I'm getting these error:
Conversion to function_handle from double is not possible.
Already searched about it and tried to change my code but without sucess. Could u help? Thanks
A=[99.23;100.05;91;107.71;104.1];
B=[3441 3441 301720.5;68750 1068750 0;170040 13085020 0;229350 229350 5729350;34194000 0 0];
N=[300000;1000000;13000000;5500000;32800000];
E=[-0.00302;-0.00261;-0.00208];
[c3,r3]=size(A);
[c4,r4]=size(B);
x=sym ('x',[1 c3]);
x=transpose(x);
for i=1:c3
Valor(i,1)=0;
for j=1:r4
Valor(i,1)=#(x){(Valor(i,1)/((1+E(j,1)+x(j,1))^j))+(B(i,j)/((1+E(j,1)+x(j,1))^j))};
end
end
What I want is to find the vector x given that I already have a vector Valorantigo that I will use to apply solve.
Valorantigo(1:c3,1)=A(1:c3,1).* N(1:c3,1) ./100;
eqn=Valor(1:c3,1)==Valorantigo(1:c3,1);
[solx, param, cond] = solve(eqn, x, 'ReturnConditions', true);
Basically x would be the solution of
Valorantigo(1,1)=3441/(1-0.00302+x1) + 3441/(1-0.00261+x1)^2 + 301720.5/(1-0.00208+x1)^3
Valorantigo(2,1)=68750/(1-0.00302+x2) + 1068750/(1-0.00261+x2)^2 + 0/(1-0.00208+x2)^3
Valorantigo(3,1)=170040/(1-0.00302+x3) + 13085020/(1-0.00261+x3)^2 + 0/(1-0.00208+x3)^3
the same fot the other lines...
Just the relevant Part:
Valor(1,1)=symfun(0,x);
for i=1:c3
Valor(i,1)=symfun(0,x);
for j=1:r4
Valor(i,1)=symfun( Valor(i,1)/(1+E(j,1)+x(j,1))^j+(B(i,j)/((1+E(j,1)+x(j,1))^j)),x);
end
end
Valor needs to be predefinded as symbolic, with Valor(i,1)=0; you made it to a double.
Oh almost forgot, your solve needs to look like this:
[solx1,solx2,solx3,solx4, solx5, param, cond] = solve(eqn, x, 'ReturnConditions', true);
Related
See the code and error. I have already tried Do, For,...and it is not working.
CODE + Error from Mathematica:
Import of survival probabilities _{k}p_x and _{k}p_y (calculated in excel)
px = Import["C:\Users\Eva\Desktop\kpx.xlsx"];
px = Flatten[Take[px, All], 1];
NOTE: The probability _{k}p_x can be found on the position px[[k+2, x -16]
i = 0.04;
v = 1/(1 + i);
JointLifeIndep[x_, y_, n_] = Sum[v^k*px[[k + 2, x - 16]]*py[[k + 2, y - 16]], {k , 0, n - 1}]
Part::pkspec1: The expression 2+k cannot be used as a part specification.
Part::pkspec1: The expression 2+k cannot be used as a part specification.
Part::pkspec1: The expression 2+k cannot be used as a part specification.
General::stop: Further output of Part::pkspec1 will be suppressed during this calculation.
Part of dataset (left corner of the dataset):
k\x 18 19 20
0 1 1 1
1 0.999478086278185 0.999363078716059 0.99927911905056
2 0.998841497412202 0.998642656911039 0.99858030519133
3 0.998121451605207 0.99794428814123 0.99788275311401
4 0.997423447323642 0.997247180349674 0.997174407432264
5 0.996726703362208 0.996539285828369 0.996437857252448
6 0.996019178300768 0.995803204773039 0.99563600297737
7 0.995283481416241 0.995001861216016 0.994823584922968
8 0.994482556091416 0.994189960607964 0.99405569519175
9 0.993671079225432 0.99342255996206 0.993339856748282
10 0.992904079096455 0.992707177451333 0.992611817294026
11 0.992189069953677 0.9919796017009 0.991832027835091
Without having the exact same data files to work with it is often easy for each of us to make mistakes that the other cannot reproduce or understand.
