I have a question on gretl and how I can compute the filter of moving avarage.
I have a time series and I want to calculate the weighted moving avarage centered in 5 with these weights: 0.15, 0.2, 0.3, 0.2, 0.15.
In the main page of gretl we have the Variabile window where I can select Filter but there's no option for what I want to do, only, for example, simple moving avarage.
In R I would do something like this:
c<-as.vector()
for (in in 3:(T-2)){
c<-rbind(c, 0.15*x[i-2]+0.2*x[i-1]+0.3*x[i]+0.2*x[i+1]+0.15*x[i+2]}
where x is my time seriee and T is the number of observations.
But my questions are:
Does it exist an user-friendly way to do it in gretl?
If not, what is the best way to do it in the console? Does it exist a specific function?
Well I don't know what exactly you call user friendly, but since you want to have those specific weights, I guess there's no way around typing in some numbers, right?
So if I understand you correctly, and given your series x (in a dataset which is declared and recognized as a time series), then you simply would need to type the formula:
series weighma = 0.15 * x(+2) + 0.2 * x(+1) + 0.3 * x + 0.2 * x(-1) + 0.15 * x(-2)
(Instead of 'series' you could also type in 'genr' or just omit it, but I recommend this explicit variant. The same goes for the + signs inside the parentheses to indicate leads instead of lags.)
The name 'weighma' is of course arbitrary.
There are at least two places where you could type in that formula: Either choose Add /Define new variable from the menus, which gives you a dialog window with a formula field, or open the gretl console (or a script editor window).
A solution which would perhaps be more flexible in a script could use a gretl list of variables and the 'lincomb' function, something like this:
maxlead = 2
matrix weights = {0.15, 0.2, 0.3, 0.2, 0.15}
list xx = lags( nelem(weights), x(maxlead + 1) )
series weighma = lincomb(xx, weights)
The correct maxlead value could also be inferred from the length of the weights vector under the assumption of a centered MA, but I leave it at that.
Related
I am trying to formulate a trajectory optimization problem for a glider, where I want to maximize the average horisontal velocity. I have formulated the system as a drakesystem, and the state vector consists of the position and velocity.
Currently, I have something like the following:
dircol = DirectCollocation(
plant,
context,
num_time_samples=N,
minimum_timestep=min_dt,
maximum_timestep=max_dt,
)
... # other constraints etc
horisontal_pos = dircol.state()[0:2] # Only (x,y)
time = dircol.time()
dircol.AddFinalCost(-w.T.dot(horisontal_pos) / time)
where AddFinalCost() should replace all instances of state() and time() with the final values, as far as I understand from the documentation. min_dt is non-zero and w is a vector of linear weights.
However, I am getting the following error message
Expression (...) is not a polynomial. ParseCost does not support non-polynomial expression.
which makes me think that there is no way of adding the type of cost function that I am looking for. Is there anything that I am missing?
Thank you in advance!
When calling AddFinalCost(e) with e being a symbolic expression, we can only handle it when e is a polynomial function of the state (more precisely, either a quadratic function or a linear function). Hence the error you see complaining that the cost is not polynomial.
You could add the cost like this
def average_speed(v):
x = v[0]
time_steps = v[1:]
return v[0] / np.sum(time_steps)
h_vars = [dircol.timestep[i] for i in range(N-1)]
dircol.AddCost(average_speed, vars=[dircol.state(N-1)[0]] + h_vars)
which uses a function average_speed to evaluate the average speed. You could find example of doing this in https://github.com/RobotLocomotion/drake/blob/e5f3c3e5f7927ef675066d97d3afac55d3481305/bindings/pydrake/solvers/test/mathematicalprogram_test.py#L590
First, the cost function should be a scalar, but you a vector-valued horisontal_pos / time, which has two entries containing both position_x / dt and position_y / dt, namely a vector as the cost. You should instead provide a scalar valued cost.
Second, it is unclear to me why you divide time in the final cost. As far as I understand it, you want the final position to be close to the origin, so something like position_x² + position_y². The code can look like
dircol.AddFinalCost(horisontal_pos[0]**2 + horisontal_pos[1]**2)
I have a python xarray dataset with time,x,y for its dimensions and value1 as its variable. I'm trying to compute annual mean of value1 for each x,y coordinate pair.
I've run into this function while reading the docs:
ds.groupby('time.year').mean()
This seems to compute a single annual mean for all x,y coordinate pairs in value1 at each given time slice
rather than the annual means of individual x,y coordinate pairs at each given time slice.
While the code snippet above produces the wrong output, I'm very interested in its oversimplified form. I would really like to figure out the "X-arrays trick" to doing annual mean for a given x,y coordinate pair rather than hacking it together myself.
Cam someone point me in the right direction? Should I temporarily turn this into a pandas object?
To avoid the default of averaging over all dimensions, you simply need to supply the dimension you want to average over explicitly:
ds.groupby('time.year').mean('time')
Note, that calling ds.groupby('time.year').mean('time') will be incorrect if you are working with monthly and not daily data. Taking the mean will place equal weight on months of different length, e.g., Feb and July, which is wrong.
