In Python we have lru_cache as a function wrapper. Add it to your function and the function will only be evaluated once per different input argument.
Example (from Python docs):
#lru_cache(maxsize=None)
def fib(n):
if n < 2:
return n
return fib(n-1) + fib(n-2)
>>> [fib(n) for n in range(16)]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610]
>>> fib.cache_info()
CacheInfo(hits=28, misses=16, maxsize=None, currsize=16)
I wonder whether a similar thing exists in Matlab? At the moment I am using cache files, like so:
function result = fib(n):
% FIB example like the Python example. Don't implement it like that!
cachefile = ['fib_', n, '.mat'];
try
load(cachefile);
catch e
if n < 2
result = n;
else
result = fib(n-1) + fib(n-2);
end
save(cachefile, 'result');
end
end
The problem I have with doing it this way, is that if I change my function, I need to delete the cachefile.
Is there a way to do this with Matlab realising when I changed the function and the cache has become invalidated?
Since matlab 2017 this is available:
https://nl.mathworks.com/help/matlab/ref/memoizedfunction.html
a = memoized(#sin)
I've created something like this for my own personal use: a CACHE class. (I haven't documented the code yet though.) It appears to be more flexible than Python's lru_cache (I wasn't aware of that, thanks) in that it has several methods for adjusting exactly what gets cached (to save memory) and how the comparisons are made. It could still use some refinement (#Daniel's suggestion to use the containers.Map class is a good one – though it would limit compatibility with old Matlab versions). The code is on GitHub so you're welcome to fork and improve it.
Here is a basic example of how it can be used:
function Output1 = CacheDemo(Input1,Input2)
persistent DEMO_CACHE
if isempty(DEMO_CACHE)
% Initialize cache object on first run
CACHE_SIZE = 10; % Number of input/output patterns to cache
DEMO_CACHE = CACHE(CACHE_SIZE,Input1,Input2);
CACHE_IDX = 1;
else
% Check if input pattern corresponds something stored in cache
% If not, return next available CACHE_IDX
CACHE_IDX = DEMO_CACHE.IN([],Input1,Input2);
if ~isempty(CACHE_IDX) && DEMO_CACHE.OUT(CACHE_IDX) > 0
[~,Output1] = DEMO_CACHE.OUT(CACHE_IDX);
return;
end
end
% Perform computation
Output1 = rand(Input1,Input2);
% Save output to cache CACHE_IDX
DEMO_CACHE.OUT(CACHE_IDX,Output1);
I created this class to cache the results from time-consuming stochastic simulations and have since used it to good effect in a few other places. If there is interest, I might be willing to spend some time documenting the code sooner as opposed to later. It would be nice if there was a way to limit memory use as well (a big consideration in my own applications), but getting the size of arbitrary Matlab datatypes is not trivial. I like your idea of caching to a file, which might be a good idea for larger data. Also, it might be nice to create a "lite" version that does what Python's lru_cache does.
Related
I am new to Julia and trying to use the Julia package DifferentialEquations to simultaneously solve for several conditions of the same set of coupled ODEs. My system is a model of an experiment and in one of the conditions, I increase the amount of one of the dependent variables at mid-way through the process.
I would like to be able to adjust the condition of this single trajectory, however so far I am only able to adjust all the trajectories at once. Is it possible to access a single one using callbacks? If not, is there a better way to do this?
Here is a simplified example using the lorentz equations for what I want to be doing:
#Differential Equations setup
function lorentz!(du,u,p,t)
a,r,b=p
du[1]= a*(u[2]-u[1])
du[2]=u[1]*(r-u[3])-u[2]
du[3]=u[1]*u[2]-b*u[3];
end
#function to cycle through inital conditions
function prob_func(prob,i,repeat)
remake(prob; u0 = u0_arr[i]);
end
#inputs
t_span=[(0.0,100.0),(0.0,100.0)];
u01=[0.0;1.0;0.0];
u02=[0.0;1.0;0.0];
u0_arr = [u01,u02];
p=[10.,28.,8/3];
#initialising the Ensemble Problem
prob = ODEProblem(lorentz!,u0_arr[1],t_span[1],p);
CombinedProblem = EnsembleProblem(prob,
prob_func = prob_func, #-> (prob),#repeat is a count for how many times the trajectories had been repeated
safetycopy = true # determines whether a safetly deepcopy is called on the prob before the prob_func (sounds best to leave as true for user-given prob_func)
);
#introducing callback
function condition(u,t,repeat)
return 50 .-t
end
function affect!(repeat)
repeat.u[1]=repeat.u[1] +50
end
callback = DifferentialEquations.ContinuousCallback(condition, affect!)
#solving
sim=solve(CombinedProblem,Rosenbrock23(),EnsembleSerial(),trajectories=2,callback=callback);
# Plotting for ease of understanding example
plot(sim[1].t,sim[1][1,:])
plot!(sim[2].t,sim[2][1,:])
I want to produce something like this:
Example_desired_outcome
But this code produces:
Example_current_outcome
Thank you for your help!
