In GPflow I have multiple time series and the sampling times are not aligned across time series, and the time series may have different length (longitudinal data). I assume that they are independent realizations from the same GP. What is the right way to handle this with svgp, and more generally with GPflow? Do i need to use coregionalization? The coregionalization notebook assumed correlated trajectories, while I want shared mean/kernel but independent.
Yes, the Coregion kernel implemented in GPflow is what you can use for your problem.
Let's set up some data from the generative model you describe, with different lengths for the timeseries:
import numpy as np
import gpflow
import matplotlib.pyplot as plt
Ns = [80, 90, 100] # number of observations for three different realizations
Xs = [np.random.uniform(0, 10, size=N) for N in Ns] # observation locations
# three different draws from the same GP:
k = gpflow.kernels.Matern52(variance=2.0, lengthscales=0.5) # kernel
Ks = [k(X[:, None]) for X in Xs]
Ls = [np.linalg.cholesky(K) for K in Ks]
vs = [np.random.randn(N, 1) for N in Ns]
fs = [(L # v).squeeze(axis=-1) for L, v in zip(Ls, vs)]
To actually set up the training data for the gpflow GP model:
# output indicator for the observations: which timeseries is this?
os = [o * np.ones(N) for o, N in enumerate(Ns)] # [0 ... 0, 1 ... 1, 2 ... 2]
# now assemble the three timeseries in single data set:
allX = np.concatenate(Xs)
allo = np.concatenate(os)
allf = np.concatenate(fs)
X = np.c_[allX, allo]
Y = allf[:, None]
assert X.shape == (sum(Ns), 2)
assert Y.shape == (sum(Ns), 1)
# now let's set up a copy of the original kernel:
k2 = gpflow.kernels.Matern52(active_dims=[0]) # the same as k above, but with different hyperparameters
# and a Coregionalization kernel that effectively says they are all independent:
kc = gpflow.kernels.Coregion(output_dim=len(Ns), rank=1, active_dims=[1])
kc.W.assign(np.zeros(kc.W.shape))
kc.kappa.assign(np.ones(kc.kappa.shape))
gpflow.set_trainable(kc, False) # we want W and kappa fixed
The Coregion kernel defines a covariance matrix B = W Wᵀ + diag(kappa), so by setting W=0 we prescribe zero correlations (independent realizations) and kappa=1 (actually the default) ensures that the variance hyperparameter of the copy of the original kernel remains interpretable.
Now construct the actual model and optimize hyperparameters:
k2c = k2 * kc
m = gpflow.models.GPR((X, Y), k2c, noise_variance=1e-5)
opt = gpflow.optimizers.Scipy()
opt.minimize(m.training_loss, m.trainable_variables, compile=False)
which recovers the initial variance and lengthscale hyperparameters pretty well.
If you want to predict, you have to provide the extra "output" column in the Xnew argument to m.predict_f(), e.g. as follows:
Xtest = np.linspace(0, 10, 100)
Xtest_augmented = np.c_[Xtest, np.zeros_like(Xtest)]
f_mean, f_var = m.predict_f(Xtest_augmented)
(whether you set the output column to 0, 1, or 2 does not matter, as we set them all to be the same with our choice of W and kappa).
If your input was more than one-dimensional, you could set
active_dims=list(range(X.shape[1] - 1)) for the first kernel(s) and active_dims=[X.shape[1]-1] for the Coregion kernel.
Related
I am not familiar with nonlinear regression and would appreciate some help with running an exponential decay model in R. Please see the graph for how the data looks like. My hunch is that an exponential model might be a good choice. I have one fixed effect and one random effect. y ~ x + (1|random factor). How to get the starting values for the exponential model (please assume that I know nothing about nonlinear regression) in R? How do I subsequently run a nonlinear model with these starting values? Could anyone please help me with the logic as well as the R code?
As I am not familiar with nonlinear regression, I haven't been able to attempt it in R.
raw plot
The correct syntax will depend on your experimental design and model but I hope to give you a general idea on how to get started.
