Hi Nick,
back to office I was eager to try by myself. Actually it seems that the
result of multiple AND or multiple layers - I didn't check this by now -
results in values slightly lower 1 (e.g. 0.9995 in my case). And
therefore maybe rounded to "0". What I have done is the folowing:
(1/("pm2-5-europe-2001-2010@1">240 AND
"pm2-5-europe-2001-2010@1"<250))*(1/("pm2-5-europe-2001-2010@2">139 AND
"pm2-5-europe-2001-2010@2"<145)) * (1/("pm2-5-europe-2001-2010@3">80 AND
"pm2-5-europe-2001-2010@3"<85))
which results in a perfektly fitting mask of my pseudo demand.
Mayby you could verfy this with your data
cheers
Stefan
Am 31.07.2015 um 08:48 schrieb Stefan Kiefer:
Hi Nick,
you are absolutely right. My thought was, that you get A layer with
distinct values to identify the road. For a mask you are on the right
way, and I either don't understand the behaviour except that you
operate over three layers, which of course should work.
Have you tryed to generate a composit of the three layers and mask the
single values resulting for road structures? (it's more or less what I
expected from my first approach.)
cheers
Stefan
> Nick Papadonis <[email protected]> hat am 31. Juli 2015 um 08:31
geschrieben:
>
>
> Hi Stefan,
>
> It’s my understanding black has a value of 0 in the resulting layer.
>
> I tried this and it results in similar image to step (a) and also
includes other colors at lower intensities mixed in with the red. The
red has the highest intensity in the greyscale. I’m looking to create
a binary image with just the colors of red in the palette I choose and
using this trace vectors over the paths.
>
> Thanks,
> Nick
>
> > On Jul 31, 2015, at 2:04 AM, Stefan Kiefer <[email protected]> wrote:
> >
> > Hi Nick,
> > I believe it is black bcause you always get a value of "1".
Unfortunately I can not verify this, because I have no QGis by this
moment. Most propably you wanted to calculate:
> >
> > (“m@1" < 238 AND “m@1" > 213 AND “m@2" < 123 AND “m@2" > 98 AND
“m@3" < 125 AND “m@3” > 99) * ((“m@1" < 238 AND “m@1" > 210) * "m@1")
+ ((“m@2" < 123 AND “m@2" > 94) * “m@2") + ((“m@3" < 130 AND “m@3" >
98) *“m@3"))
> >
> > cheers
> >
> > Stefan
> >
> > > Nick Papadonis <[email protected]> hat am 31. Juli 2015 um
07:49 geschrieben:
> > >
> > >
> > > One more comment. The resulting layer histogram is showing the
pixel range spread over frequency in floating point values. Is the
raster calculator performing floating point math with potential
rounding error?
> > >
> > > I found it also interesting that the following expression
resulted in a layer, which when inspected for band values, has integer
values of 2 and 3. 3 being the value I want for the red route.
> > >
> > > a) ((“m@1" < 238 AND “m@1" > 210) * 1) + ((“m@2" < 123 AND “m@2"
> 94) * 1) + ((“m@3" < 130 AND “m@3" > 98) * 1)
> > >
> > > I then change the expression to only use values 2 and greater
and this shows properly:
> > > b) ((“m@1" < 238 AND “m@1" > 210) * 1) + ((“m@2" < 123 AND “m@2"
> 94) * 1) + ((“m@3" < 130 AND “m@3" > 98) * 1) > 2
> > >
> > > I then changed the expression to ensure all three values are
obtained and it results in a black image of 0’s. I was expecting only
the red route to appear as it resulted in value of 3 in step (a).
> > >
> > > ((“m@1" < 238 AND “m@1" > 210) * 1) + ((“m@2" < 123 AND “m@2" >
94) * 1) + ((“m@3" < 130 AND “m@3" > 98) * 1) > 2.1
> > > ((“m@1" < 238 AND “m@1" > 210) * 1) + ((“m@2" < 123 AND “m@2" >
94) * 1) + ((“m@3" < 130 AND “m@3" > 98) * 1) >= 3
> > >
> > > I’m wondering how much testing the Raster Calculator has gone
through and if there is a possible bug here. Perhaps something to do
with floating point?
> > >
> > > Thanks again
> > >
> > > > On Jul 31, 2015, at 12:39 AM, Nick Papadonis
<[email protected]> wrote:
> > > >
> > > > Folks,
> > > >
> > > > I’m using QGIS 10.1. The following expressions result in a
black raster of 0’s, when I expected only red pixels to appears in the
binary image indicating routes on a map:
> > > >
> > > > a) (“m@1" < 238 AND “m@1" > 213 AND “m@2" < 123 AND “m@2" > 98
AND “m@3" < 125 AND “m@3” > 99) * 1
> > > > b) ((“m@1" < 238 AND “m@1" > 210) * 1) * ((“m@2" < 123 AND
“m@2" > 94) * 1) * ((“m@3" < 130 AND “m@3" > 98) * 1)
> > > >
> > > > I then tried the following individual expressions for each
band as separate steps (sanity check) and they work to cover the
pixels in range:
> > > > c) (“m@1" < 238 AND “m@1" > 213) * 1
> > > > d) (“m@2" < 123 AND “m@2" > 98) * 1
> > > > e) (“m@3" < 125 AND “m@3” > 99) * 1
> > > >
> > > > I then tried the following expression which appears to create
a proper greyscale image focusing on the red pixels. I replaced the
multiplication with addition to see what was happening:
> > > > f) ((“m@1" < 238 AND “m@1" > 210) * 1) + ((“m@2" < 123 AND
“m@2" > 94) * 1) + ((“m@3" < 130 AND “m@3" > 98) * 1)
> > > >
> > > > The resulting raster has a Min = 0 and Max = 1.998. I was
expecting it to be Min = 0 and Max = 3. The value of 3 would indicate
all 3 bands were positive on color match. I then go to the layer
properties and load calculate min/max again and it is Min = 0 and Max
= 3. I tried to change the min/max settings on they layer and these
settings will not stay set. The layer goes back to Max = 1.998. What’s
even more odd is the max being a floating point number. I suspect that
may be part of the issue. Anyone know why this is the case for integer
band values? Has anyone successfully used the Raster Calculator to
perform this sort of work before?
> > > >
> > > > Thanks again,
> > > > Nick
> > >
> > > _______________________________________________
> > > Qgis-user mailing list
> > > [email protected]
> > > http://lists.osgeo.org/mailman/listinfo/qgis-user
>
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