You can make that callback dependent on a parameter and make the parameter different between problems. For example:
function f(du,u,p,t)
if p == 0
du[1] = 2u[1]
else
du[1] = -2u[1]
end
du[2] = -u[2]
end
condition(t,u,integrator) = u[2] - 0.5
affect!(integrator) = integrator.prob.p = 1
For more information, check out the FAQ on this topic: https://diffeq.sciml.ai/stable/basics/faq/#Switching-ODE-functions-in-the-middle-of-integration
I have the data of electricity consumption of a region during the year of 2017. So I have to matrix 1x1, one with the month and other with the consumption. I want to use the command forecast to forecast the consumption of the first month of 2018, but I don't know how to do this even after reading the examples on MATLAB's help page.
Example:
data = {1166974.25000000, 1132479.36000000, 1137173.86000000, 1145853.58000000, 1118875.72000000, 1071456.85000000 ,1047171.87000000, 1071179.65000000 ,1077986.32000000 ,1112111.10000000, 1149668.47000000 ,1161649.19000000, 1175576.25000000 ,1126753.31000000 ,1204843.11000000 ,1183946.03000000, 1153080.36000000, 1120182.07000000, 1104726.03000000 ,1108110.02000000 ,1137729.28000000 ,1189699.45000000, 1252975.55000000, 1218118.20000000 ,1259580 ,1208193 ,1194430, 1244458, 1218867, 1205705 ,1177362, 1185584, 1164758, 1226991 ,1286044 ,1305312, 1360681.70000000 ,1332020 ,1306497.90000000 ,1299819.10000000 ,1316167.70000000 ,1246959.40000000 ,1256700.20000000 ,1266490.60000000, 1275642.90000000, 1358839.80000000, 1361440.10000000, 1398059.40000000};
data = [data{:}];
sys = ar(data,4)
K = 49;
p = forecast(sys,data,K);
plot(data,'b',p,'r'), legend('measured','forecasted')
Why does this not work?
I hope you found a solution to your problem. If you have not, maybe I can be of assistance.
MathWork's documentation of the function notes that the "PastData" entry (labeled "data" in your code) can either be an iddata object or an N x N_y matrix of doubles. Your implementation uses a matrix, so I decided to try out the code with an iddata object.
rawdat = [1166974.25000000, 1132479.36000000, 1137173.86000000, 1145853.58000000, 1118875.72000000, 1071456.85000000 ,1047171.87000000, 1071179.65000000 ,1077986.32000000 ,1112111.10000000, 1149668.47000000 ,1161649.19000000, 1175576.25000000 ,1126753.31000000 ,1204843.11000000 ,1183946.03000000, 1153080.36000000, 1120182.07000000, 1104726.03000000 ,1108110.02000000 ,1137729.28000000 ,1189699.45000000, 1252975.55000000, 1218118.20000000 ,1259580 ,1208193 ,1194430, 1244458, 1218867, 1205705 ,1177362, 1185584, 1164758, 1226991 ,1286044 ,1305312, 1360681.70000000 ,1332020 ,1306497.90000000 ,1299819.10000000 ,1316167.70000000 ,1246959.40000000 ,1256700.20000000 ,1266490.60000000, 1275642.90000000, 1358839.80000000, 1361440.10000000, 1398059.40000000];
data = iddata(rawdat',[]);
sys = ar(data,4);
K = 49;
p = forecast(sys,data,K);
plot(data,'b',p,'r'), legend('measured','forecasted')
Notice that I also changed the initial data's variable name and type.
The above code leads to the following figure.
Please update us. Thanks.
I am usually programming in Python, but for an assignment, I am using Simulink. I am wondering why the above elseif ladder does not generate an incremental increase of the variable [IP3] over time. What I would think it should do is return 0.01 until t = 500, then 0.03 until t = 1000, then 0.1 until 1500, 1 until 2000, and 10 from then on. Apologies for the older image btw, I updated the variables in the mean time.
In the Simulink model that you showed, elseif parts will never execute since:
if u1>0 is satisfied, none of the other conditions will be checked and thus it will always be returning 0.01 for all u1>0.
And when u1<=0, all the conditions will be checked but none of them
will be satisfied. (u1 may never be less than zero in your case as u1 is time).
This behavior is same in every programming language.
Fixing your If-elseif Statements:
You need to add this in the If block:
Under If expression (e.g. u1 ~= 0), write this:
u1>0 & u1<=500
Under Elseif expressions (comma-separated list, e.g. u2 ~= 0, u3(2) < u2):, write this:
u1>500 & u1<=1000, u1>1000 & u1<=1500, u1>1500 & u1<=2000, u1>2000
Since u1 is time in your case which cannot be negative, you may also want to use the else part. So instead of the last step you can also do this:
Under Elseif expressions (comma-separated list, e.g. u2 ~= 0, u3(2) < u2):, write this:
u1>500 & u1<=1000, u1>1000 & u1<=1500, u1>1500 & u1<=2000
and connect the output of the else part which was connected with the output of u1>2000 before.