We begin by generating some data that should match the type of data you are working with. You had mentioned a fixed factor and a random one. Here, the fixed factor is represented by the variable treatment and the random factor is represented by the variable grouping_factor.
library(nlraa)
library(nlme)
library(ggplot2)
## Setting this seed should allow you to reach the same result as me
set.seed(3232333)
example_data <- expand.grid(treatment = c("A", "B"),
grouping_factor = c('1', '2', '3'),
replication = c(1, 2, 3),
xvar = 1:15)
The next step is to create some "observations". Here, we use an exponential function y=a∗exp(c∗x) and some random noise to create some data. Also, we add a constant to treatment A just to create some treatment differences.
example_data$y <- ave(example_data$xvar, example_data[, c('treatment', 'replication', 'grouping_factor')],
FUN = function(x) {expf(x = x,
a = 10,
c = -0.3) + rnorm(1, 0, 0.6)})
example_data$y[example_data$treatment == 'A'] <- example_data$y[example_data$treatment == 'A'] + 0.8
All right, now we start fitting the model.
## Create a grouped data frame
exampleG <- groupedData(y ~ xvar|grouping_factor, data = example_data)
## Fit a separate model to each groupped level
fitL <- nlsList(y ~ SSexpf(xvar, a, c), data = exampleG)
## Grab the coefficients of the general model
fxf <- fixed.effects(fit1)
## Add treatment as a fixed effect. Also, use the coeffients from the previous
## regression model as starting values.
fit2 <- update(fit1, fixed = a + c ~ treatment,
start = c(fxf[1], 0,
fxf[2], 0))
Looking at the model output, it will give you information like the following:
Nonlinear mixed-effects model fit by maximum likelihood
Model: y ~ SSexpf(xvar, a, c)
Data: exampleG
AIC BIC logLik
475.8632 504.6506 -229.9316
Random effects:
Formula: list(a ~ 1, c ~ 1)
Level: grouping_factor
Structure: General positive-definite, Log-Cholesky parametrization
StdDev Corr
a.(Intercept) 3.254827e-04 a.(In)
c.(Intercept) 1.248580e-06 0
Residual 5.670317e-01
Fixed effects: a + c ~ treatment
Value Std.Error DF t-value p-value
a.(Intercept) 9.634383 0.2189967 264 43.99329 0.0000
a.treatmentB 0.353342 0.3621573 264 0.97566 0.3301
c.(Intercept) -0.204848 0.0060642 264 -33.77976 0.0000
c.treatmentB -0.092138 0.0120463 264 -7.64867 0.0000
Correlation:
a.(In) a.trtB c.(In)
a.treatmentB -0.605
c.(Intercept) -0.785 0.475
c.treatmentB 0.395 -0.792 -0.503
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-1.93208903 -0.34340037 0.04767133 0.78924247 1.95516431
Number of Observations: 270
Number of Groups: 3
Then, if you wanted to visualize the model fit, you could do the following.
## Here we store the model predictions for visualization purposes
predictionsDf <- cbind(example_data,
predict_nlme(fit2, interval = 'conf'))
## Here we make a graph to check it out
ggplot()+
geom_ribbon(data = predictionsDf,
aes( x = xvar , ymin = Q2.5, ymax = Q97.5, fill = treatment),
color = NA, alpha = 0.3)+
geom_point(data = example_data, aes( x = xvar, y = y, col = treatment))+
geom_line(data = predictionsDf, aes(x = xvar, y = Estimate, col = treatment), size = 1.1)
This shows the model fit.
i run scipy.signal.lsim 10 times, it seems that the x0 only be used in the first time, why?
t=np.linspace(0.0,100,100*100)
transfun=[]
for i in range(10):
transfun.append(signal.lti([1],[1+i,1]))
y=[]
for i in range(10):
y.append(np.sin(2*np.pi*300*t)+np.random.normal(0,1,10000)+50)
sensor_output=[]
for i in range(10):
tout, yout, xout =signal.lsim(transfun[i],y[i],t,X0=[50.0])
sensor_output.append(yout)
fig=plt.figure()
for i in range(10):
plt.subplot(10,1,i+1)
plt.plot(t,y[i])
plt.plot(t,sensor_output[i])
plt.show()
lsim takes initial state vector as an argument, not initial output.
