Could someone suggest the right way to eliminate constants from the code in production mode? I've already tested babel-plugin-constant-folding and babel-plugin-dead-code-elimination but they both work only with babel 5, not with babel 6.
Example:
const K=1;
const B=K + 13;
console.log("b=" + B);
I expect to get:
console.log("k=" + 1 + ", b=" + 14)
or, better (honestly, I don't need this level of optimization, replacing all ID with their values without string concat is completely enough for me):
console.log("k=1, b=13")
but get only:
var K = 1,
B = K + 13;
console.log("k=" + K + ", b=" + B);
Could someone suggest me the right sequence of babel plugins?
Related
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 have the following predefined codes that represent an index in a binary bitmap:
0 = standard
1 = special
2 = regular
3 = late
4 = early
5 = on time
6 = generic
7 = rfu
An example value I would take as an input would be 213, which becomes 11010101 in binary. Index 0, 2, 4, 6, and 7 have their bit flipped indicating that this record is:
standard + regular + early + generic + rfu.
I am trying to figure out in perl how to take that binary data and build a string, like mentioned above with code + code + code, etc.
Any help would be greatly appreciated. Thanks.
Edit: My thoughts on how I might approach this are:
Convert decimal to binary
Find length of binary string
Using substr get the value (0 or 1) index by index
If index value = 1 then add relevant code to string
Is there a better way to go about this?
You can test bits on input from 0 to 7, and take only these that are set,
my $in = 213;
my #r = ("standard","special","regular","late","early","on time","generic","rfu");
print join " + ", #r[ grep { $in & (1 << $_) } 0 .. $#r ];
# or
# print join " + ", map { $in & (1<<$_) ? $r[$_] : () } 0 .. $#r;
output
standard + regular + early + generic + rfu
I have a matrix with a bunch of unknown constants such as the one below:
a*b -c -d 0
-c e -a -b-d
-d -a d -e
0 -b-d -e a
As you may realize it is symmetric about the diagonal and therefore, the diagonal values are all positive. All constants are greater than 0.
I would like to solve this for the eigenvalues in matlab. How would I go about doing this? I do not know the values a,b,c,d, and e. I would like to do something like this:
d = eig(#getMatrix)
but the eig function does not accept function handles.
No problem in MATLAB.
>> syms a b c d e
>> M = [a*b -c -d 0
-c e -a -b-d
-d -a d -e
0 -b-d -e a];
>> eig(M)
ans =
a/4 + d/4 + e/4 + (a*b)/4 - ((51*a*d^3)/16 - (117*a^4*b)/16 + (27*a^3*d)/16 + (27*a*e^3)/16 + (57*b*d^3)/2 + (27*a^3*e)/16 + (27*d*e^3)/16 + (51*d^3*e)/16 + 6*((4*(2*b*d - (a*e)/4 - (a*d)/4 - (d*e)/4 - (a^2*b)/4 + (11*a^2)/8 + b^2 + c^2 + (19*d^2)/8 + (11*e^2)/8 + (3*a^2*b^2)/8 - (a*b*d)/4 - (a*b*e)/4)*((17*a*d^3)/64 - (39*a^4*b)/64 + (9*a^3*d)/64 + (9*a*e^3)/64 + (19*b*d^3)/8 + (9*a^3*e)/64 + (9*d*e^3)/64 + (17*d^3*e)/64 + (45*a^4)/256 + (285*d^4)/256 + (45*e^4)/256 - (a^2*b^2)/16 + (a^2*b^3)/8 + (3*a^2*b^4)/16 + (31*a^4*b^2)/128 + (a^4*b^3)/64 - (3*a^4*b^4)/256 + (3*a^2*c^2)/16 + (15*a^2*d^2)/128 - (9*a^2*e^2)/128 + (19*b^2*d^2)/16 - (b^2*e^2)/16 + (3*c^2*d^2)/16 + (15*c^2*e^2)/16 +
...
(a*b*c^2*e)/8 + (3*a*b*d*e^2)/64 + (11*a*b*d^2*e)/64 + (a*b^2*d*e)/4 - (33*a^2*b*d*e)/32 - (5*a^2*b^2*d*e)/64 + (a*b*d*e)/4 + (a*c*d*e)/2 - 2*b*c*d*e) - 256*((17*a*d^3)/64 - (39*a^4*b)/64 + (9*a^3*d)/64 + (9*a*e^3)/64 + (19*b*d^3)/8 + (9*a^3*e)/64 + (9*d*e^3)/64 + (17*d^3*e)/64 + (45*a^4)/256 + (285*d^4)/256 + (45*e^4)/256 - (a^2*b^2)/16 + (a^2*b^3)/8 + (3*a^2*b^4)/16 + (31*a^4*b^2)/128 + (a^4*b^3)/64 - (3*a^4*b^4)/256 + (3*a^2*c^2)/16 + (15*a^2*d^2)/128 - (9*a^2*e^2)/128 + (19*b^2*d^2)/16 - (b^2*e^2)/16 + (3*c^2*d^2)/16 + (15*c^2*e^2)/16 + (15*d^2*e^2)/1...
Output truncated. Text exceeds maximum line length of 25,000 characters for Command Window display.
I deleted a lot there. Admittedly, its rather messy and lengthy, but can you really expect better?
Edit: I should comment that such a long extended formula may be dangerous in terms of computational accuracy. I've seen people blindly use such a mess of an expression, evaluating it in Fortran or MATLAB. They think that because it is "symbolic" that it is also exact. This is a total fallacy when numerical computations are done.
There may well be immense subtractive cancellation in those terms, with huge positive and negative terms almost canceling each other out, leaving a tiny result that is essentially worthless because of the limited dynamic range of floating point computations. BEWARE. At the very least, compare single and double precision computations done with the same expression. If they differ by any significant amount, try an extended precision version to verify there is not a problem for the doubles. If you have not tested such an expression and verified it extensively, don't trust it.
for Temp = 1000:10:6000
cp_CO2 = ((2e-18)*Temp.^5) - ((4e-14)*Temp.^4) + ((3e-10)*Temp.^3) - ((8e-07)*Temp.^2) + (0.0013*Temp) + 0.5126;
cp_CO = ((5e-12)*Temp.^3) - ((7e-08)*Temp.^2) + (0.0003*Temp) + 0.9657;
cp_H2O = ((7e-12)*Temp.^3) - ((1e-07)*Temp.^2) + (0.0008*Temp) + 1.6083;
cp_N2 = ((-1e-18)*Temp.^5) + ((2e-14)*Temp.^4) - ((8e-11)*Temp.^3) + ((1e-07)*Temp.^2) + (0.0001*Temp) + 0.9985;
D_H = (y(1)*cp_CO2*44*(25-Temp)) + (y(2)*cp_CO*28*(25-Temp)) + (y(3)*cp_H2O*18*(25-Temp)) + (percent_air*x_final(2)*3.76*28*(25-Temp));
DELTA_H = round(D_H);
if DELTA_H == delta_h
break
end
end
The 'for' loop in my code is above, the variables delta_h, y and percent_air have been defined and calculated/input earlier. If I work on the loop as a cell and manually increase Temp then the values of D_H etc. all change. But for some reason when I try and execute the loop the 'if' statement doesn't seem to come into effect and the final values where Temp = 6000 are displayed in the workspace instead of the value of Temp where it produces a DELTA_H equal to that of delta_h. It's the first time I've used MATLAB for about 2 years (I'm a 3rd Year Mech Eng student) so please forgive me if it's a simple error to fix.
If either of the variables are floating-point, doing an exact compare like that is problematic. A <= or >= comparison might work better.