I have a large Float64 array x and often treat a slice of it as matrix before changing exactly that slice. Can I somehow refer to this slice as y which already has the correct shape. To
x=zeros(10000)
y=x[10:18]
reshape!(y,(3,3))
y=y+eye(3) # this doesn't change x
This does not work as x[10:18] creates a copy. I had a look at pointer_to_array but I couldn't work it out.
The usual tool of producing no-copy slices in julia is SubArrays:
x = zeros(10000)
y = sub(x, 10:18)
now changing y would change the corresponding elements of x. (you can also use slice, which behaves differently with higher-dimensional arrays).
You can also use y = sub(x, A) where A is an array of indices. Unfortunately reshape(y, 3, 3) produces a copy of the values by converting the SubArray to an Array (currently discussed in Julia issue 9874, so that may change in the future). Using y = sub(x, A) with A a matrix of indices of the right shape does not work.
Maybe for your particular application, it is sufficient to do
x = zeros(10000)
y = sub(x, 10:18)
y[:] += vec(eye(3))
which changes x as you wanted it, without producing a slice of different shape.
I guess the following will do the job:
y = pointer_to_array( pointer( x, 10 ), (3,3) ) # make a slice starting from the 10th element
This can be tested, for example, as
x = zeros( 8 )
p = pointer_to_array( pointer( x, 3 ), (3,2) )
p[:,1] = 100.0
p[:,2] = 200.0
#show x # => [ 0.0, 0.0, 100.0, 100.0, 100.0, 200.0, 200.0, 200.0 ]
If the size of x is a multiple of the slice size, reshape() can also be used directly for modifying a slice. For example,
x = [ i for i=1:8 ]
s = reshape( x, (2,2,2) )
s[:,:,2] = 1000
#show x # => [1,2,3,4,1000,1000,1000,1000]
Related
I was working with Scilab and I decide to work with Julia however there are some errors which I didn't arrive to solve. For instance, I would like to fill out a vector using values of a given function but I got this error. Here is the code that I used:
using LinearAlgebra
A = [5/12 -1/12; 3/4 1/4]; c=[1/3;1]; b=[3/4; 1/4];
N = 10; T = 4; ts = (0:N)*T/N;
dt = T/N; λ = 10^(-14/(2*N+1));
m=length(c) ;
em0=b'/A # b^t * inv(A)
em1 = 1 .-em0*ones(m,1)
γ(z) =#. z/(1.0 -z*em1)
u_hat=complex(zeros(1,N+1));
u_hat[1]=γ(im)
The over-arching issue you are facing is that, coming from Scilab, you are probably not used to distinguishing scalars, vectors and matrices. Like in Matlab, Scilab scalars are really 1x1 matrices, and vectors are really Nx1 or 1xN matrices.
This is very different in Julia. A scalar is not the same as a 1x1 matrix, and a vector is not the same as a Nx1 matrix. You should therefore take care to distinguish them. In particular, you should avoid creating a matrix, zeros(M, 1), when what you really need is a vector, zeros(M).
The direct reason for the error message is that γ(im) is a matrix, because em1 is a matrix:
julia> γ(im)
1×1 Matrix{ComplexF64}:
0.0 + 1.0im
u_hat is also a matrix of ComplexF64, and you are trying to assign a matrix as one of its elements, which naturally won't work, only scalar values can be elements of a Matrix{ComplexF64}.
I took the liberty of writing a cleaned up version of your code:
A = [5/12 -1/12; 3/4 1/4]
# use commas when defining vectors (this is just about style)
b = [3/4, 1/4]
N = 10
## None of the below variables are used. Try to make your example minimal
c = [1/3, 1]
T = 4
dt = T/N;
ts = (0:N) .* dt
λ = 10^(-14/(2*N+1))
m = length(c)
############### <- not used
# prefer vectors over 1xN or Nx1 matrices
em0 = A' \ b
# dot product of a vector and a vector of ones is just a sum, but super-wasterful and slow.
em1 = 1 - sum(em0)
# don't use global variables(!!!), and remove the `#.`
γ(z, a) = z / (1 - z * a)
# use vectors, not 1xN matrices, and directly create a complex matrix instead of converting a real one.
û = zeros(ComplexF64, N+1)
# Now this works
û[1] = γ(im, em1)
I renamed u_hat to û for fun.
Also: remember to put your code in a function, always.