Transfer functions don't really have state vectors, but under the hood lsim is converting the transfer function to a state-space realization (which does have a state vector), and using that to simulate the system.
One problem is that, for a given transfer function, there's no unique realization. lsim doesn't say how it converts transfer functions to state-space realizations, but given your results I took a guess which happened to work (see below), but it's not robust.
To solve this for general transfer functions (i.e., not just first-order), you'd need to work with a specific state-space realization, and also specify more than just initial output, or the problem is under-constrained (I guess a typical approach would be to require d(y)/dt = 0, and similarly for all higher derivatives).
Below is a quick-and-dirty fix for your problem, and a sketch of how to do this for first-order state-space realizations.
import numpy as np
import matplotlib.pyplot as plt
from scipy import signal
nex = 3
t = np.linspace(0, 40, 1001)
taus = [1, 5, 10]
transfun = [signal.lti([1],[tau,1])
for tau in taus]
u = np.tile(50, t.shape)
yinit = 49
sensor_output = [signal.lsim(tf,u,t,X0=[yinit*tau])[1]
for tf, tau in zip(transfun, taus)]
fig=plt.figure()
for i in range(nex):
plt.subplot(nex,1,i+1)
plt.plot(t,u)
plt.plot(t,sensor_output[i])
plt.savefig('img1.png')
# different SS realizations of the same TF need different x0
# to get the same initial y0
g = signal.tf2ss([1], [10, 1])
# create SS system with the same TF
k = 1.234
g2 = (g[0], k*g[1], g[2]/k, g[3])
# desired initial value
y0 = 321
#solve for initial state for the two SS systems
x0 = y0 / g[2]
x0_2 = y0 / g2[2]
output = [signal.lsim(g,u,t,X0=x0)[1],
signal.lsim(g2,u,t,X0=x0_2)[1]]
fig=plt.figure()
for i,out in enumerate(output):
plt.subplot(len(output),1,i+1)
plt.plot(t,u)
plt.plot(t,out)
plt.savefig('img2.png')
plt.show()
I would like to perform some multivariant regression using gaussian process regression as implemented in GPflow using version 2.
Installed with pip install gpflow==2.0.0rc1
Below is some example code that generates some 2D data and then attempts to fit it with using GPR and the finally computes the difference
between the true input data and the GPR prediction.
Eventually I would like to extend to higher dimensions
and do tests against a validation set to check for over-fitting
and experiment with other kernels and "Automatic Relevance Determination"
but understanding how to get this to work is the first step.
Thanks!
The following code snippet will work in a jupyter notebook.
import gpflow
import numpy as np
import matplotlib
from gpflow.utilities import print_summary
%matplotlib inline
matplotlib.rcParams['figure.figsize'] = (12, 6)
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
def gen_data(X, Y):
"""
make some fake data.
X, Y are np.ndarrays with shape (N,) where
N is the number of samples.
"""
ys = []
for x0, x1 in zip(X,Y):
y = x0 * np.sin(x0*10)
y = x1 * np.sin(x0*10)
y += 1
ys.append(y)
return np.array(ys)
# generate some fake data
x = np.linspace(0, 1, 20)
X, Y = np.meshgrid(x, x)
X = X.ravel()
Y = Y.ravel()
z = gen_data(X, Y)
#note X.shape, Y.shape and z.shape
#are all (400,) for this case.
