inexact error in Julia — matrix calculation
The error happens after else condition. I want to replace A's column by new one. The following is the datatype of variables.It seems that it is datatype issue.
A = [1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
b = [20 24 16 20]'
c = [-1 -2 -3 -4 0 0 0]'
C = [1 5 6 4]'
## convert the datatype of original data
C = vec(C)
# initial basis matrix
B = A[:, C]
# initial basic solution
x = inv(B)*b
c_B = c[C]
# initial reduced costs
c_r = (c' - c_B'*inv(B)*A)'
# if basic matrix is not identity (e.g., take other variables other than slack
# variables as starting point), convert B=I and N=inv(B)
ind = eye(length(C))
j = 1
if B != ind
for i in length(c)
if i in C
A[:,i] = ind[:,j]
j += 1
else
A[:,i] = inv(B)*A[:,i]
end
end
end
julia-lang
asked Nov 14 '18 at 11:54
Jiayan YangJiayan Yang
184
add a comment |
The error happens after else condition. I want to replace A's column by new one. The following is the datatype of variables.It seems that it is datatype issue.
A = [1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
b = [20 24 16 20]'
c = [-1 -2 -3 -4 0 0 0]'
C = [1 5 6 4]'
## convert the datatype of original data
C = vec(C)
# initial basis matrix
B = A[:, C]
# initial basic solution
x = inv(B)*b
c_B = c[C]
# initial reduced costs
c_r = (c' - c_B'*inv(B)*A)'
# if basic matrix is not identity (e.g., take other variables other than slack
# variables as starting point), convert B=I and N=inv(B)
ind = eye(length(C))
j = 1
if B != ind
for i in length(c)
if i in C
A[:,i] = ind[:,j]
j += 1
else
A[:,i] = inv(B)*A[:,i]
end
end
end
julia-lang
asked Nov 14 '18 at 11:54
Jiayan YangJiayan Yang
184
2
please edit your post and paste console text rather a screenshot image. The text will be formatted as code if preceded by four spaces. You can also use Ctrl+K for that
– Przemyslaw Szufel
Nov 14 '18 at 12:05
Note thateye() has been deprecated in favor of I and Matrix constructorsin versions 0.7 and 1.0.
– daycaster
Nov 14 '18 at 13:09
add a comment |
The error happens after else condition. I want to replace A's column by new one. The following is the datatype of variables.It seems that it is datatype issue.
A = [1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
b = [20 24 16 20]'
c = [-1 -2 -3 -4 0 0 0]'
C = [1 5 6 4]'
## convert the datatype of original data
C = vec(C)
# initial basis matrix
B = A[:, C]
# initial basic solution
x = inv(B)*b
c_B = c[C]
# initial reduced costs
c_r = (c' - c_B'*inv(B)*A)'
# if basic matrix is not identity (e.g., take other variables other than slack
# variables as starting point), convert B=I and N=inv(B)
ind = eye(length(C))
j = 1
if B != ind
for i in length(c)
if i in C
A[:,i] = ind[:,j]
j += 1
else
A[:,i] = inv(B)*A[:,i]
end
end
end
julia-lang
asked Nov 14 '18 at 11:54
Jiayan YangJiayan Yang
184
The error happens after else condition. I want to replace A's column by new one. The following is the datatype of variables.It seems that it is datatype issue.
A = [1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
b = [20 24 16 20]'
c = [-1 -2 -3 -4 0 0 0]'
C = [1 5 6 4]'
## convert the datatype of original data
C = vec(C)
# initial basis matrix
B = A[:, C]
# initial basic solution
x = inv(B)*b
c_B = c[C]
# initial reduced costs
c_r = (c' - c_B'*inv(B)*A)'
# if basic matrix is not identity (e.g., take other variables other than slack
# variables as starting point), convert B=I and N=inv(B)
ind = eye(length(C))
j = 1
if B != ind
for i in length(c)
if i in C
A[:,i] = ind[:,j]
j += 1
else
A[:,i] = inv(B)*A[:,i]
end
end
end
julia-lang
julia-lang
asked Nov 14 '18 at 11:54
Jiayan YangJiayan Yang
184
asked Nov 14 '18 at 11:54
Jiayan YangJiayan Yang
184
edited Nov 14 '18 at 12:45
Jiayan Yang
asked Nov 14 '18 at 11:54
Jiayan YangJiayan Yang
184
asked Nov 14 '18 at 11:54
Jiayan YangJiayan Yang
184
asked Nov 14 '18 at 11:54
Jiayan YangJiayan Yang
184
184
2
please edit your post and paste console text rather a screenshot image. The text will be formatted as code if preceded by four spaces. You can also use Ctrl+K for that
– Przemyslaw Szufel
Nov 14 '18 at 12:05
Note thateye() has been deprecated in favor of I and Matrix constructorsin versions 0.7 and 1.0.
