Finding Nearest Colour

[pepito72]

Hi

A problem with trying to find the nearest colour. I have an image
which I use GetPixel to return the colour of various points.
However .... not all colours returned will be a System/Known/Pre-
defined colour. So they wont have corresponding names - this is a
problem because my app needs to return the colour name of each pixel.
So I have looked at GetNearestColor but this just seems to return the
original color I passed it in the first place!?

Any pointers would be appreciated.

Thanks
Tad

[Andrew Morton]

Jeremy Cowles wrote:

Dim dist As Integer

dist = Math.Abs(CInt(c .R) - known.R) _
+ Math.Abs(CInt(c .G) - known.G) _
+ Math.Abs(CInt(c .B) - known.B)

Wouldn't that be better measured as the root mean square distance?

<snip>
Dim known As Color
Dim dist As Double
For Each colorName As String In _
[Enum].GetNames(GetTy pe(KnownColor))
Color = Color.FromName( colorName)

dist=math.sqrt( (CInt(c.R) - known.R)^2 + (CInt(c.G) - known.G)^2 +
(CInt(c.B) - known.B)^2))

If best.Name Is Nothing OrElse dist < best.Distance Then
best.Color = known
best.Name = colorName
best.Distance = dist
End If
Next
<snip>

(OK, so not any need to take the square root; just use the square of the
distance. The comparisons remain valid.)

Andrew

[Jeremy Cowles]

On Jan 9, 9:18 am, "Andrew Morton" <[e-mail] alid>
wrote:

Jeremy Cowles wrote:

Dim dist As Integer

dist = Math.Abs(CInt(c .R) - known.R) _
+ Math.Abs(CInt(c .G) - known.G) _
+ Math.Abs(CInt(c .B) - known.B)

Wouldn't that be better measured as the root mean square distance?
Andrew

Yes, the RMS method is definitely a better match - good suggestion. I
ran several tests, and consistently about 160 out of 1000 samples
(~16%) are more accurate with RMS. So I guess it depends on weather
you want performance or accuracy, although, the performance gain is
probably going to be negligible.

Jeremy

[Jeremy Cowles]

On Jan 10, 12:49 pm, Jeremy Cowles <jeremy.cow...@ gmail.comwrote:

On Jan 9, 9:18 am, "Andrew Morton" <[e-mail] alid>
wrote:

Jeremy Cowles wrote:

Dim dist As Integer

dist = Math.Abs(CInt(c .R) - known.R) _
+ Math.Abs(CInt(c .G) - known.G) _
+ Math.Abs(CInt(c .B) - known.B)

Wouldn't that be better measured as the root mean square distance?
Andrew

Yes, the RMS method is definitely a better match - good suggestion. I
ran several tests, and consistently about 160 out of 1000 samples
(~16%) are more accurate with RMS. So I guess it depends on weather
you want performance or accuracy, although, the performance gain is
probably going to be negligible.

Jeremy

One thing, I think your RMS calculation is missing something:

dist=math.sqrt( (CInt(c.R) - known.R)^2 + (CInt(c.G) - known.G)^2 +
(CInt(c.B) - known.B)^2))
Shouldn't it be:

dist=math.sqrt(
((CInt(c.R) - known.R)^2 +
(CInt(c.G) - known.G)^2 +
(CInt(c.B) - known.B)^2))
/ 3
)
?

[Andrew Morton]

Jeremy Cowles wrote:

One thing, I think your RMS calculation is missing something:
Shouldn't it be:

dist=math.sqrt(
((CInt(c.R) - known.R)^2 +
(CInt(c.G) - known.G)^2 +
(CInt(c.B) - known.B)^2))
/ 3
)

Hey, it was only wrong by a factor of 1.7-ish :-)

Dim known As Color
Dim dist As Double ' temporary distance measure
For Each colorName As String In _
[Enum].GetNames(GetTy pe(KnownColor))
Color = Color.FromName( colorName)

dist=(CInt(c.R) - known.R)^2 + (CInt(c.G) - known.G)^2 + (CInt(c.B) -
known.B)^2)

If best.Name Is Nothing OrElse dist < best.Distance Then
best.Color = known
best.Name = colorName
best.Distance = Math.Sqrt(dist/3)
End If
Next

Also, the OP may want to consider the idea of a "nearest" colour in a
different colour-space, for example the hue may be more important than the
brightness.

Andrew

[Jeremy Cowles]

Also, the OP may want to consider the idea of a "nearest" colour in a

different colour-space, for example the hue may be more important than the
brightness.

Andrew

Actually, I was thinking the same thing. I created a test with various
combinations, Hue, RGB, Brightness, Saturation, and it didn't seem to
help much. I think it is because there is such large gaps between the
known colors that it can only get so accurate. Maybe if this were a
different application it would have a bigger impact (i.e. using more
colors than the "known" colors).

Jeremy