From your snapshot of your data set I used Export in Mathematica to try to reproduce your .xlsx file. Then I tried the following
px = Import["kpx.xlsx"];
px = Flatten[Take[px, All], 1];
py = px; (* fake some py data *)
i = 0.04;
v = 1/(1 + i);
JointLifeIndep[x_, y_, n_] := Sum[v^k*px[[k+2,x-16]]*py[[k+2,y-16]], {k,0,n-1}];
JointLifeIndep[17, 17, 12]
and it displays 362.402
Notice I used := instead of = in my definition of JointLifeIndep. := and = do different things in Mathematica. = will immediately evaluate the right hand side of that definition. This is possibly the reason that you are getting the error that you do.
You should also be careful with your subscript values and make sure that every subscript is between 1 and the number of rows (or columns) in your matrix.
So see if you can try this example with an Excel sheet containing only the snapshot of data that you showed and see if you get the same result that I do.
Hopefully that will be enough for you to make progress.
my question is about vectorize my for loop in Matlab using CUDA.
This arrays are gpuArrays.
The main problem is that the array isn't completed at start and it's getting bigger with every k iteration. Also it is creating by previous probes from array. Is it possbile to make this loop parallel/vectorize or any other way to optimize the code for CUDA to make it faster?
for k=3:5000
q_=gWmacierz((1:N_/2+1),k-1);
p_d1_=[p0_;gWmacierz((N_/2+2:N_+1),k-1)];
p_d2_=[gWmacierz((N_/2+2:N_+1),k-1);pN_];
F_=arrayfun(#funkcja,q_,p_d1_,p_d2_);
C1_=diag(F_+4*Gc);
C_=[C1_ C2_;C3_ C4_];
tic;
if k>3000
Wwektor = ginvA_* ((gB_*gWmacierz(:,k-2)) + (C_*gWmacierz(:,k-1)) + (gD_*gu01_) + (gE_*gu00_) - gQ*gleak);
else
Wwektor = ginvA_* ((gB_*gWmacierz(:,k-2)) + (C_*gWmacierz(:,k-1)) + (gD_*gu01_) + (gE_*gu00_));
end
toc;
gWmacierz(:,k)= Wwektor;
end
Pin=(gWmacierz(N_/2+2,:)+gWmacierz(N_/2+3,:))/2;
Pout=(gWmacierz(N_,:)+gWmacierz(N_+1,:))/2;
for k=3:5000
q=gmacierz((1:N/2+1),k-1);
p_d1=[Pin(k);gmacierz((N/2+2:N+1),k-1)];
p_d2=[gmacierz((N/2+2:N+1),k-1);Pout(k)];
F=arrayfun(#funkcja,q,p_d1,p_d2);
C1=diag(F+4*Gc);
C=[C1 C2;C3 C4];
wektor = ginvA* ((gB*gmacierz(:,k-2)) + (C*gmacierz(:,k-1)) + (gD*[Pin(k-1);Pout(k-1)]) + (gE*[Pin(k);Pout(k)]));
gmacierz=[gmacierz wektor];
end
Qin=(gWmacierz(2,:));
Qout=(gWmacierz(N/2,:));
Qin_ref=(gmacierz(1,:));
Qout_ref=(gmacierz(N/2+1,:));
deltaQin=Qin-Qin_ref;
deltaQout=Qout-Qout_ref;
let's say I have s=g(1,2,0)+g(1,3,0)+u(1,3)+g(1,1,0) where g, u are functions; I want to replace all 3rd arguments of g to something I choose without going through my script and doing it manually.
