I want to test if
model_1 <- feols(auth ~ dummy_past1 + dummy_past2 + dummy_past3
| region + date_f, # Fixed Effects
data=df)
is just as good as
model_2 <- feols(auth ~ i(dummy_past1,specific_regions) + i(dummy_past2,specific_regions) + i(dummy_past3, specific_regions) # interaction(dummies * specific regions)
| region + date_f, # Fixed Effects
data=df)
If this was a normal linear regression I would conduct a lrtest (likelihood ratio test). As I heard that a waldtest can also be applied to panel data (?) I conducted a waldtest
waldtest(model_1, model_2). Yet, when doing that I get the warning: "Error in solve.default(vc[ovar, ovar]) : system is computationally singular: reciprocal condition number = 2.10749e-44"
Does someone know if doing a waldtest is the correct approach here, or if there are any other tests that I could do on plm or feols regressions? Or it would also be very helpful if you have ideas on how I could get the waldtest working.
Related
I'm a student teaching myself Drake, specifically pydrake with Dr. Russ Tedrake's excellent Underactuated Robotics course. I am trying to write a combined energy shaping and lqr controller for keeping a cartpole system balanced upright. I based the diagram on the cartpole example found in Chapter 3 of Underactuated Robotics [http://underactuated.mit.edu/acrobot.html], and the SwingUpAndBalanceController on Chapter 2: [http://underactuated.mit.edu/pend.html].
I have found that due to my use of the cart_pole.sdf model I have to create an abstract input port due receive FramePoseVector from the cart_pole.get_output_port(0). From there I know that I have to create a control signal output of type BasicVector to feed into a Saturation block before feeding into the cartpole's actuation port.
The problem I'm encountering right now is that I'm not sure how to get the system's current state data in the DeclareVectorOutputPort's callback function. I was under the assumption I would use the LeafContext parameter in the callback function, OutputControlSignal, obtaining the BasicVector continuous state vector. However, this resulting vector, x_bar is always NaN. Out of desperation (and testing to make sure the rest of my program worked) I set x_bar to the controller's initialization cart_pole_context and have found that the simulation runs with a control signal of 0.0 (as expected). I can also set output to 100 and the cartpole simulation just flies off into endless space (as expected).
TL;DR: What is the proper way to obtain the continuous state vector in a custom controller extending LeafSystem with a DeclareVectorOutputPort?
Thank you for any help! I really appreciate it :) I've been teaching myself so it's been a little arduous haha.
# Combined Energy Shaping (SwingUp) and LQR (Balance) Controller
# with a simple state machine
class SwingUpAndBalanceController(LeafSystem):
def __init__(self, cart_pole, cart_pole_context, input_i, ouput_i, Q, R, x_star):
LeafSystem.__init__(self)
self.DeclareAbstractInputPort("state_input", AbstractValue.Make(FramePoseVector()))
self.DeclareVectorOutputPort("control_signal", BasicVector(1),
self.OutputControlSignal)
(self.K, self.S) = BalancingLQRCtrlr(cart_pole, cart_pole_context,
input_i, ouput_i, Q, R, x_star).get_LQR_matrices()
(self.A, self.B, self.C, self.D) = BalancingLQRCtrlr(cart_pole, cart_pole_context,
input_i, ouput_i,
Q, R, x_star).get_lin_matrices()
self.energy_shaping = EnergyShapingCtrlr(cart_pole, x_star)
self.energy_shaping_context = self.energy_shaping.CreateDefaultContext()
self.cart_pole_context = cart_pole_context
def OutputControlSignal(self, context, output):
#xbar = copy(self.cart_pole_context.get_continuous_state_vector())
xbar = copy(context.get_continuous_state_vector())
xbar_ = np.array([xbar[0], xbar[1], xbar[2], xbar[3]])
xbar_[1] = wrap_to(xbar_[1], 0, 2.0*np.pi) - np.pi
# If x'Sx <= 2, then use LQR ctrlr. Cost-to-go J_star = x^T * S * x
threshold = np.array([2.0])
if (xbar_.dot(self.S.dot(xbar_)) < 2.0):
#output[:] = -self.K.dot(xbar_) # u = -Kx
output.set_value(-self.K.dot(xbar_))
else:
self.energy_shaping.get_input_port(0).FixValue(self.energy_shaping_context,
self.cart_pole_context.get_continuous_state_vector())
output_val = self.energy_shaping.get_output_port(0).Eval(self.energy_shaping_context)
output.set_value(output_val)
print(output)
Here are two things that might help:
If you want to get the state of the cart-pole from MultibodyPlant, you probably want to be connecting to the continuous_state output port, which gives you a normal vector instead of the abstract-type FramePoseVector. In that case, your call to get_input_port().Eval(context) should work just fine.