# if you would like to plot the data you can do the following
fig = plt.figure()
ax = Axes3D(fig)
ax.scatter(X, Y, z, s=100, c='k')
# had to set this
# to avoid the following error
# tensorflow.python.framework.errors_impl.InvalidArgumentError: Cholesky decomposition was not successful. The input might not be valid. [Op:Cholesky]
gpflow.config.set_default_positive_minimum(1e-7)
# setup the kernel
k = gpflow.kernels.Matern52()
# set up GPR model
# I think the shape of the independent data
# should be (400, 2) for this case
XY = np.column_stack([[X, Y]]).T
print(XY.shape) # this will be (400, 2)
m = gpflow.models.GPR(data=(XY, z), kernel=k, mean_function=None)
# optimise hyper-parameters
opt = gpflow.optimizers.Scipy()
def objective_closure():
return - m.log_marginal_likelihood()
opt_logs = opt.minimize(objective_closure,
m.trainable_variables,
options=dict(maxiter=100)
)
# predict training set
mean, var = m.predict_f(XY)
print(mean.numpy().shape)
# (400, 400)
# I would expect this to be (400,)
# If it was then I could compute the difference
# between the true data and the GPR prediction
# `diff = mean - z`
# but because the shape is not as expected this of course
# won't work.
The shape of z must be (N, 1), whereas in your case it is (N,). However, this is a missing check in GPflow and not your fault.
I'm trying to convolve two 1D tensors in Keras.
I get two inputs from other models:
x - of length 100
ker - of length 5
I would like to get the 1D convolution of x using the kernel ker.
I wrote a Lambda layer to do it:
import tensorflow as tf
def convolve1d(x):
y = tf.nn.conv1d(value=x[0], filters=x[1], padding='VALID', stride=1)
return y
x = Input(shape=(100,))
ker = Input(shape=(5,))
y = Lambda(convolve1d)([x,ker])
model = Model([x,ker], [y])
I get the following error:
ValueError: Shape must be rank 4 but is rank 3 for 'lambda_67/conv1d/Conv2D' (op: 'Conv2D') with input shapes: [?,1,100], [1,?,5].
Can anyone help me understand how to fix it?
It was much harder than I expected because Keras and Tensorflow don't expect any batch dimension in the convolution kernel so I had to write the loop over the batch dimension myself, which requires to specify batch_shape instead of just shape in the Input layer. Here it is :
import numpy as np
import tensorflow as tf
import keras
from keras import backend as K
from keras import Input, Model
from keras.layers import Lambda
def convolve1d(x):
input, kernel = x
output_list = []
if K.image_data_format() == 'channels_last':
kernel = K.expand_dims(kernel, axis=-2)
else:
kernel = K.expand_dims(kernel, axis=0)
for i in range(batch_size): # Loop over batch dimension
output_temp = tf.nn.conv1d(value=input[i:i+1, :, :],
filters=kernel[i, :, :],
padding='VALID',
stride=1)
output_list.append(output_temp)
print(K.int_shape(output_temp))
return K.concatenate(output_list, axis=0)
batch_input_shape = (1, 100, 1)
batch_kernel_shape = (1, 5, 1)
x = Input(batch_shape=batch_input_shape)
ker = Input(batch_shape=batch_kernel_shape)
y = Lambda(convolve1d)([x,ker])
model = Model([x, ker], [y])
a = np.ones(batch_input_shape)
b = np.ones(batch_kernel_shape)
c = model.predict([a, b])
In the current state :
It doesn't work for inputs (x) with multiple channels.
If you provide several filters, you get as many outputs, each being the convolution of the input with the corresponding kernel.
From given code it is difficult to point out what you mean when you say
is it possible
But if what you mean is to merge two layers and feed merged layer to convulation, yes it is possible.
x = Input(shape=(100,))
ker = Input(shape=(5,))
merged = keras.layers.concatenate([x,ker], axis=-1)
y = K.conv1d(merged, 'same')
model = Model([x,ker], y)
EDIT:
#user2179331 thanks for clarifying your intention. Now you are using Lambda Class incorrectly, that is why the error message is showing.
But what you are trying to do can be achieved using keras.backend layers.