– daycaster
Nov 14 '18 at 13:09
add a comment |
2
please edit your post and paste console text rather a screenshot image. The text will be formatted as code if preceded by four spaces. You can also use Ctrl+K for that
– Przemyslaw Szufel
Nov 14 '18 at 12:05
Note thateye() has been deprecated in favor of I and Matrix constructorsin versions 0.7 and 1.0.
– daycaster
Nov 14 '18 at 13:09
2
2
please edit your post and paste console text rather a screenshot image. The text will be formatted as code if preceded by four spaces. You can also use Ctrl+K for that
– Przemyslaw Szufel
Nov 14 '18 at 12:05
please edit your post and paste console text rather a screenshot image. The text will be formatted as code if preceded by four spaces. You can also use Ctrl+K for that
– Przemyslaw Szufel
Nov 14 '18 at 12:05
Note that
eye() has been deprecated in favor of I and Matrix constructors in versions 0.7 and 1.0.– daycaster
Nov 14 '18 at 13:09
Note that
eye() has been deprecated in favor of I and Matrix constructors in versions 0.7 and 1.0.– daycaster
Nov 14 '18 at 13:09
add a comment |
1 Answer
1
active
oldest
votes
You have the following problems the code. First A is Matrix{Int64} and it should be Matrix{Float64}. You can fix it by writing:
A = Float64[1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
Second - you probably want index i to range from 1 to length(c) so you should write your loop as:
for i in 1:length(c)
Finally - your code will not work under Julia 1.0 not only because eye is not defined, but also because you update variable j inside a loop. I would recommend you to wrap the whole code in a function or write global j += 1 for it to work under Julia 1.0.
EDIT:
The problem - in a nutshell is the following:
julia> A = [1,2,3]
3-element Array{Int64,1}:
1
2
3
julia> A[1] = 0.5
ERROR: InexactError: Int64(Int64, 0.5)
Array A can hold only integers, so you cannot assign a float to it. By writing Float64 in front of array literal you force it to have another type of element, so this works:
julia> A = Float64[1,2,3]
3-element Array{Float64,1}:
1.0
2.0
3.0
julia> A[1] = 0.5
0.5
julia> A
3-element Array{Float64,1}:
0.5
2.0
3.0
In short Julia knows if your array holds integers or floats and checks it. Sometimes type promotion is allowed (e.g. you can assign an integer to an array of floats as this normally does not lead to the loss of precision) as is explained here https://docs.julialang.org/en/latest/manual/conversion-and-promotion/#Conversion-1.
answered Nov 14 '18 at 13:52
Bogumił KamińskiBogumił Kamiński
13k11220
Hi, thank you for your solution. The function eye works well in Julia on my laptop though I changes it to another. However the problem has been solved I still don't understand why it affects A[:,i] = inv(B)*A[:,i].
– Jiayan Yang
Nov 14 '18 at 14:54
I have added an explanation. I would highly recommend you to read the Julia manual as it contains many useful examples and explanations.
– Bogumił Kamiński
Nov 14 '18 at 15:38
add a comment |
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1 Answer
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oldest
votes
You have the following problems the code. First A is Matrix{Int64} and it should be Matrix{Float64}. You can fix it by writing:
A = Float64[1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
Second - you probably want index i to range from 1 to length(c) so you should write your loop as:
for i in 1:length(c)
Finally - your code will not work under Julia 1.0 not only because eye is not defined, but also because you update variable j inside a loop. I would recommend you to wrap the whole code in a function or write global j += 1 for it to work under Julia 1.0.