Just in the case of locating the root of the problem:
The problem is where you declared the em1 as em1 = 1 .-em0*ones(m,1). Since the output of the em0*ones(m,1) is expected to be a scalar, you can grasp it using the only function (I don't argue with your approach, and that's out of the interest of this answer):
julia> using LinearAlgebra
# Note that with this modification, there isn't any need for `#.` anymore.
julia> γ(z) = z/(1.0 -z*em1)
γ (generic function with 1 method)
julia> A = [5/12 -1/12; 3/4 1/4]; c=[1/3;1]; b=[3/4; 1/4];
N = 10; T = 4; ts = (0:N)*T/N;
dt = T/N; λ = 10^(-14/(2*N+1));
m=length(c);
em0=b'/A;
#This is where the problem can be solved
em1 = 1 - only(em0*ones(m,1));
u_hat=complex(zeros(1,N+1));
u_hat[1]=γ(im)
0.0 + 1.0im
julia> u_hat
1×11 Matrix{ComplexF64}:
0.0+1.0im 0.0+0.0im 0.0+0.0im 0.0+0.0im 0.0+0.0im … 0.0+0.0im 0.0+0.0im 0.0+0.0im 0.0+0.0im 0.0+0.0im
Good time of the day!
Here is the code:
eq:'diff(x,t)=(exp(cos(t))-1)*x;
ode2(eq,x,t);
sol:ic1(%,t=1,x=-1);
/*---------------------*/
plot2d(
rhs(sol),
[t,-4*%pi, 4*%pi],
[y,-5,5],
[xtics,-4*%pi, 1*%pi, 4*%pi],
[ytics, false],
/*[yx_ratio , 0.6], */
[legend,"Solution."],
[xlabel, "t"], [ylabel, "x(t)"],
[style, [lines,1]],
[color, blue]
);
and here is the errors:
integrate: variable must not be a number; found: -12.56637061435917
What went wrong?
Thanks.
Here's a way to plot the solution sol which was found by ode2 and ic2 as you showed. First replace the integrate nouns with calls to quad_qags, a numerical quadrature function. I'll introduce a made-up variable name (a so-called gensym) to avoid confusion with the variable t.
(%i59) subst (nounify (integrate) =
lambda ([e, xx],
block ([u: gensym(string(xx))],
quad_qags (subst (xx = u, e), u, -4*%pi, xx)[1])),
rhs(sol));
(%o59) -%e^((-t)-quad_qags(%e^cos(t88373),t88373,-4*%pi,t,
epsrel = 1.0E-8,epsabs = 0.0,
limit = 200)[
1]
+quad_qags(%e^cos(t88336),t88336,-4*%pi,t,
epsrel = 1.0E-8,epsabs = 0.0,
limit = 200)[
1]+1)
Now I'll define a function foo1 with that result. I'll make a list of numerical values to see if it works right.
(%i60) foo1(t) := ''%;
(%o60) foo1(t):=-%e
^((-t)-quad_qags(%e^cos(t88373),t88373,-4*%pi,t,
epsrel = 1.0E-8,epsabs = 0.0,
limit = 200)[
1]
+quad_qags(%e^cos(t88336),t88336,-4*%pi,t,
epsrel = 1.0E-8,epsabs = 0.0,
limit = 200)[
1]+1)
(%i61) foo1(0.5);
(%o61) -1.648721270700128
(%i62) makelist (foo1(t), t, makelist (k, k, -10, 10));
(%o62) [-59874.14171519782,-22026.46579480672,
-8103.083927575384,-2980.957987041728,
-1096.633158428459,-403.4287934927351,
-148.4131591025766,-54.59815003314424,
-20.08553692318767,-7.38905609893065,-2.71828182845904,
-1.0,-0.3678794411714423,-0.1353352832366127,
-0.04978706836786394,-0.01831563888873418,
-0.006737946999085467,-0.002478752176666358,
-9.118819655545163E-4,-3.354626279025119E-4,
-1.234098040866796E-4]
Does %o62 look right to you? I'll assume it is okay. Next I'll define a function foo which calls foo1 defined before when the argument is a number, otherwise it just returns 0. This is a workaround for a bug in plot2d, which incorrectly determines that foo1 is not a function of t alone. Usually that workaround isn't needed, but it is needed in this case.
(%i63) foo(t) := if numberp(t) then foo1(t) else 0;
(%o63) foo(t):=if numberp(t) then foo1(t) else 0
Okay, now the function foo can be plotted!
(%i64) plot2d (foo, [t, -4*%pi, 4*%pi], [y, -5, 5]);
plot2d: some values were clipped.