Though be noted that when using lower level layers you will lose some higher level abstraction. E.g when using keras.backend.conv1d you need to have input shape of (BATCH_SIZE,width, channels) and kernel with shape of (kernel_size,input_channels,output_channels). So in your case let as assume the x has channels of 1(input channels ==1) and y also have the same number of channels(output channels == 1).
So your code now can be refactored as follows
from keras import backend as K
def convolve1d(x,kernel):
y = K.conv1d(x,kernel, padding='valid', strides=1,data_format="channels_last")
return y
input_channels = 1
output_channels = 1
kernel_width = 5
input_width = 100
ker = K.variable(K.random_uniform([kernel_width,input_channels,output_channels]),K.floatx())
x = Input(shape=(input_width,input_channels)
y = convolve1d(x,ker)
I guess I have understood what you mean. Given the wrong example code below:
input_signal = Input(shape=(L), name='input_signal')
input_h = Input(shape=(N), name='input_h')
faded= Lambda(lambda x: tf.nn.conv1d(input, x))(input_h)
You want to convolute each signal vector with different fading coefficients vector.
The 'conv' operation in TensorFlow, etc. tf.nn.conv1d, only support a fixed value kernel. Therefore, the code above can not run as you want.
I have no idea, too. The code you given can run normally, however, it is too complex and not efficient. In my idea, another feasible but also inefficient way is to multiply with the Toeplitz matrix whose row vector is the shifted fading coefficients vector. When the signal vector is too long, the matrix will be extremely large.
I am looking at the KalmanFilter from pykalman shown in examples:
pykalman documentation
Example 1
Example 2
and I am wondering
observation_covariance=100,
vs
observation_covariance=1,
the documentation states
observation_covariance R: e(t)^2 ~ Gaussian (0, R)
How should the value be set here correctly?
Additionally, is it possible to apply the Kalman filter without intercept in the above module?
The observation covariance shows how much error you assume to be in your input data. Kalman filter works fine on normally distributed data. Under this assumption you can use the 3-Sigma rule to calculate the covariance (in this case the variance) of your observation based on the maximum error in the observation.
The values in your question can be interpreted as follows:
Example 1
observation_covariance = 100
sigma = sqrt(observation_covariance) = 10
max_error = 3*sigma = 30
Example 2
observation_covariance = 1
sigma = sqrt(observation_covariance) = 1
max_error = 3*sigma = 3
So you need to choose the value based on your observation data. The more accurate the observation, the smaller the observation covariance.
Another point: you can tune your filter by manipulating the covariance, but I think it's not a good idea. The higher the observation covariance value the weaker impact a new observation has on the filter state.
Sorry, I did not understand the second part of your question (about the Kalman Filter without intercept). Could you please explain what you mean?
You are trying to use a regression model and both intercept and slope belong to it.
---------------------------
UPDATE
I prepared some code and plots to answer your questions in details. I used EWC and EWA historical data to stay close to the original article.
First of all here is the code (pretty the same one as in the examples above but with a different notation)
from pykalman import KalmanFilter
import numpy as np
import matplotlib.pyplot as plt
# reading data (quick and dirty)
Datum=[]
EWA=[]
EWC=[]
for line in open('data/dataset.csv'):
f1, f2, f3 = line.split(';')
Datum.append(f1)
EWA.append(float(f2))
EWC.append(float(f3))
n = len(Datum)
# Filter Configuration
# both slope and intercept have to be estimated
# transition_matrix
F = np.eye(2) # identity matrix because x_(k+1) = x_(k) + noise
# observation_matrix
# H_k = [EWA_k 1]
H = np.vstack([np.matrix(EWA), np.ones((1, n))]).T[:, np.newaxis]
# transition_covariance
Q = [[1e-4, 0],
[ 0, 1e-4]]
# observation_covariance
R = 1 # max error = 3
# initial_state_mean
X0 = [0,
0]
# initial_state_covariance
P0 = [[ 1, 0],
[ 0, 1]]
# Kalman-Filter initialization
kf = KalmanFilter(n_dim_obs=1, n_dim_state=2,
transition_matrices = F,
observation_matrices = H,
transition_covariance = Q,
observation_covariance = R,
initial_state_mean = X0,
initial_state_covariance = P0)
# Filtering
state_means, state_covs = kf.filter(EWC)
# Restore EWC based on EWA and estimated parameters
EWC_restored = np.multiply(EWA, state_means[:, 0]) + state_means[:, 1]
# Plots
plt.figure(1)
ax1 = plt.subplot(211)
plt.plot(state_means[:, 0], label="Slope")
plt.grid()
plt.legend(loc="upper left")
ax2 = plt.subplot(212)
plt.plot(state_means[:, 1], label="Intercept")
plt.grid()
plt.legend(loc="upper left")
# check the result
plt.figure(2)
plt.plot(EWC, label="EWC original")
plt.plot(EWC_restored, label="EWC restored")
plt.grid()
plt.legend(loc="upper left")
plt.show()
I could not retrieve data using pandas, so I downloaded them and read from the file.
