I am looking for a way to simulate the output of a signal for various input signals. To be more precise, I have a system defined by its transfer function H that takes one input and has one output. I generated several signals (stored in a numpy array). What I would like to do, is get the response of the system, to each input signal whithout using a for loop. Is there a way to proceed? Below is the code I wrote so far.
from __future__ import division
import numpy as np
from scipy import signal
nbr_inputs = 5
t_in = np.arange(0,10,0.2)
dim = (nbr_inputs, len(t_in))
x = np.cumsum(np.random.normal(0,2e-3, dim), axis=1)
H = signal.TransferFunction([1, 3, 3], [1, 2, 1])
t_out, y, _ = signal.lsim(H, x[0], t_in) # here, I would just like to simply write x
thanks for your help
This is not a MIMO system, it is a SISO system but you have multiple inputs.
You can create a MIMO system and apply your inputs all at once which will be computed channel by channel but simultaneously. Moreover, you can't use scipy.signal.lsim for MIMO systems yet. You can use other options such as python-control (if you have slycot extension otherwise again no MIMO) or harold if you have Python 3.6 or greater (disclaimer: I'm the author).
import numpy as np
from harold import *
import matplotlib.pyplot
nbr_inputs = 5
t_in = np.arange(0,10,0.2)
dim = (nbr_inputs, len(t_in))
x = np.cumsum(np.random.normal(0,2e-3, dim), axis=1)
# Forming a 1x5 system, common denominator will be completed automatically
H = Transfer([[[1, 3, 3]]*nbr_inputs], [1, 2, 1])
The keyword per_channel=True applies first input to first channel, second input to second and so on. Otherwise combined response is returned. You can check the shapes by playing around with it to see what I mean.
# Notice it is x.T below -> input shape = <num samples>, <num inputs>
y, t = simulate_linear_system(H, x.T, t_in, per_channel=True)
plt.plot(t, y)
This gives
Related
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 am using the following code to run uniform filter on my data:
from scipy.ndimage.filters import uniform_filter
a = np.arange(1000)
b = uniform_filter(a, size=10)
The filter right now semms to work as if a stride was set to size // 2.
How to adjust the code so that the stride of the filter is not half of the size?
You seem to be misunderstanding what uniform_filter is doing.
In this case, it creates an array b that replaces every a[i] with the mean of a block of size 10 centered at a[i]. So, something like:
for i in range(0, len(a)): # for the 1D case
b[i] = mean(a[i-10//2:i+10//2]
Note that this tries to access values with indices outside the range 0..1000. In the default case, uniform_filter supposes that the data before position 0 is just a reflection of the data thereafter. And similarly at the end.
Also note that b uses the same type as a. In the example where a is of integer type, the mean will also be calculated at integer, which can cause some loss of precision.
Here is some code and plot to illustrate what's happening:
import matplotlib.pyplot as plt
import numpy as np
from scipy.ndimage.filters import uniform_filter
fig, axes = plt.subplots(ncols=2, figsize=(15,4))
for ax in axes:
if ax == axes[1]:
a = np.random.uniform(-1,1,50).cumsum()
ax.set_title('random curve')
else:
a = np.arange(50, dtype=float)
ax.set_title('values from 0 to 49')
b = uniform_filter(a, size=10)
ax.plot(a, 'b-')
ax.plot(-np.arange(0, 10)-1, a[:10], 'b:') # show the reflection at the start
ax.plot(50 + np.arange(0, 10), a[:-11:-1], 'b:') # show the reflection at the end
ax.plot(b, 'r-')
plt.show()
I have a set of about 33K (x,y,z) points in a csv file and would like to convert this to a grid of density values using scipy.stats.gaussian_kde. I have not been able to find a way to convert this point cloud array into an appropriate input format for the gaussian_kde function (and then take the output of this and convert it into a density value grid). Can anyone provide sample code?