(%o64) false
That takes about 30 seconds to plot -- calling quad_qags is relatively expensive.
it looks like ode2 does not know how to completely solve the problem, so the result contains an integral:
(%i6) display2d: false $
(%i7) eq:'diff(x,t)=(exp(cos(t))-1)*x;
(%o7) 'diff(x,t,1) = (%e^cos(t)-1)*x
(%i8) ode2(eq,x,t);
(%o8) x = %c*%e^('integrate(%e^cos(t),t)-t)
(%i9) sol:ic1(%,t=1,x=-1);
(%o9) x = -%e^((-%at('integrate(%e^cos(t),t),t = 1))
+'integrate(%e^cos(t),t)-t+1)
I tried it with contrib_ode also:
(%i12) load (contrib_ode);
(%o12) "/Users/dodier/tmp/maxima-code/share/contrib/diffequations/contrib_ode.mac"
(%i13) contrib_ode (eq, x, t);
(%o13) [x = %c*%e^('integrate(%e^cos(t),t)-t)]
So contrib_ode did not solve it completely either.
However the solution returned by ode2 (same for contrib_ode) appears to be a valid solution. I'll post a separate answer describing how to evaluate it numerically for plotting.
I'm struggling to amend the Julia-specific tutorial on NLopt to meet my needs and would be grateful if someone could explain what I'm doing wrong or failing to understand.
I wish to:
Minimise the value of some objective function myfunc(x); where
x must lie in the unit hypercube (just 2 dimensions in the example below); and
the sum of the elements of x must be one.
Below I make myfunc very simple - the square of the distance from x to [2.0, 0.0] so that the obvious correct solution to the problem is x = [1.0,0.0] for which myfunc(x) = 1.0. I have also added println statements so that I can see what the solver is doing.
testNLopt = function()
origin = [2.0,0.0]
n = length(origin)
#Returns square of the distance between x and "origin", and amends grad in-place
myfunc = function(x::Vector{Float64}, grad::Vector{Float64})
if length(grad) > 0
grad = 2 .* (x .- origin)
end
xOut = sum((x .- origin).^2)
println("myfunc: x = $x; myfunc(x) = $xOut; ∂myfunc/∂x = $grad")
return(xOut)
end
#Constrain the sums of the x's to be 1...
sumconstraint =function(x::Vector{Float64}, grad::Vector{Float64})
if length(grad) > 0
grad = ones(length(x))
end
xOut = sum(x) - 1
println("sumconstraint: x = $x; constraint = $xOut; ∂constraint/∂x = $grad")
return(xOut)
end
opt = Opt(:LD_SLSQP,n)
lower_bounds!(opt, zeros(n))
upper_bounds!(opt,ones(n))
equality_constraint!(opt,sumconstraint,0)
#xtol_rel!(opt,1e-4)
xtol_abs!(opt,1e-8)
min_objective!(opt, myfunc)
maxeval!(opt,20)#to ensure code always terminates, remove this line when code working correctly?
optimize(opt,ones(n)./n)
end
I have read this similar question and documentation here and here, but still can't figure out what's wrong. Worryingly, each time I run testNLopt I see different behaviour, as in this screenshot including occasions when the solver uselessly evaluates myfunc([NaN,NaN]) many times.
You aren't actually writing to the grad parameters in-place, as you write in the comments;
grad = 2 .* (x .- origin)
just overrides the local variable, not the array contents -- and I guess that's why you see these df/dx = [NaN, NaN] everywhere. The simplest way to fix that would be with broadcasting assignment (note the dot):
grad .= 2 .* (x .- origin)
and so on. You can read about that behaviour here and here.
So I have a simple example of what I want to do:
restart;
assume(can, real);
f := {g = x+can*x*y, t = x+x*y};
assign(f[1]); g;
can := 2;
plot3d(g, x = 0 .. 100, y = 0 .. 100);
while this works:
restart;
f := {g = x+can*x*y, t = x+x*y};
assign(f[1]);
can := 2;
plot3d(g, x = 0 .. 100, y = 0 .. 100);
But that assumptions are really important for my real life case (for some optimisations with complex numbers) so I cant just leve can not preassumed.
Why it plots nothuing for me and how to make it plot?
The expression (or procedure) to be plotted must evaluate to a numeric, floating-point quantity. And so, for your expression g, the name can must have a specific numeric value at the time any plot of g is generated.
But you can produce a sequence of 3D plots, for various values of can, and display them. You can display them all at once, overlaid. Or you can display them in an animated sequence. And you can color or shade them each differently, to give a visual cue that can is changing and different for each.
restart;
f := {g = x+can*x*y, t = x+x*y};
eval(g,f);
N:=50:
Pseq := seq(
plot3d(eval(g,f),
x=0..10,y=0..10,
color=RGB(0.5,0.5,can/(2*N)),
transparency=0.5*(can/(N+1))),
can=1 .. N):
plots:-display(Pseq, axes=box);
plots:-display([Pseq],insequence=true,axes=box);
By the way, you don't have to assign to g just for the sake of using the equation for g that appears inside f. Doing that assignment (using assign, say, like you did) makes it more awkward for you subsequently to create other equations in terms of the pure name g unless you first unassign the name g. Some people find it easier to not make the assignment to g at all for such tasks, and to simply use eval as I've done above.