If you do really want to read the FramePoseVector, then you have to evaluate the input port slightly differently. You can find an example of that here.
So I've got a fairly large optimization problem and I'm trying to solve it within a sensible amount of time.
Ive set it up as:
import pulp as pl
my_problem = LpProblem("My problem",LpMinimize)
# write to problem file
my_problem.writeLP("MyProblem.lp")
And then alternatively
solver = CPLEX_CMD(timeLimit=1, gapRel=0.1)
status = my_problem .solve(solver)
solver = pl.apis.CPLEX_CMD(timeLimit=1, gapRel=0.1)
status = my_problem .solve(solver)
path_to_cplex = r'C:\Program Files\IBM\ILOG\CPLEX_Studio1210\cplex\bin\x64_win64\cplex.exe' # and yes this is the actual path on my machine
solver = pl.apis.cplex_api.CPLEX_CMD(timeLimit=1, gapRel=0.1, path=path_to_cplex)
status = my_problem .solve(solver)
solver = pl.apis.cplex_api.CPLEX_CMD(timeLimit=1, gapRel=0.1, path=path_to_cplex)
status = my_problem .solve(solver)
It runs in each case.
However, the solver does not repond to the timeLimit or gapRel instructions.
If I use timelimit it does warn this is depreciated for timeLimit. Same for fracgap: it tells me I should use relGap. So somehow I am talking to the solver.
However, nor matter what values i pick for timeLimit and relGap, it always returns the exact same answer and takes the exact same amount of time (several minutes).
Also, I have tried alternative solvers, and I cannot get any one of them to accept their variants of time limits or optimization gaps.
In each case, the problem solves and returns an status: optimal message. But it just ignores the time limit and gap instructions.
Any ideas?
out of the zoo example:
import pulp
import cplex
bus_problem = pulp.LpProblem("bus", pulp.LpMinimize)
nbBus40 = pulp.LpVariable('nbBus40', lowBound=0, cat='Integer')
nbBus30 = pulp.LpVariable('nbBus30', lowBound=0, cat='Integer')
# Objective function
bus_problem += 500 * nbBus40 + 400 * nbBus30, "cost"
# Constraints
bus_problem += 40 * nbBus40 + 30 * nbBus30 >= 300
solver = pulp.CPLEX_CMD(options=['set timelimit 40'])
bus_problem.solve(solver)
print(pulp.LpStatus[bus_problem.status])
for variable in bus_problem.variables():
print ("{} = {}".format(variable.name, variable.varValue))
Correct way to pass solver option as dictionary
pulp.CPLEX_CMD(options={'timelimit': 40})
#Alex Fleisher has it correct with pulp.CPLEX_CMD(options=['set timelimit 40']). This also works for CBC using the following syntax:
prob.solve(COIN_CMD(options=['sec 60','Presolve More','Multiple 15', 'Node DownFewest','HEUR on', 'Round On','PreProcess Aggregate','PassP 10','PassF 40','Strong 10','Cuts On', 'Gomory On', 'CutD -1', 'Branch On', 'Idiot -1', 'sprint -1','Reduce On','Two On'],msg=True)).
It is important to understand that the parameters, and associated options, are specific to a solver. PuLP seems to be calling CBC via the command line so an investigation of those things is required. Hope that helps
Is there any way to detect the highest peaks using a python library without setting any parameter?. I'm developing a user interface and I want the algorithm to be able to detect highest peaks automatically...
I want it to be able to detect these peaks in picture below:
graph here
Data looks like this:
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-inf
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The task you are describing could be treated like anomaly/outlier detection.
One possible solution is to use a Z-score transformation and treat every value with a z score above a certain threshold as an outlier. Because there is no clear definition of an outlier it won't be able to detect such peaks without setting any parameters (threshold).
One possible solution could be:
import numpy as np
def detect_outliers(data):
outliers = []
d_mean = np.mean(data)
d_std = np.std(data)
threshold = 3 # this defines what you would consider a peak (outlier)
for point in data:
z_score = (point - d_mean)/d_std
if np.abs(z_score) > threshold:
outliers.append(point)
return outliers
# create normal data
data = np.random.normal(size=100)
# create outliers
outliers = np.random.normal(100, size=3)
# combine normal data and outliers
full_data = data.tolist() + outliers.tolist()
# print outliers
print(detect_outliers(full_data))
If you only want to detect peaks, remove the np.abs function call from the code.
