Minimize distance between two lists
up vote
5
down vote
favorite
Writing:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
On the other hand, if I write:
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4} k;
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
where it is clear that, compared to the previous case, in some bars the gap has decreased and in others it has increased.
Question: How can I determine the best value of k
to get the smallest possible gap?
Writing:
h = -0.35;
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = h + k {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
Question 2: is it possible to determine the pair of values h
, k
that minimize the gap?
mathematical-optimization charts
add a comment |
up vote
5
down vote
favorite
Writing:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
On the other hand, if I write:
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4} k;
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
where it is clear that, compared to the previous case, in some bars the gap has decreased and in others it has increased.
Question: How can I determine the best value of k
to get the smallest possible gap?
Writing:
h = -0.35;
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = h + k {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
Question 2: is it possible to determine the pair of values h
, k
that minimize the gap?
mathematical-optimization charts
This is a related question: How to find the distance of two lists?
– Artes
Nov 29 at 16:03
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Writing:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
On the other hand, if I write:
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4} k;
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
where it is clear that, compared to the previous case, in some bars the gap has decreased and in others it has increased.
Question: How can I determine the best value of k
to get the smallest possible gap?
Writing:
h = -0.35;
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = h + k {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
Question 2: is it possible to determine the pair of values h
, k
that minimize the gap?
mathematical-optimization charts
Writing:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
On the other hand, if I write:
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4} k;
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
where it is clear that, compared to the previous case, in some bars the gap has decreased and in others it has increased.
Question: How can I determine the best value of k
to get the smallest possible gap?
Writing:
h = -0.35;
k = 0.83;
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = h + k {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
I get:
Question 2: is it possible to determine the pair of values h
, k
that minimize the gap?
mathematical-optimization charts
mathematical-optimization charts
edited Nov 28 at 21:27
asked Nov 28 at 20:19
TeM
1,777619
1,777619
This is a related question: How to find the distance of two lists?
– Artes
Nov 29 at 16:03
add a comment |
This is a related question: How to find the distance of two lists?
– Artes
Nov 29 at 16:03
This is a related question: How to find the distance of two lists?
– Artes
Nov 29 at 16:03
This is a related question: How to find the distance of two lists?
– Artes
Nov 29 at 16:03
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Update: Using two parameters:
lmf2 = LinearModelFit[data, t, t];
Normal@lmf2
1.76563 + 0.546875 t
lmf2["BestFitParameters"]
{1.76563, 0.546875}
Fit[data, {1, t}, t]
1.76563 + 0.546875 t
ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]
{43.3594, {h -> 1.76562, k -> 0.546875}}
N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]
{1.76563, 0.546875}
Original answer:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];
You can use LinearModelFit
or Fit
or NMinimize
or LeastSquares
to get the value of k
that minimizes the sum of squared distances between expectedresults
and k achievedresults
:
lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]
Normal@lmf
0.828025 t
Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]
0.828025 #1
Fit[data, {t}, t]
0.828025 t
ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]
{49.7134, {k -> 0.828025}}
N@LeastSquares[Thread[{achievedresults}], expectedresults]
{0.828025}
k = lmf["BestFitParameters"][[1]]
0.828025
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
BarChart[Transpose@{expectedresults, achievedresults, k achievedresults},
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]
For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42
@TeM, please see the update.
– kglr
Nov 28 at 21:47
Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48
1
@TeM, If you compare theNMinimize
result adding the intercept parameter improves the squared loss from49.7134
to43.3594
.
– kglr
Nov 28 at 21:54
add a comment |
up vote
7
down vote
{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults
{35/64, 113/64}
2
Why do you add a zero column? I thinkPseudoInverse[Transpose[{achievedresults}]].expectedresults
will do
– MeMyselfI
Nov 28 at 20:56
1
@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05
Really great!!!
– TeM
Nov 28 at 21:50
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Update: Using two parameters:
lmf2 = LinearModelFit[data, t, t];
Normal@lmf2
1.76563 + 0.546875 t
lmf2["BestFitParameters"]
{1.76563, 0.546875}
Fit[data, {1, t}, t]
1.76563 + 0.546875 t
ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]
{43.3594, {h -> 1.76562, k -> 0.546875}}
N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]
{1.76563, 0.546875}
Original answer:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];
You can use LinearModelFit
or Fit
or NMinimize
or LeastSquares
to get the value of k
that minimizes the sum of squared distances between expectedresults
and k achievedresults
:
lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]
Normal@lmf
0.828025 t
Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]
0.828025 #1
Fit[data, {t}, t]
0.828025 t
ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]
{49.7134, {k -> 0.828025}}
N@LeastSquares[Thread[{achievedresults}], expectedresults]
{0.828025}
k = lmf["BestFitParameters"][[1]]
0.828025
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
BarChart[Transpose@{expectedresults, achievedresults, k achievedresults},
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]
For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42
@TeM, please see the update.