Instead use below from NCAR:
def weighted_temporal_mean(ds, var):
"""
weight by days in each month
"""
# Determine the month length
month_length = ds.time.dt.days_in_month
# Calculate the weights
wgts = month_length.groupby("time.year") / month_length.groupby("time.year").sum()
# Make sure the weights in each year add up to 1
np.testing.assert_allclose(wgts.groupby("time.year").sum(xr.ALL_DIMS), 1.0)
# Subset our dataset for our variable
obs = ds[var]
# Setup our masking for nan values
cond = obs.isnull()
ones = xr.where(cond, 0.0, 1.0)
# Calculate the numerator
obs_sum = (obs * wgts).resample(time="AS").sum(dim="time")
# Calculate the denominator
ones_out = (ones * wgts).resample(time="AS").sum(dim="time")
# Return the weighted average
return obs_sum / ones_out
average_weighted_temp = weighted_temporal_mean(ds_first_five_years, 'TEMP')
I have a convolutional neural network whose output is a 4-channel 2D image. I want to apply sigmoid activation function to the first two channels and then use BCECriterion to computer the loss of the produced images with the ground truth ones. I want to apply squared loss function to the last two channels and finally computer the gradients and do backprop. I would also like to multiply the cost of the squared loss for each of the two last channels by a desired scalar.
So the cost has the following form:
cost = crossEntropyCh[{1, 2}] + l1 * squaredLossCh_3 + l2 * squaredLossCh_4
The way I'm thinking about doing this is as follow:
criterion1 = nn.BCECriterion()
criterion2 = nn.MSECriterion()
error = criterion1:forward(model.output[{{}, {1, 2}}], groundTruth1) + l1 * criterion2:forward(model.output[{{}, {3}}], groundTruth2) + l2 * criterion2:forward(model.output[{{}, {4}}], groundTruth3)
However, I don't think this is the correct way of doing it since I will have to do 3 separate backprop steps, one for each of the cost terms. So I wonder, can anyone give me a better solution to do this in Torch?
SplitTable and ParallelCriterion might be helpful for your problem.
Your current output layer is followed by nn.SplitTable that splits your output channels and converts your output tensor into a table. You can also combine different functions by using ParallelCriterion so that each criterion is applied on the corresponding entry of output table.
For details, I suggest you read documentation of Torch about tables.
After comments, I added the following code segment solving the original question.
M = 100
C = 4
H = 64
W = 64
dataIn = torch.rand(M, C, H, W)
layerOfTables = nn.Sequential()
-- Because SplitTable discards the dimension it is applied on, we insert
-- an additional dimension.
layerOfTables:add(nn.Reshape(M,C,1,H,W))
-- We want to split over the second dimension (i.e. channels).
layerOfTables:add(nn.SplitTable(2, 5))
-- We use ConcatTable in order to create paths accessing to the data for
-- numereous number of criterions. Each branch from the ConcatTable will
-- have access to the data (i.e. the output table).
criterionPath = nn.ConcatTable()
-- Starting from offset 1, NarrowTable will select 2 elements. Since you
-- want to use this portion as a 2 dimensional channel, we need to combine
-- then by using JoinTable. Without JoinTable, the output will be again a
-- table with 2 elements.
criterionPath:add(nn.Sequential():add(nn.NarrowTable(1, 2)):add(nn.JoinTable(2)))
-- SelectTable is simplified version of NarrowTable, and it fetches the desired element.
criterionPath:add(nn.SelectTable(3))
criterionPath:add(nn.SelectTable(4))
layerOfTables:add(criterionPath)
-- Here goes the criterion container. You can use this as if it is a regular
-- criterion function (Please see the examples on documentation page).
criterionContainer = nn.ParallelCriterion()
criterionContainer:add(nn.BCECriterion())
criterionContainer:add(nn.MSECriterion())
criterionContainer:add(nn.MSECriterion())
Since I used almost every possible table operation, it looks a little bit nasty. However, this is the only way I could solve this problem. I hope that it helps you and others suffering from the same problem. This is how the result looks like:
dataOut = layerOfTables:forward(dataIn)
print(dataOut)
{
1 : DoubleTensor - size: 100x2x64x64
2 : DoubleTensor - size: 100x1x64x64
3 : DoubleTensor - size: 100x1x64x64
}
This has become quite a frustrating question, but I've asked in the Coursera discussions and they won't help. Below is the question:
I've gotten it wrong 6 times now. How do I normalize the feature? Hints are all I'm asking for.
I'm assuming x_2^(2) is the value 5184, unless I am adding the x_0 column of 1's, which they don't mention but he certainly mentions in the lectures when talking about creating the design matrix X. In which case x_2^(2) would be the value 72. Assuming one or the other is right (I'm playing a guessing game), what should I use to normalize it? He talks about 3 different ways to normalize in the lectures: one using the maximum value, another with the range/difference between max and mins, and another the standard deviation -- they want an answer correct to the hundredths. Which one am I to use? This is so confusing.