This code snippet is based on a Medium Post, which also provides another way of detecting outliers.
I am trying to learn linearK estimates on a small linnet object from the CRC spatstat book (chapter 17) and when I use the linearK function, spatstat throws an error. I have documented the process in the comments in the r code below. The error is as below.
Error in seq.default(from = 0, to = right, length.out = npos + 1L) : 'to' cannot be NA, NaN or infinite
I do not understand how to resolve this. I am following this process:
# I have data of points for each data of the week
# d1 is district 1 of the city.
# I did the step below otherwise it was giving me tbl class
d1_data=lapply(split(d1, d1$openDatefactor),as.data.frame)
# I previously create a linnet and divided it into districts of the city
d1_linnet = districts_linnet[["d1"]]
# I create point pattern for each day
d1_ppp = lapply(d1_data, function(x) as.ppp(x, W=Window(d1_linnet)))
plot(d1_ppp[[1]], which.marks="type")
# I am then converting the point pattern to a point pattern on linear network
d1_lpp <- as.lpp(d1_ppp[[1]], L=d1_linnet, W=Window(d1_linnet))
d1_lpp
Point pattern on linear network
3 points
15 columns of marks: ‘status’, ‘number_of_’, ‘zip’, ‘ward’,
‘police_dis’, ‘community_’, ‘type’, ‘days’, ‘NAME’,
‘DISTRICT’, ‘openDatefactor’, ‘OpenDate’, ‘coseDatefactor’,
‘closeDate’ and ‘instance’
Linear network with 4286 vertices and 6183 lines
Enclosing window: polygonal boundary
enclosing rectangle: [441140.9, 448217.7] x [4640080, 4652557] units
# the errors start from plotting this lpp object
plot(d1_lpp)
"show.all" is not a graphical parameter
Show Traceback
Error in plot.window(...) : need finite 'xlim' values
coords(d1_lpp)
x y seg tp
441649.2 4649853 5426 0.5774863
445716.9 4648692 5250 0.5435492
444724.6 4646320 677 0.9189631
3 rows
And then consequently, I also get error on linearK(d1_lpp)
Error in seq.default(from = 0, to = right, length.out = npos + 1L) : 'to' cannot be NA, NaN or infinite
I feel lpp object has the problem, but I find it hard to interpret the errors and how to resolve them. Could someone please guide me?
Thanks
I can confirm there is a bug in plot.lpp when trying to plot the marked point pattern on the linear network. That will hopefully be fixed soon. You can plot the unmarked point pattern using
plot(unmark(d1_lpp))
I cannot reproduce the problem with linearK. Which version of spatstat are you running? In the development version on my laptop spatstat_1.51-0.073 everything works. There has been changes to this code recently, so it is likely that this will be solved by updating to development version (see https://github.com/spatstat/spatstat).
I hope you can help me with a very simple model I'm running right now in Rjags.
The data I have are as follows:
> print(data)
$R
225738 184094 66275 24861 11266
228662 199379 70308 27511 12229
246808 224814 78255 30447 13425
254823 236063 83099 33148 13961
263772 250706 89182 35450 14750
272844 262707 96918 37116 15715
280101 271612 102604 38692 16682
291493 283018 111125 40996 18064
310474 299315 119354 44552 19707
340975 322054 126901 47757 21510
347597 332946 127708 49103 21354
354252 355994 130561 51925 22421
366818 393534 140628 56562 23711
346430 400629 146037 59594 25313
316438 399545 150733 62414 26720
303294 405876 161793 67060 29545
$N
9597000 8843000 9154000 9956000 11329000
9854932 9349814 9532373 10195193 11357751
9908897 9676950 9303113 10263930 11141510
9981879 9916245 9248586 10270193 10903446
10086567 10093723 9307104 10193818 10660101
10242793 10190641 9479080 10041145 10453320
10434789 10222806 9712544 9835154 10411620
10597293 10238784 10014422 9611918 10489448
10731326 10270163 10229259 9559334 10502839
10805148 10339566 10393532 9625879 10437809
10804571 10459413 10466871 9800559 10292169
10696317 10611599 10477448 10030407 10085603
10540942 10860363 10539271 10245334 9850488
10411836 11053751 10569913 10435763 9797028