– kglr
Nov 28 at 21:47
Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48
1
@TeM, If you compare theNMinimize
result adding the intercept parameter improves the squared loss from49.7134
to43.3594
.
– kglr
Nov 28 at 21:54
add a comment |
up vote
4
down vote
accepted
Update: Using two parameters:
lmf2 = LinearModelFit[data, t, t];
Normal@lmf2
1.76563 + 0.546875 t
lmf2["BestFitParameters"]
{1.76563, 0.546875}
Fit[data, {1, t}, t]
1.76563 + 0.546875 t
ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]
{43.3594, {h -> 1.76562, k -> 0.546875}}
N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]
{1.76563, 0.546875}
Original answer:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];
You can use LinearModelFit
or Fit
or NMinimize
or LeastSquares
to get the value of k
that minimizes the sum of squared distances between expectedresults
and k achievedresults
:
lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]
Normal@lmf
0.828025 t
Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]
0.828025 #1
Fit[data, {t}, t]
0.828025 t
ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]
{49.7134, {k -> 0.828025}}
N@LeastSquares[Thread[{achievedresults}], expectedresults]
{0.828025}
k = lmf["BestFitParameters"][[1]]
0.828025
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
BarChart[Transpose@{expectedresults, achievedresults, k achievedresults},
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]
For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42
@TeM, please see the update.
– kglr
Nov 28 at 21:47
Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48
1
@TeM, If you compare theNMinimize
result adding the intercept parameter improves the squared loss from49.7134
to43.3594
.
– kglr
Nov 28 at 21:54
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Update: Using two parameters:
lmf2 = LinearModelFit[data, t, t];
Normal@lmf2
1.76563 + 0.546875 t
lmf2["BestFitParameters"]
{1.76563, 0.546875}
Fit[data, {1, t}, t]
1.76563 + 0.546875 t
ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]
{43.3594, {h -> 1.76562, k -> 0.546875}}
N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]
{1.76563, 0.546875}
Original answer:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];
You can use LinearModelFit
or Fit
or NMinimize
or LeastSquares
to get the value of k
that minimizes the sum of squared distances between expectedresults
and k achievedresults
:
lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]
Normal@lmf
0.828025 t
Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]
0.828025 #1
Fit[data, {t}, t]
0.828025 t
ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]
{49.7134, {k -> 0.828025}}
N@LeastSquares[Thread[{achievedresults}], expectedresults]
{0.828025}
k = lmf["BestFitParameters"][[1]]
0.828025
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
BarChart[Transpose@{expectedresults, achievedresults, k achievedresults},
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]
Update: Using two parameters:
lmf2 = LinearModelFit[data, t, t];
Normal@lmf2
1.76563 + 0.546875 t
lmf2["BestFitParameters"]
{1.76563, 0.546875}
Fit[data, {1, t}, t]
1.76563 + 0.546875 t
ClearAll[h, k]
NMinimize[Total[Subtract[expectedresults, h + k achievedresults]^2], {h, k}]
{43.3594, {h -> 1.76562, k -> 0.546875}}
N @ LeastSquares[Thread[{1, achievedresults}], expectedresults]
{1.76563, 0.546875}
Original answer:
expectedresults = {4, 8, 5, 1, 4, 6, 4, 1, 9, 3};
achievedresults = {3, 6, 4, 2, 10, 7, 2, 4, 8, 4};
data = Transpose[{ achievedresults,expectedresults}];
You can use LinearModelFit
or Fit
or NMinimize
or LeastSquares
to get the value of k
that minimizes the sum of squared distances between expectedresults
and k achievedresults
:
lmf = LinearModelFit[data, t, t, IncludeConstantBasis -> False]
Normal@lmf
0.828025 t
Normal @ LinearModelFit[{Transpose[{achievedresults}], expectedresults}]
0.828025 #1
Fit[data, {t}, t]
0.828025 t
ClearAll[k]
NMinimize[Total[Subtract[expectedresults, k achievedresults]^2], k]
{49.7134, {k -> 0.828025}}
N@LeastSquares[Thread[{achievedresults}], expectedresults]
{0.828025}
k = lmf["BestFitParameters"][[1]]
0.828025
p1 = BarChart[expectedresults, ChartStyle -> Directive[Opacity[0.1], Blue]];
p2 = BarChart[k achievedresults, ChartStyle -> Directive[Opacity[0.1], Red]];
Show[p1, p2]
BarChart[Transpose@{expectedresults, achievedresults, k achievedresults},
ChartStyle -> {Blue, Red, Green}, ChartLayout -> "Grouped",
ChartLegends -> {"expectedresults", "achievedresults", "k achievedresults"}]
edited Nov 28 at 21:52
answered Nov 28 at 20:29
kglr
175k9197402
175k9197402
For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42
@TeM, please see the update.