I want to design logics similar to a counter in Z3py.
If writing python script, we usually define a variable "counter" and then keep incrementing it when necessary. However, in Z3, there is no variant. Therefore, instead of defining an variant, I define a trace of that variant.
This is a sample code. Suppose there is an array "myArray" of size 5, and the elements in the array are 1 or 2. I want to assert a constraint that there must be two '2's in "myArray"
from z3 import *
s = Solver()
myArray = IntVector('myArray',5)
for i in range(5):
s.add(Or(myArray[i]==1,myArray[i]==2))
counterTrace = IntVector('counterTrace',6)
s.add(counterTrace[0]==0)
for i in range(5):
s.add(If(myArray[i]==2,counterTrace[i+1]==counterTrace[i]+1,counterTrace[i+1]==counterTrace[i]))
s.add(counterTrace[5]==2)
print s.check()
print s.model()
My question is that is this an efficient way of implementing the concept of counter in Z3? In my real problem, which is more complicated, this is really inefficient.
You can do this but it is much easier to create the sum over myArray[i] == 2 ? 1 : 0. That way you don't need to assert anything and you are dealing with normal expressions.
I am new to Mathematica and I am having difficulties with one thing. I have this Table that generates 10 000 times 13 numbers (12 numbers + 1 that is a starting number). I need to create a Histogram from all 10 000 13th numbers. I hope It's quite clear, quite tricky to explain.
This is the table:
F = Table[(Xi = RandomVariate[NormalDistribution[], 12];
Mu = -0.00644131;
Sigma = 0.0562005;
t = 1/12; s = 0.6416;
FoldList[(#1*Exp[(Mu - Sigma^2/2)*t + Sigma*Sqrt[t]*#2]) &, s,
Xi]), {SeedRandom[2]; 10000}]
The result for the following histogram could be a table that will take all the 13th numbers to one table - than It would be quite easy to create an histogram. Maybe with "select"? Or maybe you know other ways to solve this.
You can access different parts of a list using Part or (depending on what parts you need) some of the more specialised commands, such as First, Rest, Most and (the one you need) Last. As noted in comments, Histogram[Last/#F] or Histogram[F[[All,-1]]] will work fine.
Although it wasn't part of your question, I would like to note some things you could do for your specific problem that will speed it up enormously. You are defining Mu, Sigma etc 10,000 times, because they are inside the Table command. You are also recalculating Mu - Sigma^2/2)*t + Sigma*Sqrt[t] 120,000 times, even though it is a constant, because you have it inside the FoldList inside the Table.
On my machine:
F = Table[(Xi = RandomVariate[NormalDistribution[], 12];
Mu = -0.00644131;
Sigma = 0.0562005;
t = 1/12; s = 0.6416;
FoldList[(#1*Exp[(Mu - Sigma^2/2)*t + Sigma*Sqrt[t]*#2]) &, s,
Xi]), {SeedRandom[2]; 10000}]; // Timing
{4.19049, Null}
This alternative is ten times faster:
F = Module[{Xi, beta}, With[{Mu = -0.00644131, Sigma = 0.0562005,
t = 1/12, s = 0.6416},
beta = (Mu - Sigma^2/2)*t + Sigma*Sqrt[t];
Table[(Xi = RandomVariate[NormalDistribution[], 12];
FoldList[(#1*Exp[beta*#2]) &, s, Xi]), {SeedRandom[2];
10000}] ]]; // Timing
{0.403365, Null}
I use With for the local constants and Module for the things that are other redefined within the Table (Xi) or are calculations based on the local constants (beta). This question on the Mathematica StackExchange will help explain when to use Module versus Block versus With. (I encourage you to explore the Mathematica StackExchange further, as this is where most of the Mathematica experts are hanging out now.)
For your specific code, the use of Part isn't really required. Instead of using FoldList, just use Fold. It only retains the final number in the folding, which is identical to the last number in the output of FoldList. So you could try:
FF = Module[{Xi, beta}, With[{Mu = -0.00644131, Sigma = 0.0562005,
t = 1/12, s = 0.6416},
beta = (Mu - Sigma^2/2)*t + Sigma*Sqrt[t];
Table[(Xi = RandomVariate[NormalDistribution[], 12];
Fold[(#1*Exp[beta*#2]) &, s, Xi]), {SeedRandom[2];
10000}] ]];
Histogram[FF]
Calculating FF in this way is even a little faster than the previous version. On my system Timing reports 0.377 seconds - but such a difference from 0.4 seconds is hardly worth worrying about.
Because you are setting the seed with SeedRandom, it is easy to verify that all three code examples produce exactly the same results.
Making my comment an answer:
Histogram[Last /# F]