x = ... % assign some value beforehand
s = g(1,2,x) + g(1,3,x) + u(1,3) + g(1,1,x)
What follows is an ugly hack and I don't recommend using it:
g = #(a,b,c) g(a,b,0)
This redefines g function in a way that executing after that:
s = g(1,2,5) + g(1,3,3) + u(1,3) + g(1,1,2)
actually executes:
s = g(1,2,0) + g(1,3,0) + u(1,3) + g(1,1,0)
My question is if there is a good way to use MuPAD functions in a Matlab script. The background is that I have a problem where I need to find all solutions to a set of non-linear equations. The previous solution was to use solve in Matlab, which works for some of my simulations (i.e., some of the sets of input T) but not always. So instead I'm using MuPAD in the following way:
function ut1 = testMupadSolver(T)
% # Input T should be a vector of 15 elements
mupadCommand = ['numeric::polysysroots({' eq1(T) ' = 0,' ...
eq2(T) '= 0},[u, v])'];
allSolutions = evalin(symengine, mupadCommand);
ut1 = allSolutions;
end
function strEq = eq1(T)
sT = #(x) ['(' num2str(T(x)) ')'];
strEq = [ '-' sT(13) '*u^4 + (4*' sT(15) '-2*' sT(10) '-' sT(11) '*v)*u^3 + (3*' ...
sT(13) '-3*' sT(6) '+v*(3*' sT(14) '-2*' sT(7) ')-' sT(8) '*v^2)*u^2 + (2*' ...
sT(10) '-4*' sT(1) '+v*(2*' sT(11) '-3*' sT(2) ')+v^2*(2*' sT(12) ' - 2*' ...
sT(3) ')-' sT(4) '*v^3)*u + v*(' sT(7) '+' sT(8) '*v+' sT(9) '*v^2)+' sT(6)];
end
function strEq = eq2(T)
sT = #(x) ['(' num2str(T(x)) ')'];
strEq = ['(' sT(14) '-' sT(13) '*v)*u^3 + u^2*' '(' sT(11) '+(2*' sT(12) '-2*' sT(10) ...
')*v-' sT(11) '*v^2) + u*(' sT(7) '+v*(2*' sT(8) '-3*' sT(6) ')+v^2*(3*' sT(9) ...
'-2*' sT(7) ') - ' sT(8) '*v^3) + v*(2*' sT(3) '-4*' sT(1) '+v*(3*' sT(4) ...
'-3*' sT(2) ')+v^2*(4*' sT(5) ' - 2*' sT(3) ')-' sT(4) '*v^3)+' sT(2)];
end
I have two queries:
1) In order to use MuPAD I need to rewrite my two equations for the equation-system as strings, as you can see above. Is there a better way to do this, preferably without the string step?
2) And regarding the format output; when
T = [0 0 0 0 0 0 0 0 0 0 1 0 1 0 1];
the output is:
testMupadSolver(T)
ans =
matrix([[u], [v]]) in {matrix([[4.4780323328249527319374854327354], [0.21316518769990291263811232040432]]), matrix([[- 0.31088044854742790561428736573347 - 0.67937835289645431373983117422178*i], [1.1103383836576028262792542770062 + 0.39498445715599777249947213893789*i]]), matrix([[- 0.31088044854742790561428736573347 + 0.67937835289645431373983117422178*i], [1.1103383836576028262792542770062 - 0.39498445715599777249947213893789*i]]), matrix([[0.47897094942962218512261248590261], [-1.26776233072168360314707025141]]), matrix([[-0.83524238515971910583152318717102], [-0.66607962429342496204955062300669]])} union solvelib::VectorImageSet(matrix([[0], [z]]), z, C_)
Can MuPAD give the solutions as a set of vectors or similarly? In order to use the answer above I need to sort out the solutions from that string-set of solutions. Is there a clever way to do this? My solution so far is to find the signs I know will be present in the solution, such as '([[' and pick the numbers following, which is really ugly, and if the solution for some reason looks a little bit different than the cases I've covered it doesn't work.