...use both feature scaling (dividing by the
"max-min", or range, of a feature) and mean normalization.
So for any individual feature f:
f_norm = (f - f_mean) / (f_max - f_min)
e.g. for x2,(midterm exam)^2 = {7921, 5184, 8836, 4761}
> x2 <- c(7921, 5184, 8836, 4761)
> mean(x2)
6676
> max(x2) - min(x2)
4075
> (x2 - mean(x2)) / (max(x2) - min(x2))
0.306 -0.366 0.530 -0.470
Hence norm(5184) = 0.366
(using R language, which is great at vectorizing expressions like this)
I agree it's confusing they used the notation x2 (2) to mean x2 (norm) or x2'
EDIT: in practice everyone calls the builtin scale(...) function, which does the same thing.
It's asking to normalize the second feature under second column using both feature scaling and mean normalization. Therefore,
(5184 - 6675.5) / 4075 = -0.366
Usually we normalize all of them to have zero mean and go between [-1, 1].
You can do that easily by dividing by the maximum of the absolute value and then remove the mean of the samples.
"I'm assuming x_2^(2) is the value 5184" is this because it's the second item in the list and using the subscript _2? x_2 is just a variable identity in maths, it applies to all rows in the list. Note that the highest raw mid-term exam result (i.e. that which is not squared) goes down on the final test and the lowest raw mid-term result increases the most for the final exam result. Theta is a fixed value, a coefficient, so somewhere your normalisation of x_1 and x_2 values must become (EDIT: not negative, less than 1) in order to allow for this behaviour. That should hopefully give you a starting basis, by identifying where the pivot point is.
I had the same problem, in my case the thing was that I was using as average the maximum x2 value (8836) minus minimum x2 value (4761) divided by two, instead of the sum of each x2 value divided by the number of examples.
For the same training set, I got the question as
Q. What is the normalized feature x^(3)_1?
Thus, 3rd training ex and 1st feature makes out to 94 in above table.
Now, normalized form is
x = (x - mean(x's)) / range(x)
Values are :
x = 94
mean(89+72+94+69) / 4 = 81
range = 94 - 69 = 25
Normalized x = (94 - 81) / 25 = 0.52
I'm taking this course at the moment and a really trivial mistake I made first time I answered this question was using comma instead of dot in the answer, since I did by hand and in my country we use comma to denote decimals. Ex:(0,52 instead of 0.52)
So in the second time I tried I used dot and works fine.
We have a boolean variable X which is either true or false and alternates at each time step with a probability p. I.e. if p is 0.2, X would alternate once every 5 time steps on average. We also have a time line and observations of the value of this variable at various non-uniformly sampled points in time.
How would one learn, from observations, the probability that after t+n time steps where t is the time X is observed and n is some time in the future that X has alternated/changed value at t+n given that p is unknown and we only have observations of the value of X at previous times? Note that I count changing from true to false and back to true again as changing value twice.
I'm going to approach this problem as if it were on a test.
First, let's name the variables.
Bx is value of the boolean variable after x opportunities to flip (and B0 is the initial state). P is the chance of changing to a different value every opportunity.
Given that each flip opportunity is not related to other flip opportunities (there is, for example, no minimum number of opportunities between flips) the math is extremely simple; since events are not affected by the events of the past, we can consolidate them into a single computation, which works best when considering Bx not as a boolean value, but as itself a probability.
Here is the domain of the computations we will use: Bx is a probability (with a value between 0 and 1 inclusive) representing the likelyhood of truth. P is a probability (with a value between 0 and 1 inclusive) representing the likelyhood of flipping at any given opportunity.
The probability of falseness, 1 - Bx, and the probability of not flipping, 1 - P, are probabilistic identities which should be quite intuitive.
Assuming these simple rules, the general probability of truth of the boolean value is given by the recursive formula Bx+1 = Bx*(1-P) + (1-Bx)*P.
Code (in C++, because it's my favorite language and you didn't tag one):
int max_opportunities = 8; // Total number of chances to flip.
float flip_chance = 0.2; // Probability of flipping each opportunity.
float probability_true = 1.0; // Starting probability of truth.
// 1.0 is "definitely true" and 0.0 is
// "definitely false", but you can extend this
// to situations where the initial value is not
// certain (say, 0.8 = 80% probably true) and
// it will work just as well.
for (int opportunities = 0; opportunities < max_opportunities; ++opportunities)
{
probability_true = probability_true * (1 - flip_chance) +
(1 - probability_true) * flip_chance;
}
Here is that code on ideone (the answer for P=0.2 and B0=1 and x=8 is B8=0.508398). As you would expect, given that the value becomes less and less predictable as more and more opportunities pass, the final probability will approach Bx=0.5. You will also observe oscillations between more and less likely to be true, if your chance of flipping is high (for instance, with P=0.8, the beginning of the sequence is B={1.0, 0.2, 0.68, 0.392, 0.46112, ...}.
For a more complete solution that will work for more complicated scenarios, consider using a stochastic matrix (page 7 has an example).