– kglr
Nov 28 at 21:47
Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48
1
@TeM, If you compare theNMinimize
result adding the intercept parameter improves the squared loss from49.7134
to43.3594
.
– kglr
Nov 28 at 21:54
add a comment |
For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42
@TeM, please see the update.
– kglr
Nov 28 at 21:47
Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48
1
@TeM, If you compare theNMinimize
result adding the intercept parameter improves the squared loss from49.7134
to43.3594
.
– kglr
Nov 28 at 21:54
For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42
For general data you can always find several values of $k$ that eliminate the difference between whichever bars you like.
– David G. Stork
Nov 28 at 20:42
@TeM, please see the update.
– kglr
Nov 28 at 21:47
@TeM, please see the update.
– kglr
Nov 28 at 21:47
Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48
Perfect, mathematically it is clear to me! But I wonder if so "improve" the minimization or less than before!
– TeM
Nov 28 at 21:48
1
1
@TeM, If you compare the
NMinimize
result adding the intercept parameter improves the squared loss from 49.7134
to 43.3594
.– kglr
Nov 28 at 21:54
@TeM, If you compare the
NMinimize
result adding the intercept parameter improves the squared loss from 49.7134
to 43.3594
.– kglr
Nov 28 at 21:54
add a comment |
up vote
7
down vote
{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults
{35/64, 113/64}
2
Why do you add a zero column? I thinkPseudoInverse[Transpose[{achievedresults}]].expectedresults
will do
– MeMyselfI
Nov 28 at 20:56
1
@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05
Really great!!!
– TeM
Nov 28 at 21:50
add a comment |
up vote
7
down vote
{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults
{35/64, 113/64}
2
Why do you add a zero column? I thinkPseudoInverse[Transpose[{achievedresults}]].expectedresults
will do
– MeMyselfI
Nov 28 at 20:56
1
@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05
Really great!!!
– TeM
Nov 28 at 21:50
add a comment |
up vote
7
down vote
up vote
7
down vote
{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults
{35/64, 113/64}
{k, h} = PseudoInverse[{#, 1} & /@ achievedresults].expectedresults
{35/64, 113/64}
edited Nov 28 at 21:45
answered Nov 28 at 20:36
Chris
54116
54116
2
Why do you add a zero column? I thinkPseudoInverse[Transpose[{achievedresults}]].expectedresults
will do
– MeMyselfI
Nov 28 at 20:56
1
@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05
Really great!!!
– TeM
Nov 28 at 21:50
add a comment |
2
Why do you add a zero column? I thinkPseudoInverse[Transpose[{achievedresults}]].expectedresults
will do
– MeMyselfI
Nov 28 at 20:56
1
@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05
Really great!!!
– TeM
Nov 28 at 21:50
2
2
Why do you add a zero column? I think
PseudoInverse[Transpose[{achievedresults}]].expectedresults
will do– MeMyselfI
Nov 28 at 20:56
Why do you add a zero column? I think
PseudoInverse[Transpose[{achievedresults}]].expectedresults
will do– MeMyselfI
Nov 28 at 20:56
1
1
@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05
@MeMyselfI Nice! Even better
– Chris
Nov 28 at 21:05
Really great!!!
– TeM
Nov 28 at 21:50
Really great!!!
– TeM
Nov 28 at 21:50
add a comment |
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This is a related question: How to find the distance of two lists?
– Artes
Nov 29 at 16:03