EDIT:
The problem - in a nutshell is the following:
julia> A = [1,2,3]
3-element Array{Int64,1}:
1
2
3
julia> A[1] = 0.5
ERROR: InexactError: Int64(Int64, 0.5)
Array A can hold only integers, so you cannot assign a float to it. By writing Float64 in front of array literal you force it to have another type of element, so this works:
julia> A = Float64[1,2,3]
3-element Array{Float64,1}:
1.0
2.0
3.0
julia> A[1] = 0.5
0.5
julia> A
3-element Array{Float64,1}:
0.5
2.0
3.0
In short Julia knows if your array holds integers or floats and checks it. Sometimes type promotion is allowed (e.g. you can assign an integer to an array of floats as this normally does not lead to the loss of precision) as is explained here https://docs.julialang.org/en/latest/manual/conversion-and-promotion/#Conversion-1.
answered Nov 14 '18 at 13:52
Bogumił KamińskiBogumił Kamiński
13k11220
Hi, thank you for your solution. The function eye works well in Julia on my laptop though I changes it to another. However the problem has been solved I still don't understand why it affects A[:,i] = inv(B)*A[:,i].
– Jiayan Yang
Nov 14 '18 at 14:54
I have added an explanation. I would highly recommend you to read the Julia manual as it contains many useful examples and explanations.
– Bogumił Kamiński
Nov 14 '18 at 15:38
add a comment |
You have the following problems the code. First A is Matrix{Int64} and it should be Matrix{Float64}. You can fix it by writing:
A = Float64[1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
Second - you probably want index i to range from 1 to length(c) so you should write your loop as:
for i in 1:length(c)
Finally - your code will not work under Julia 1.0 not only because eye is not defined, but also because you update variable j inside a loop. I would recommend you to wrap the whole code in a function or write global j += 1 for it to work under Julia 1.0.
EDIT:
The problem - in a nutshell is the following:
julia> A = [1,2,3]
3-element Array{Int64,1}:
1
2
3
julia> A[1] = 0.5
ERROR: InexactError: Int64(Int64, 0.5)
Array A can hold only integers, so you cannot assign a float to it. By writing Float64 in front of array literal you force it to have another type of element, so this works:
julia> A = Float64[1,2,3]
3-element Array{Float64,1}:
1.0
2.0
3.0
julia> A[1] = 0.5
0.5
julia> A
3-element Array{Float64,1}:
0.5
2.0
3.0
In short Julia knows if your array holds integers or floats and checks it. Sometimes type promotion is allowed (e.g. you can assign an integer to an array of floats as this normally does not lead to the loss of precision) as is explained here https://docs.julialang.org/en/latest/manual/conversion-and-promotion/#Conversion-1.
answered Nov 14 '18 at 13:52
Bogumił KamińskiBogumił Kamiński
13k11220
Hi, thank you for your solution. The function eye works well in Julia on my laptop though I changes it to another. However the problem has been solved I still don't understand why it affects A[:,i] = inv(B)*A[:,i].
– Jiayan Yang
Nov 14 '18 at 14:54
I have added an explanation. I would highly recommend you to read the Julia manual as it contains many useful examples and explanations.
– Bogumił Kamiński
Nov 14 '18 at 15:38
add a comment |
You have the following problems the code. First A is Matrix{Int64} and it should be Matrix{Float64}. You can fix it by writing:
A = Float64[1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
Second - you probably want index i to range from 1 to length(c) so you should write your loop as:
for i in 1:length(c)
Finally - your code will not work under Julia 1.0 not only because eye is not defined, but also because you update variable j inside a loop. I would recommend you to wrap the whole code in a function or write global j += 1 for it to work under Julia 1.0.
EDIT:
The problem - in a nutshell is the following:
julia> A = [1,2,3]
3-element Array{Int64,1}:
1
2
3
julia> A[1] = 0.5
ERROR: InexactError: Int64(Int64, 0.5)
Array A can hold only integers, so you cannot assign a float to it. By writing Float64 in front of array literal you force it to have another type of element, so this works:
julia> A = Float64[1,2,3]
3-element Array{Float64,1}:
1.0
2.0
3.0
julia> A[1] = 0.5
0.5
julia> A
3-element Array{Float64,1}:
0.5
2.0
3.0
In short Julia knows if your array holds integers or floats and checks it. Sometimes type promotion is allowed (e.g. you can assign an integer to an array of floats as this normally does not lead to the loss of precision) as is explained here https://docs.julialang.org/en/latest/manual/conversion-and-promotion/#Conversion-1.