10336667 11152428 10652017 10613341 9850533
10283624 11172747 10826549 10719741 9981814
$n
[1] 16
$na
[1] 5
$pbeta
[1] 0.70 0.95
and the model is as follows:
cat('model{
## likelihoods ##
for(k in 1:na){ for(w in 1:n){ R[w,k] ~ dbin( theta[w,k], N[w,k] ) }}
for(k in 1:na){ for(w in 1:n){ theta[w,k] <- 0.5*beta[w,k]*0.5 }}
for(k in 1:na){
beta[1,k] ~ dunif(pbeta[1], pbeta[2])
beta.plus[1,k] <- beta[1,k]
for (w in 2:n){
beta.plus[w,k] ~ dunif(beta[(w-1),k], 0.95)
beta[w,k] <- beta.plus[w,k]} } }',
file='model1.bug')
######## initial random values for beta
bbb=bb.plus=matrix(rep(NA, 16*5), byrow=T, ncol=5);
for(k in 1:5){
bbb[1,k]=runif(1, 0.7,0.95);
for (w in 2:16){
bb.plus[w,k] = runif(1, bbb[w-1,k], 0.95);
bbb[w,k]=bb.plus[w,k]} }
## data & initial values
inits1 <- list('beta'= bbb )
jags_mod <- jags.model('model1.bug', data=data, inits=inits1, n.chains=1, n.adapt=1000)
update(jags_mod, n.iter=1000)
posts=coda.samples(model=jags_mod,variable.names=c('beta','theta'), n.iter=niter, thin=1000)
Super easy. This is actually a scaled down model from a more complex one which gives me exactly the same error message I get here.
Whenever I run this model, no problems at all.
You will notice that the priors for parameter beta are written in such a way to be increasing from 0.7 to 0.95.
Now I would like to "shut off" the likelihood for R by commenting out the first line of the model. I'd like to do so, to see how the parameter theta gets estimated in any case (basically I should find theta=beta/4 in this case, but that would be fine with me)
When I do that, I get an "Invalid parent" error for parameter beta, generally in the bottom rows (rows 15 or 16) of the matrix.
Actually it's more sophisticated than this: sometimes I get an error, and sometimes I don't (mostly, I do).
I don' t understand why this happens: shouldn't the values of beta generated independently from the presence/absence of a likelihood?
Sorry if this is a naive question, I really hope you can help me sort it out.
Thanks, best!
Emanuele
After playing around with the model a bit more I think I found the cause of your problem. One necessary aspect of the uniform distribution (i.e., unif(a,b)) is that a<b. When you are making the uniform distribution smaller and smaller within your model you are bringing a closer and closer to b. At times, it does not reach it, but other times a equals b and you get the invalid parent values error. For example, in your model if you include:
example ~ dunif(0.4,0.4)
You will get "Error in node example, Invalid parent values".
So, to solve this I think it will be easier to adjust how you specify your priors instead of assigning them randomly. You could do this with the beta distribution. At the first step, beta(23.48, 4.98) covers most of the range from 0.7 to 0.95, but we could truncate it to make sure it lies between that range. Then, as n increases you can lower 4.98 so that the prior shrinks towards 0.95. The model below will do this. After inspecting the priors, it does turn out that theta does equal beta/4.
data.list <- list( n = 16, na = 5,
B = rev(seq(0.1, 4.98, length.out = 16)))
cat('model{
## likelihoods ##
#for(k in 1:na){ for(w in 1:n){ R[w,k] ~ dbin( theta[w,k], N[w,k] ) }}
for(k in 1:na){ for(w in 1:n){ theta[w,k] <- 0.5*beta[w,k]*0.5 }}
for(k in 1:na){
for(w in 1:n){
beta[w,k] ~ dbeta(23.48, B[w]) T(0.7,0.95)
} } }',
file='model1.bug')
jags_mod <- jags.model('model1.bug', data=data.list,
inits=inits1, n.chains=1, n.adapt=1000)
update(jags_mod, n.iter=1000)
posts=coda.samples(model=jags_mod,
variable.names=c('beta','theta'), n.iter=10000, thin=10)
Looking at some of the output from this model we get
beta[1,1] theta[1,1]
[1,] 0.9448125 0.2362031
[2,] 0.7788794 0.1947198
[3,] 0.9498806 0.2374702
0.9448125/4
[1] 0.2362031
Since I don't really know what you are trying to use the model for I do not know if the beta distribution would suit your needs, but the above method will mimic what you are trying to do.