EDIT
When I'm using the solution suggested in the answer below by #horchler, I get the same solution as with my previous implementation. But for some cases (not all) it takes much longer time. Eg. for the T below the solution suggested below takes more than a minute whilst using evalin (my previous implementation) takes one second.
T = [2.4336 1.4309 0.5471 0.0934 9.5838 -0.1013 -0.2573 2.4830 ...
36.5464 0.4898 -0.5383 61.5723 1.7637 36.0816 11.8262]
The new function:
function ut1 = testMupadSolver(T)
% # Input T should be a vector of 15 elements
allSolutions = feval(symengine,'numeric::polysysroots', ...
[eq1(T),eq2(T)],'[u,v]');
end
function eq = eq1(T)
syms u v
eq = -T(13)*u^4 + (4*T(15) - 2*T(10) - T(11)*v)*u^3 + (3*T(13) - 3*T(6) ...
+ v*(3*T(14) -2*T(7)) - T(8)*v^2)*u^2 + (2*T(10) - 4*T(1) + v*(2*T(11) ...
- 3*T(2)) + v^2*(2*T(12) - 2*T(3)) - T(4)*v^3)*u + v*(T(7) + T(8)*v ...
+ T(9)*v^2) + T(6);
end
function eq = eq2(T)
syms u v
eq = (T(14) - T(13)*v)*u^3 + u^2*(T(11) + (2*T(12) - 2*T(10))*v ...
- T(11)*v^2) + u*(T(7) + v*(2*T(8) - 3*T(6) ) + v^2*(3*T(9) - 2*T(7)) ...
- T(8)*v^3) + v*(2*T(3) - 4*T(1) + v*(3*T(4) - 3*T(2)) + v^2*(4*T(5) ...
- 2*T(3)) - T(4)*v^3) + T(2);
end
Is there a good reason to why it takes so much longer time?
Firstly, Matlab communicates with MuPAD via string commands so ultimately there is no way of getting around the use of strings. And because it's the native format, if you're passing large amounts of data into MuPAD, the best approach will be to convert everything to strings fast and efficiently (sprintf is usually best). However, in your case, I think that you can use feval instead of evalin which allows you to pass in regular Matlab datatypes (under the hood sym/feval does the string conversion and calls evalin). This method is discussed in this MathWorks article. The following code could be used:
T = [0 0 0 0 0 0 0 0 0 0 1 0 1 0 1];
syms u v;
eq1 = -T(13)*u^4 + (4*T(15) - 2*T(10) - T(11)*v)*u^3 + (3*T(13) - 3*T(6) ...
+ v*(3*T(14) -2*T(7)) - T(8)*v^2)*u^2 + (2*T(10) - 4*T(1) + v*(2*T(11) ...
- 3*T(2)) + v^2*(2*T(12) - 2*T(3)) - T(4)*v^3)*u + v*(T(7) + T(8)*v ...
+ T(9)*v^2) + T(6);
eq2 = (T(14) - T(13)*v)*u^3 + u^2*(T(11) + (2*T(12) - 2*T(10))*v ...
- T(11)*v^2) + u*(T(7) + v*(2*T(8) - 3*T(6) ) + v^2*(3*T(9) - 2*T(7)) ...
- T(8)*v^3) + v*(2*T(3) - 4*T(1) + v*(3*T(4) - 3*T(2)) + v^2*(4*T(5) ...
- 2*T(3)) - T(4)*v^3) + T(2);
allSolutions = feval(symengine, 'numeric::polysysroots',[eq1,eq2],'[u,v]');
The last argument still needed to be a string (or omitted) and adding ==0 to the equations also doesn't work, but the zero is implicit anyways.