answered Nov 14 '18 at 13:52
Bogumił KamińskiBogumił Kamiński
13k11220
You have the following problems the code. First A is Matrix{Int64} and it should be Matrix{Float64}. You can fix it by writing:
A = Float64[1 1 1 1 1 0 0;
1 2 0 3 0 1 0;
2 1 2 0 0 0 -1;
2 3 1 1 0 0 0]
Second - you probably want index i to range from 1 to length(c) so you should write your loop as:
for i in 1:length(c)
Finally - your code will not work under Julia 1.0 not only because eye is not defined, but also because you update variable j inside a loop. I would recommend you to wrap the whole code in a function or write global j += 1 for it to work under Julia 1.0.
EDIT:
The problem - in a nutshell is the following:
julia> A = [1,2,3]
3-element Array{Int64,1}:
1
2
3
julia> A[1] = 0.5
ERROR: InexactError: Int64(Int64, 0.5)
Array A can hold only integers, so you cannot assign a float to it. By writing Float64 in front of array literal you force it to have another type of element, so this works:
julia> A = Float64[1,2,3]
3-element Array{Float64,1}:
1.0
2.0
3.0
julia> A[1] = 0.5
0.5
julia> A
3-element Array{Float64,1}:
0.5
2.0
3.0
In short Julia knows if your array holds integers or floats and checks it. Sometimes type promotion is allowed (e.g. you can assign an integer to an array of floats as this normally does not lead to the loss of precision) as is explained here https://docs.julialang.org/en/latest/manual/conversion-and-promotion/#Conversion-1.
answered Nov 14 '18 at 13:52
Bogumił KamińskiBogumił Kamiński
13k11220
edited Nov 14 '18 at 15:37
answered Nov 14 '18 at 13:52
Bogumił KamińskiBogumił Kamiński
13k11220
answered Nov 14 '18 at 13:52
Bogumił KamińskiBogumił Kamiński
13k11220
answered Nov 14 '18 at 13:52
Bogumił KamińskiBogumił Kamiński
13k11220
13k11220
Hi, thank you for your solution. The function eye works well in Julia on my laptop though I changes it to another. However the problem has been solved I still don't understand why it affects A[:,i] = inv(B)*A[:,i].
– Jiayan Yang
Nov 14 '18 at 14:54
I have added an explanation. I would highly recommend you to read the Julia manual as it contains many useful examples and explanations.
– Bogumił Kamiński
Nov 14 '18 at 15:38
add a comment |
Hi, thank you for your solution. The function eye works well in Julia on my laptop though I changes it to another. However the problem has been solved I still don't understand why it affects A[:,i] = inv(B)*A[:,i].
– Jiayan Yang
Nov 14 '18 at 14:54
I have added an explanation. I would highly recommend you to read the Julia manual as it contains many useful examples and explanations.
– Bogumił Kamiński
Nov 14 '18 at 15:38
Hi, thank you for your solution. The function eye works well in Julia on my laptop though I changes it to another. However the problem has been solved I still don't understand why it affects A[:,i] = inv(B)*A[:,i].
– Jiayan Yang
Nov 14 '18 at 14:54
Hi, thank you for your solution. The function eye works well in Julia on my laptop though I changes it to another. However the problem has been solved I still don't understand why it affects A[:,i] = inv(B)*A[:,i].
– Jiayan Yang
Nov 14 '18 at 14:54
I have added an explanation. I would highly recommend you to read the Julia manual as it contains many useful examples and explanations.
– Bogumił Kamiński
Nov 14 '18 at 15:38
I have added an explanation. I would highly recommend you to read the Julia manual as it contains many useful examples and explanations.
– Bogumił Kamiński
Nov 14 '18 at 15:38
add a comment |
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2
please edit your post and paste console text rather a screenshot image. The text will be formatted as code if preceded by four spaces. You can also use Ctrl+K for that
– Przemyslaw Szufel
Nov 14 '18 at 12:05
Note that
eye() has been deprecated in favor of I and Matrix constructorsin versions 0.7 and 1.0.– daycaster
Nov 14 '18 at 13:09