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How can an equation having ColorDistance function be solved?
The 2019 Stack Overflow Developer Survey Results Are InCan't solve equation having complex coefficientsRecursive functions taking list argumentsThe system can not be solvedHow to interpolate colors on scattered points for a smooth visualizationHow to add a common color-bar to a set of 2d maps produced from evaluating a function along three projections in Mathematica 8.0?How to imitate the opacity plot in EditColorFunctionHandling oscilloscope bitmap dataHow can I combine the ChromaticityPlot3D with a Sphere?Can this be solved even faster?
$begingroup$
I wish to extract the colors on the L
axis in the CIELab
color space such that the CIE2000
distance between a corresponding color and LABColor[0., 0., 0.]
will be gradually increasing by a factor 0.01
. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0]
are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)
Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0],
DistanceFunction -> "CIE2000"] == 0.01, l]
I got an error message.
How can I automate this process to complete the series.
list-manipulation equation-solving color
$endgroup$
add a comment |
$begingroup$
I wish to extract the colors on the L
axis in the CIELab
color space such that the CIE2000
distance between a corresponding color and LABColor[0., 0., 0.]
will be gradually increasing by a factor 0.01
. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0]
are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)
Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0],
DistanceFunction -> "CIE2000"] == 0.01, l]
I got an error message.
How can I automate this process to complete the series.
list-manipulation equation-solving color
$endgroup$
add a comment |
$begingroup$
I wish to extract the colors on the L
axis in the CIELab
color space such that the CIE2000
distance between a corresponding color and LABColor[0., 0., 0.]
will be gradually increasing by a factor 0.01
. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0]
are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)
Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0],
DistanceFunction -> "CIE2000"] == 0.01, l]
I got an error message.
How can I automate this process to complete the series.
list-manipulation equation-solving color
$endgroup$
I wish to extract the colors on the L
axis in the CIELab
color space such that the CIE2000
distance between a corresponding color and LABColor[0., 0., 0.]
will be gradually increasing by a factor 0.01
. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0]
are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)
Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0],
DistanceFunction -> "CIE2000"] == 0.01, l]
I got an error message.
How can I automate this process to complete the series.
list-manipulation equation-solving color
list-manipulation equation-solving color
asked Apr 5 at 8:13
MajisMajis
1,477415
1,477415
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
f[l_?NumericQ] :=
ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]
FindRoot[f[l] == 0.01, l, .5]
(* l -> 0.0173396 *)
ColorDistance
does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.
Solve
is meant for exact symbolic computations. Use FindRoot
instead.
$endgroup$
$begingroup$
Perfect. Now I can get the complete list. Thanks.
$endgroup$
– Majis
Apr 5 at 8:45
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
f[l_?NumericQ] :=
ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]
FindRoot[f[l] == 0.01, l, .5]
(* l -> 0.0173396 *)
ColorDistance
does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.
Solve
is meant for exact symbolic computations. Use FindRoot
instead.
$endgroup$
$begingroup$
Perfect. Now I can get the complete list. Thanks.
$endgroup$
– Majis
Apr 5 at 8:45
add a comment |
$begingroup$
f[l_?NumericQ] :=
ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]
FindRoot[f[l] == 0.01, l, .5]
(* l -> 0.0173396 *)
ColorDistance
does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.
Solve
is meant for exact symbolic computations. Use FindRoot
instead.
$endgroup$
$begingroup$
Perfect. Now I can get the complete list. Thanks.
$endgroup$
– Majis
Apr 5 at 8:45
add a comment |
$begingroup$
f[l_?NumericQ] :=
ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]
FindRoot[f[l] == 0.01, l, .5]
(* l -> 0.0173396 *)
ColorDistance
does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.
Solve
is meant for exact symbolic computations. Use FindRoot
instead.
$endgroup$
f[l_?NumericQ] :=
ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]
FindRoot[f[l] == 0.01, l, .5]
(* l -> 0.0173396 *)
ColorDistance
does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.
Solve
is meant for exact symbolic computations. Use FindRoot
instead.
answered Apr 5 at 8:26
SzabolcsSzabolcs
164k14448946
164k14448946
$begingroup$
Perfect. Now I can get the complete list. Thanks.
$endgroup$
– Majis
Apr 5 at 8:45
add a comment |
$begingroup$
Perfect. Now I can get the complete list. Thanks.
$endgroup$
– Majis
Apr 5 at 8:45
$begingroup$
Perfect. Now I can get the complete list. Thanks.
$endgroup$
– Majis
Apr 5 at 8:45
$begingroup$
Perfect. Now I can get the complete list. Thanks.
$endgroup$
– Majis
Apr 5 at 8:45
add a comment |
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