I am trying to make a plot of a simple mutation accumulation process on a binary tree...
My technical problem in gnuplot, is that is that I want to plot the possibility of getting 2 mutations on a specific lineage on the graph, here is the equation which determines it:
P_{2 mutation} = sum[k=0:n] (m/(2**(k+1)/(1-(1/2)**k)))(1-exp(-muk))
(dont bother with the formula im not sure that this is the correct one yet :))
where n is the number of levels of the binary tree, mu is the mutation rate, and m is the number of previously randomly thrown mutations onto the graphs edges...
I want to make a plot which is this possibility depending on the levels of the binary tree...
Therefore I wrote a script which is something like this:
set term pngcairo size 800,600
set title "Két mutáció megjelenésének valószínűsége, egy n szintű bináris fa egyik sejtvonalában"
set xlabel"szintek száma (n)"
set ylabel"Két mutáció megjelenésének valószínűsége (P_{2^{lin})"
set xrange[1:10]
set yrange[0:1]
set output '2mutvalsz.png'
set multiplot
do for[i=1:10]{
mu = 0.1+(i*0.1)
m = 4
f(x)=(x/((2**(x+1))*(1-(0.5)**x)))
if(m<floor(f(x)))
{
p(x)=sum [k=0:floor(x)](m*(1/((2**(x+1))*(1-(0.5)**x))))*(1-exp(-mu*k))
}
else
{
p(x)=1
}
plot p(x) lt i lw 1
}
unset multiplot
set output
So my problem is, that I dont know if it is correct to do what I do in the
if statement...
What I want is to behold the statement m< f(x) where f(x) is the number of edges in respect of n, which is an integer value therefore I use floor(f(x)), and sum through the x values (which are the number of levels what has to be an integer too... so floor(x), like a heavyside function to make the x axis discrete) in the sum...
And also I get an error message:
gnuplot> load '2mutvalsz.plt'
line 27: undefined variable: x
where line 27 is the end of the do for loop...
So my question is that is it a correct way to make a summation integer the x values and of course why I get the error message...
Thank you, and I hope everything is clear...
The error message is produced because the if statement in your script is interpreted when Gnuplot loads the script - it tries to evaluate the condition of the if statement and since the variable x is not defined, it produces the mentioned message.
You could put everything together using the ternary operator as:
p(x)=( m<floor(f(x)) )?( sum [k=0:floor(x)](m*(1/((2**(x+1))*(1-(0.5)**x))))*(1-exp(-mu*k)) ):1;
However, since the function f(x) is on the imposed x-range of [0,1] less than 1, the condition m<floor(f(x)) will be always false.
Related
I'm trying to solve a nonlinear system of 7 equations with least_squares, as some variables are molar fractions and go between 0 and 1 (fsolve doesnt allow constraints as far as I've read). Heres the main part of the code:
def fun_t(w,x_2):
x_1=1-x_2;
z_A=w[0];
z_B=w[1];
z_AB=w[2];
gamma_A=w[3];
gamma_B=w[4];
gamma_AB=w[5];
eps=w[6]
K_gamma=gamma_AB/(gamma_A*gamma_B)
K=K_gamma*z_AB/(z_A*z_B)
phi_A=z_A*v_A/(z_A*v_A+z_B*v_B+z_AB*v_AB);
phi_B=z_B*v_B/(z_A*v_A+z_B*v_B+z_AB*v_AB);
phi_AB=z_AB*v_AB/(z_A*v_A+z_B*v_B+z_AB*v_AB);
return [z_AB-eps/(1-eps),
z_A-(x_2-eps)/(1-eps),
z_B-(x_1-eps)/(1-eps),
eps-(1-(1-4*K/(K+K_gamma)*x_1*x_2)**0.5)/2,
R*T*np.log(gamma_A)-v_A*(alphaT_AB*phi_B**2+alphaT_AAB*phi_AB**2+(alphaT_AB+alphaT_AAB-alphaT_BAB)*phi_B*phi_AB),
R*T*np.log(gamma_B)-v_B*(alphaT_AB*phi_A**2+alphaT_BAB*phi_AB**2+(alphaT_AB+alphaT_BAB+alphaT_AAB)*phi_A*phi_AB),
R*T*np.log(gamma_AB)-v_AB*(alphaT_AAB*phi_A**2+alphaT_BAB*phi_B**2+(alphaT_AAB+alphaT_BAB-alphaT_AB)*phi_A*phi_B)
]
x_2=np.linspace(0,1,11)
guess=[0.5,0.5,0.5,1,1,1,0.25];
roots_t=np.zeros(shape=(len(x_2),len(guess)))
for i in range(len(x_2)):
roots_t[i] = least_squares(fun_t, guess, bounds = ((0,0,0,-5,-5,-5,0),(1,1,1,5,5,5,0.5)), args=x_2[i])
I get the error TypeError: fun_t() argument after * must be an iterable, not numpy.float64. My goal is to give the elements of the array x_2 as the fuction's second argument in order to get the rest of the variables for a set of compositions. I dont understand why it cant iterate over x_2, as it worked when I was using fsolve.
I'm just starting using python for this kind of calculations, so any help will be much appreciated!
I am new to LabVIEW and I am trying to read a code written in LabVIEW. The block diagram is this:
This is the program to input x and y functions into the voltage input. It is meant to give an input voltage in different forms (sine, heartshape , etc.) into the fast-steering mirror or galvano mirror x and y axises.
x and y function controls are for inputting a formula for a function, and then we use "evaluation single value" function to input into a daq assistant.