For the second question, the result returned by numeric::polysysroots is very inconvenient and not easy to work with. It's a set (DOM_SET) of matrices. I tried using coerce to convert the result to something else to no avail. I think you best bet it to convert the output to a string (using char) and parse the result. I do this for simpler output formats. I'm not sure if it will be helpful, but feel free to look at my sym2float which just handles symbolic matrices (the 'matrix([[ ... ]])' part go your output) using a few optimizations.
A last thing. Is there a reason your helper function includes superfluous parentheses? This seems sufficient
sT = #(x)num2str(T(x),17);
or
sT = #(x)sprintf('%.17g',T(x));
Note that num2str only converts to four decimal places by default. int2str (or %d should be used if T(x) is always an integer).
I am trying to do something rather simple, but can't seem to get it...
I have 3 cell-arrays with strings,
A = {'ConditionA'; 'ConditionB'; 'ConditionC'; 'ConditionD'};
B = {'Case1'; 'Case2'; 'Case3'; 'Case4'};
C = {'Rice'; 'Beans'; 'Carrots'; 'Cereal';'Tomato'; 'Cabbage';...
'Sugar'}
I want to produce a vector with the concatenated (strcat?) combinations, as it this were a "tree diagram", like:
strcat(A(1),B(1),C(1))
strcat(A(1),B(1),C(2))
strcat(A(1),B(1),C(3))
strcat(A(1),B(1),C(4))
strcat(A(1),B(1),C(5))
strcat(A(1),B(1),C(6))
strcat(A(1),B(1),C(7))
strcat(A(1),B(2),C(1))
So what the first elements I am trying to get are (in a column ideally):
ConditionACase1Rice
ConditionACase1Beans
ConditionACase1Carrots
ConditionACase1Cereal
ConditionACase1Tomato
ConditionACase1Cabbage
ConditionACase1Sugar
ConditionACase2Rice
etc etc etc...
I know that:
for i=1:length(A)
E(i) = strcat(A(i),B(1),C(1))
end
Works for one "level". I have tried:
for i=1:length(A)
for j=1:length(B)
for k=1:length(C)
P(i) = strcat(A(i),B(j),C(k));
end
end
end
But this doesn't work...
I would be really grateful if I could be helped with this.
Thanks in advance!
From what I understood, you want all possible combinations of the strings of the input arrays as specified. If so, simply replace your nested loops with the following:
P = cell(length(A)*length(B)*length(C),1);
t=1;
for i=1:length(A)
for j=1:length(B)
for k=1:length(C)
P(t) = strcat(A(i),B(j),C(k));
t = t+1;
end
end
end
For the input arrays,
>> A = {'ConditionA'; 'ConditionB'; 'ConditionC'; 'ConditionD'};
>> B = {'Case1'; 'Case2'; 'Case3'; 'Case4'};
>> C = {'Rice'; 'Beans'; 'Carrots'; 'Cereal';'Tomato'; 'Cabbage';'Sugar'};
The value of P would be:
>> P
P =
'ConditionACase1Rice'
'ConditionACase1Beans'
'ConditionACase1Carrots'
'ConditionACase1Cereal'
'ConditionACase1Tomato'
'ConditionACase1Cabbage'
'ConditionACase1Sugar'
'ConditionACase2Rice'
'ConditionACase2Beans'
'ConditionACase2Carrots'
'ConditionACase2Cereal'
'ConditionACase2Tomato'
'ConditionACase2Cabbage'
'ConditionACase2Sugar'