Now on to your deeper problem. You create an expression containing a local, assumed name. and then later on you want to use the same expression but with the global, unassumed version of that name. You can create the expression, with it containing the global, unassumed name instead of the local, assumed name, buy performing a substitution.
restart;
assume(can, real);
f := {g = x+can*x*y, t = x+x*y};
{g = x + can~ x y, t = x + x y}
assign(f[1]);
g;
x + can~ x y
can := 2:
g;
x + can~ x y
# This fails, because g contains the local name can~
plot3d(g, x=0..100, y=0..100);
# A procedure to make the desired substitution
revert:=proc(nm::name)
local len, snm;
snm:=convert(nm,string);
len:=length(snm);
if snm[-1]="~" then
return parse(snm[1..-2]);
else return parse(nm);
end if;
end proc:
# This is the version of the expression, but with global name can
subsindets(g,`local`,revert);
x + can x y
# This should work
plot3d(subsindets(g,`local`,revert),
x=0..100,y=0..100);
I have two p*n arrays, y and ymiss. y contains real numbers and NA's. ymiss contains 1's and 0's, so that if y(i,j)==NA, ymiss(i,j)==0, and 1 otherwise. I also have 1*n array ydim which tells how many real numbers there is at y(1:p,n), so ydim has values 0 to p
In R programming language, I can do following:
if(ydim!=p && ydim!=0)
y(1:ydim(t), t) = y(ymiss(,t), t)
That code arranges all real numbers of y(,t) in like this
first there's for example
y(,t) = (3,1,NA,6,2,NA)
after the code it's
y(,t) = (3,1,6,2,2,NA)
Now I will only need those first 1:ydim(t), so it doesn't matter what those rest are.
The question is, how can I do something like that in Fortran?
Thanks,
Jouni
The "where statement" and the "merge" intrinsic function are powerful, operating on selected positions in arrays, but they don't move items to the front of an array. With old-fashioned code with explicit indexing (could be packaged into a function) e.g.:
k=1
do i=1, n
if (ymiss (i) == 1) then
y(k) = y(i)
k = k + 1
end if
end do
What you want could be done with array intrinsics using the "pack" intrinsic. Convert ymiss into a logical array: 0 --> .false., 1 --> .true.. Then use code like (tested without the second index):
y(1:ydim(t), t) = pack (y (:,t), ymiss (:,t))
Edit to add example code, showing use of Fortran intrinsics "where", "count" and "pack". "where" alone can't solve the problem, but "pack" can. I used "< -90" as NaN for this example. The step "y (ydim+1:LEN) = -99.0" isn't required by the OP, who doesn't need to use these elements.
program test1
integer, parameter :: LEN = 6
real, dimension (1:LEN) :: y = [3.0, 1.0, -99.0, 6.0, 2.0, -99.0 ]
real, dimension (1:LEN) :: y2
logical, dimension (1:LEN) :: ymiss
integer :: ydim
y2 = y
write (*, '(/ "The input array:" / 6(F6.1) )' ) y
where (y < -90.0)
ymiss = .false.
elsewhere
ymiss = .true.
end where
ydim = count (ymiss)
where (ymiss) y2 = y
write (*, '(/ "Masking with where does not rearrange:" / 6(F6.1) )' ) y2
y (1:ydim) = pack (y, ymiss)
y (ydim+1:LEN) = -99.0
write (*, '(/ "After using pack, and ""erasing"" the end:" / 6(F6.1) )' ) y
stop
end program test1
Output is:
The input array:
3.0 1.0 -99.0 6.0 2.0 -99.0
Masking with where does not rearrange:
3.0 1.0 -99.0 6.0 2.0 -99.0
After using pack, and "erasing" the end:
3.0 1.0 6.0 2.0 -99.0 -99.0
In Fortran you can't store na in an array of real numbers, you can only store real numbers. So you'll probably want to replace na's with some value not likely to be present in your data: huge() might be suitable. 2D arrays are no problem at all for Fortan. You might want to use a 2D array of logicals to replace ymiss rather than a 2D array of 1s and 0s.
There is no simple, intrinsic to achieve what you want, you'd need to write a function. However, a more Fortran way of doing things would be to use the array of logicals as a mask for the operations you want to carry out.
So, here's some fragmentary Fortran code, not tested:
! Declarations
real(8), dimension(m,n) :: y, ynew
logical, dimension(m,n) :: ymiss
! Executable
where (ymiss) ynew = func(y) ! here func() is whatever your function is