I understand that { 2*(|-Mpi|)/N }*i + -Mpi*pi goes into the x value. However, I dont understand why we use this kind of formula. Why we need to assign a negative value and then do the absolute value of -M*pi. Also, I don`t understand why we need to divide to N and then multiply by i. And finally, why need to add -Mpi again? If you provide any hints about this I would really appreciate it.
This is just a complicated way to write the code/formula. Given what the code looks like (unnecessary wire bends, duplicate loop-input-tunnels, hidden wires, unnecessary coercion dots, failure to use appropriate built-in 'negate' function) not much care has been given in writing it. So while it probably yields the correct results you should not expect it to do so in the most readable way.
To answer you specific questions:
Why we need to assign a negative value and then do the absolute value
We don't. We can just move the negation immediately before the last addition or change that to a subtraction:
{ 2*(|Mpi|)/N }*i - Mpi*pi
And as #yair pointed out: We are not assigning a value here, we are basically flipping the sign of whatever value the user entered.
Why we need to divide to N and then multiply by i
This gives you a fraction between 0 and 1, no matter how many steps you do in your for-loop. Think of N as a sampling rate. I.e. your mirrors will always do the same movement, but a larger N just produces more steps in between.
Why need to add -Mpi again
I would strongly assume this is some kind of quick-and-dirty workaround for a bug that has not been fixed properly. Looking at the code it seems this +Mpi*pi has been added later on in the development process. And while I don't know what the expected values are I would believe that multiplying only one of the summands by Pi is probably wrong.
Through trial and error, I have found that the GetPixel function takes two arguments, one for X and one for Y, even if used on a 1D image. On a 1D image, the second index must be set to zero.
image list := [3]: {1,2,3}
list.GetPixel(0,0) // Gets 1
GetPixel(list, 0, 0) // Equivalent
How am I supposed to know this? I can't see anything clearly specifying this in the documentation.
This is best done by using the script function with an incorrect parameter list, running the script, and observing the error output:
I write a vector in Dymola mos script in a simple manner like this:
x_axis = cell.spatialSummary.x_cell;
output: x_axis={1,2,3,4,5} // row vector
I want to do the same thing in a function.'x_cell' has 5 values which I want to store in a row vector. I use DymolaCommands.Trajectories.readTrajectory function to read x_cell values one by one in for loop (I use for loop because, readTrajectory throws an error when I try to read entire x_cell)
Real x_axis[:],axis_value[:,:];
Integer len=5;
for i in 1:len loop
axis_value:=readTrajectory(result,{"cell.spatialSummary.x_cell["+String(i)+"]"},1); //This intermediate variable returns [1,1] matrix
x_axis[i]:=scalar(axis_value);
end for;
I get an error:
Assignment failed x_axis[i] = scalar(axis_value);
what's wrong here? All I want to do is read all values of x_cell and write it into a vector. How can I do this in dymola function?
Thank you!
Solution: Initialize the vector with a certain value. In this case,
x_axis :=fill(0, len);
This solved the above problem for me.
Pre filling as in the other solution works, and is generally the best solution. However, in some cases you might have to append to the vector as follows:
x_axis=fill(0.0, 0);
for i in 1:len loop
axis_value:=readTrajectory(result,{"cell.spatialSummary.x_cell["+String(i)+"]"},1); //This intermediate variable returns [1,1] matrix
x_axis:=cat(1, x_axis, {scalar(axis_value)});
end for;
(This takes x_axis and concatenates a new element at the end. It is generally slower.)
I have written an optimization problem in pyomo and need a constraint, which contains a summation that has a variable length:
u_i_t[i, t]*T_min_run - sum (tnewnew in (t-T_min_run+1)..t-1) u_i_t[i,tnewnew] <= sum (tnew in t..(t+T_min_run-1)) u_i_t[i,tnew]
T is my actual timeline and N my machines
usually I iterate over t, but I need to guarantee the machines are turned on for certain amount of time.
def HP_on_rule(model, i, t):
return model.u_i_t[i, t]*T_min_run - sum(model.u_i_t[i, tnewnew] for tnewnew in range((t-T_min_run+1), (t-1))) <= sum(model.u_i_t[i, tnew] for tnew in range(t, (t+T_min_run-1)))
model.HP_on_rule = Constraint(N, rule=HP_on_rule)
I hope you can provide me with the correct formulation in pyomo/python.
The problem is that t is a running variable and I do not know how to implement this in Python. tnew is only a help variable. E.g. t=6 (variable), T_min_run=3 (constant) and u_i_t is binary [00001111100000...] then I get:
1*3 - 1 <= 3
As I said, I do not know how to implement this in my code and the current version is not running.
TypeError: HP_on_rule() missing 1 required positional argument: 't'
It seems like you didn't provide all your arguments to the function rule.
Since t is a parameter of your function, I assume that it corresponds to an element of set T (your timeline).
Then, your last line of your code example should include not only the set N, but also the set T. Try this:
model.HP_on_rule = Constraint(N, T, rule=HP_on_rule)
Please note: Building a Constraint with a "for each" part, you must provide the Pyomo Sets that you want to iterate over at the begining of the call for Constraint construction. As a rule of thumb, your constraint rule function should have 1 more argument than the number of Pyomo Sets specified in the Constraint initilization line.