'ConditionACase3Rice'
'ConditionACase3Beans'
'ConditionACase3Carrots'
'ConditionACase3Cereal'
'ConditionACase3Tomato'
'ConditionACase3Cabbage'
'ConditionACase3Sugar'
'ConditionACase4Rice'
'ConditionACase4Beans'
'ConditionACase4Carrots'
'ConditionACase4Cereal'
'ConditionACase4Tomato'
'ConditionACase4Cabbage'
'ConditionACase4Sugar'
'ConditionBCase1Rice'
'ConditionBCase1Beans'
'ConditionBCase1Carrots'
'ConditionBCase1Cereal'
'ConditionBCase1Tomato'
'ConditionBCase1Cabbage'
'ConditionBCase1Sugar'
'ConditionBCase2Rice'
'ConditionBCase2Beans'
'ConditionBCase2Carrots'
'ConditionBCase2Cereal'
'ConditionBCase2Tomato'
'ConditionBCase2Cabbage'
'ConditionBCase2Sugar'
'ConditionBCase3Rice'
'ConditionBCase3Beans'
'ConditionBCase3Carrots'
'ConditionBCase3Cereal'
'ConditionBCase3Tomato'
'ConditionBCase3Cabbage'
'ConditionBCase3Sugar'
'ConditionBCase4Rice'
'ConditionBCase4Beans'
'ConditionBCase4Carrots'
'ConditionBCase4Cereal'
'ConditionBCase4Tomato'
'ConditionBCase4Cabbage'
'ConditionBCase4Sugar'
'ConditionCCase1Rice'
'ConditionCCase1Beans'
'ConditionCCase1Carrots'
'ConditionCCase1Cereal'
'ConditionCCase1Tomato'
'ConditionCCase1Cabbage'
'ConditionCCase1Sugar'
'ConditionCCase2Rice'
'ConditionCCase2Beans'
'ConditionCCase2Carrots'
'ConditionCCase2Cereal'
'ConditionCCase2Tomato'
'ConditionCCase2Cabbage'
'ConditionCCase2Sugar'
'ConditionCCase3Rice'
'ConditionCCase3Beans'
'ConditionCCase3Carrots'
'ConditionCCase3Cereal'
'ConditionCCase3Tomato'
'ConditionCCase3Cabbage'
'ConditionCCase3Sugar'
'ConditionCCase4Rice'
'ConditionCCase4Beans'
'ConditionCCase4Carrots'
'ConditionCCase4Cereal'
'ConditionCCase4Tomato'
'ConditionCCase4Cabbage'
'ConditionCCase4Sugar'
'ConditionDCase1Rice'
'ConditionDCase1Beans'
'ConditionDCase1Carrots'
'ConditionDCase1Cereal'
'ConditionDCase1Tomato'
'ConditionDCase1Cabbage'
'ConditionDCase1Sugar'
'ConditionDCase2Rice'
'ConditionDCase2Beans'
'ConditionDCase2Carrots'
'ConditionDCase2Cereal'
'ConditionDCase2Tomato'
'ConditionDCase2Cabbage'
'ConditionDCase2Sugar'
'ConditionDCase3Rice'
'ConditionDCase3Beans'
'ConditionDCase3Carrots'
'ConditionDCase3Cereal'
'ConditionDCase3Tomato'
'ConditionDCase3Cabbage'
'ConditionDCase3Sugar'
'ConditionDCase4Rice'
'ConditionDCase4Beans'
'ConditionDCase4Carrots'
'ConditionDCase4Cereal'
'ConditionDCase4Tomato'
'ConditionDCase4Cabbage'
'ConditionDCase4Sugar'
Let me know if you need further assistance.
I am not really familiar with matlab.. but maybe try something like this?
for A = {'ConditionA'; 'ConditionB'; 'ConditionC'; 'ConditionD'};
for B = {'Case1'; 'Case2'; 'Case3'; 'Case4'};
for C = {'Rice'; 'Beans'; 'Carrots'; 'Cereal';'Tomato'; 'Cabbage'; 'Sugar'}
P(i) = strcat(A(i),B(j),C(k));
end
end
end
{
x = 1;
for i=1:length(A)
for j=1:length(B)
for k=1:length(C)
P(x) = strcat(A(i),B(j),C(k));
x = x + 1;
end
end
end
}
Please basically check your code before posting to PO as this is a very simple debugging