Here you can see the estimated slope and intercept:
To test the estimated data I restored the EWC value from the EWA using the estimated parameters:
About the observation covariance value
By varying the observation covariance value you tell the Filter how accurate the input data is (normally you just describe your confidence in the observation using some datasheets or your knowledge about the system).
Here are estimated parameters and the restored EWC values using different observation covariance values:
You can see the filter follows the original function better with a bigger confidence in observation (smaller R). If the confidence is low (bigger R) the filter leaves the initial estimate (slope = 0, intercept = 0) very slowly and the restored function is far away from the original one.
About the frozen intercept
If you want to freeze the intercept for some reason, you need to change the whole model and all filter parameters.
In the normal case we had:
x = [slope; intercept] #estimation state
H = [EWA 1] #observation matrix
z = [EWC] #observation
Now we have:
x = [slope] #estimation state
H = [EWA] #observation matrix
z = [EWC-const_intercept] #observation
Results:
Here is the code:
from pykalman import KalmanFilter
import numpy as np
import matplotlib.pyplot as plt
# only slope has to be estimated (it will be manipulated by the constant intercept) - mathematically incorrect!
const_intercept = 10
# reading data (quick and dirty)
Datum=[]
EWA=[]
EWC=[]
for line in open('data/dataset.csv'):
f1, f2, f3 = line.split(';')
Datum.append(f1)
EWA.append(float(f2))
EWC.append(float(f3))
n = len(Datum)
# Filter Configuration
# transition_matrix
F = 1 # identity matrix because x_(k+1) = x_(k) + noise
# observation_matrix
# H_k = [EWA_k]
H = np.matrix(EWA).T[:, np.newaxis]
# transition_covariance
Q = 1e-4
# observation_covariance
R = 1 # max error = 3
# initial_state_mean
X0 = 0
# initial_state_covariance
P0 = 1
# Kalman-Filter initialization
kf = KalmanFilter(n_dim_obs=1, n_dim_state=1,
transition_matrices = F,
observation_matrices = H,
transition_covariance = Q,
observation_covariance = R,
initial_state_mean = X0,
initial_state_covariance = P0)
# Creating the observation based on EWC and the constant intercept
z = EWC[:] # copy the list (not just assign the reference!)
z[:] = [x - const_intercept for x in z]
# Filtering
state_means, state_covs = kf.filter(z) # the estimation for the EWC data minus constant intercept
# Restore EWC based on EWA and estimated parameters
EWC_restored = np.multiply(EWA, state_means[:, 0]) + const_intercept
# Plots
plt.figure(1)
ax1 = plt.subplot(211)
plt.plot(state_means[:, 0], label="Slope")
plt.grid()
plt.legend(loc="upper left")
ax2 = plt.subplot(212)
plt.plot(const_intercept*np.ones((n, 1)), label="Intercept")
plt.grid()
plt.legend(loc="upper left")
# check the result
plt.figure(2)
plt.plot(EWC, label="EWC original")
plt.plot(EWC_restored, label="EWC restored")
plt.grid()
plt.legend(loc="upper left")
plt.show()