Here's an example with some comments which may be of use. gaussian_kde wants the data and points to be row stacked, ie. (# ndim, # num values), as per the docs. In your case you would row_stack([x, y, z]) such that the shape is (3, 33000).
from scipy.stats import gaussian_kde
import numpy as np
import matplotlib.pyplot as plt
# simulate some data
n = 33000
x = np.random.randn(n)
y = np.random.randn(n) * 2
# data must be stacked as (# ndim, # n values) as per docs.
data = np.row_stack((x, y))
# perform KDE
kernel = gaussian_kde(data)
# create grid over which to evaluate KDE
s = np.linspace(-8, 8, 128)
grid = np.meshgrid(s, s)
# again KDE needs points to be row_stacked
grid_points = np.row_stack([g.ravel() for g in grid])
# evaluate KDE and reshape result correctly
Z = kernel(grid_points)
Z = Z.reshape(grid[0].shape)
# plot KDE as image and overlay some data points
fig, ax = plt.subplots()
ax.matshow(Z, extent=(s.min(), s.max(), s.min(), s.max()))
ax.plot(x[::10], y[::10], 'w.', ms=1, alpha=0.3)
ax.set_xlim(s.min(), s.max())
ax.set_ylim(s.min(), s.max())
I'd like to build a GP with marginalized hyperparameters.
I have seen that this is possible with the HMC sampler provided in gpflow from this notebook
However, when I tried to run the following code as a first step of this (NOTE this is on gpflow 0.5, an older version), the returned samples are negative, even though the lengthscale and variance need to be positive (negative values would be meaningless).
import numpy as np
from matplotlib import pyplot as plt
import gpflow
from gpflow import hmc
X = np.linspace(-3, 3, 20)
Y = np.random.exponential(np.sin(X) ** 2)
Y = (Y - np.mean(Y)) / np.std(Y)
k = gpflow.kernels.Matern32(1, lengthscales=.2, ARD=False)
m = gpflow.gpr.GPR(X[:, None], Y[:, None], k)
m.kern.lengthscales.prior = gpflow.priors.Gamma(1., 1.)
m.kern.variance.prior = gpflow.priors.Gamma(1., 1.)
# dont want likelihood be a hyperparam now so fixed
m.likelihood.variance = 1e-6
m.likelihood.variance.fixed = True
m.optimize(maxiter=1000)
samples = m.sample(500)
print(samples)
Output:
[[-0.43764571 -0.22753325]
[-0.50418501 -0.11070128]
[-0.5932655 0.00821438]
[-0.70217714 0.05077999]
[-0.77745654 0.09362291]
[-0.79404456 0.13649446]
[-0.83989415 0.27118385]
[-0.90355789 0.29589641]
...
I don't know too much in detail about HMC sampling but I would expect that the sampled posterior hyperparameters are positive, I've checked the code and it seems maybe related to the Log1pe transform, though I failed to figure it out myself.
Any hint on this?
It would be helpful if you specified which GPflow version you are using - especially given that from the output you posted it looks like you are using a really old version of GPflow (pre-1.0), and this is actually something that got improved since. What is happening here (in old GPflow) is that the sample() method returns a single array S x P, where S is the number of samples, and P is the number of free parameters [e.g. for a M x M matrix parameter with lower-triangular transform (such as the Cholesky of the covariance of the approximate posterior, q_sqrt), only M * (M - 1)/2 parameters are actually stored and optimised!]. These are the values in the unconstrained space, i.e. they can take any value whatsoever. Transforms (see gpflow.transforms module) provide the mapping between this value (between plus/minus infinity) and the constrained value (e.g. gpflow.transforms.positive for lengthscales and variances). In old GPflow, the model provides a get_samples_df() method that takes the S x P array returned by sample() and returns a pandas DataFrame with columns for all the trainable parameters which would be what you want. Or, ideally, you would just use a recent version of GPflow, in which the HMC sampler directly returns the DataFrame!
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.