.
I created a new vector using v.digit -n map=test_lines in Spearfish. I
digitized three separate lines, making sure that each line was assigned to
layers 1,2 and 3, respectively. I then ran v.to.points on test_lines 3 times,
each time making llayer equal to 1, 2, and 3 respectively.
Here are the results I found by querying each output vector in turn:
1) The layer value given in llayer selects which layer's attributes in the
input lines vector get assigned to layer one of the output points vector.
2) If there are more than one layers in the input lines vector, all of the
layers NOT selected by llayer get assigned a layer number of "-1" in the
output points vector.
3) All lines from the input vector are converted to points, regardless of
which layer they're on. So llayer doesn't act as a geometry filter. It only
affects which layer's attributes get assigned to layer one of the output vector.
So, when I run:
v.to.points input=test_lines output=llayer_one_out llayer=1 dmax=300
all of the attributes I created in v.digit for layer one can now be queried on
layer one of the output vector points. If I now do:
v.to.points input=test_lines output=llayer_two_out llayer=2 dmax=300
Querying the points that came from lines on layer one of the input vector now
give a cat of -1, and points that came from lines on layer two of input vector
now show all of input vector attributes from layer two.
In all cases, layer two of output points vector show unique categories for
each point, as well as the cat of the input line, and the along distance.
Can you verify this before I update the docs?
~ Eric. |
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Tue, Sep 26 2006
19:30:51
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Mail sent by guest
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Correction: Item #2 above should read:
...all of the layers NOT selected by llayer get assigned a category number of
"-1" in the output points vector.
instead of:
..all of the layers NOT selected by llayer get assigned a layer number of "-1"
in the output points vector.
~ Eric.
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Sat, Nov 4 2006
18:23:50
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Mail sent by mneteler
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Was this ever added to the docs?
Markus
https://intevation.de/rt/webrt?serial_num=2822 |
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