How to fit data to model with analytical gradient in basinhopping or with another gradient descent method? - tensorflow
I'd like to fit experimental data to a model and extract the optimal model parameters, the parameters that result in minimal error between model function and experimental data. To get the optimal parameters, I'd like to use a gradient descent method, tensorflow, Bayesian inference or basinhopping or something that deals well with bad initial estimates and is rigid. To speed things up, I'd like to use the analytical gradient for example in basinhopping. How do I do that with the basinghopping routine from scipy. In the following example code, I have some example function and I'd like to use the analytical Jacobian instead of the numerical one, but I get an error. Do I have to sum up the Jacobian components?
Example code (my actual function is much more complex)
import random
import matplotlib.pyplot as plt
import numpy as np
# symbolic math
from sympy import lambdify, symbols, cos
from sympy.tensor.array import derive_by_array
# fitting
from scipy.optimize import basinhopping
# symbolic math with sympy ---
s_lst = x, a, b, c, d = symbols('x, a, b, c d', positive=True)
# mathematical function
y = a*x + cos(b*x)**2 * c*x**2 + d
# jacobian (derivatives after model parameters)
params = s_lst[1:]
jac_y = derive_by_array(y, params)
# translate sympy expression to python function
# function
get_y = lambdify(s_lst, y)
# jacobian (derivatives in a, b, c, d)
get_jac_y = [lambdify(s_lst, element) for element in jac_y]
#print(len(get_jac_y))
# data ---
x = np.linspace(0, 1, 500)
# measurement data
a = [random.randrange(4, 6, 1) for i in range(len(x))]
b = [random.randrange(3190, 3290, 1) for i in range(len(x))]
c = [random.randrange(90, 109, 1) for i in range(len(x))]
d = [0.1*random.randrange(0, 2, 1) for i in range(len(x))]
y_measured = get_y(x, a, b, c, d)
# exemplary model data
a, b, c, d = 5, 3200, 100, 1
y_model = get_y(x, a, b, c, d)
# plot
plt.plot(x, y_measured)
plt.plot(x, y_model)
plt.title('exemplary model and measured data')
plt.show()
# functions for fitting
def func(params, args1, args2=None):
a, b, c, d = params
y = get_y(args1, a, b, c, d)
if args2 is None:
return y
return np.sum((y - args2)**2)
# derivatives
def dfunc(params, args1, args2):
a, b, c, d = params
jac = [jac(args1, a, b, c, d) for jac in get_jac_y]
# because derviative in d is one
jac[-1] = np.ones(len(args1))
return np.asarray(jac)
# function and derivatives
def objective_func(params, args1, args2):
f = func(params, args1, args2)
df = dfunc(params, args1, args2)
return f, df
# fit with basinhopping and scipy ---
# initial model parameters
x0 = [1, 2, 33, 4]
# minimization with numerical jacobian, gives a result
minimizer_kwargs = {"args":(x, y_measured), 'method':'L-BFGS-B'}
ret = basinhopping(func, x0, minimizer_kwargs=minimizer_kwargs)
# minimization with analytical jacobian, fails,
# error: failed in converting 7th argument `g' of _lbfgsb.setulb to C/Fortran array
minimizer_kwargs = {"args":(x, y_measured), 'method':'L-BFGS-B', 'jac':True}
ret = basinhopping(objective_func, x0, minimizer_kwargs=minimizer_kwargs)
If I put in dfunc something like return [np.sum((j)) for j in jac] the program runs but fails. What would be the correct expression?
Related
Make an LSTM prediction written in NumPy faster
I have written a bidirectional-LSTM prediction function using NumPy (and not Tensorflow nor PyTorch), and I need to make it faster. The network has three layers, but for the sake of simplicity, I will just show (and time) the first layer. This bi-LSTM layer is called by calling the subfunctions LSTMf() and LSTMb() to process the input data (array of 500 points) forward and backwards. The LSTMf() and LSTMb() have loops which I suspect take the most time. Here is the prediction function: import numpy as np def predict(xt, ht, c, u, t, whff, wxff, bff, whif, wxif, bif, whlf, wxlf, blf, whof, wxof, bof, whfb, wxfb, bfb, whib, wxib, bib, whlb, wxlb, blb, whob, wxob, bob): def tanh(a): return np.tanh(a) def sig(a): return 1 / (1 + np.exp(-a)) def cell(x, h, c, wh1, wx1, b1, wh2, wx2, b2, wh3, wx3, b3, wh4, wx4, b4): new_c = c * sig(h # wh1 + x # wx1 + b1) + sig(h # wh2 + x # wx2 + b2) * tanh(h # wh3 + x # wx3 + b3) new_h = tanh(new_c) * sig(h # wh4 + x # wx4 + b4) return new_c, new_h def LSTMf(xt, ht, c, t, whf, wxf, bf, whi, wxi, bi, whl, wxl, bl, who, wxo, bo): h = ht[t - 1:t] for i in range(t): c, h = cell(xt[i:i + 1], h, c, whf, wxf, bf, whi, wxi, bi, whl, wxl, bl, who, wxo, bo) ht[i] = h return ht def LSTMb(xt, ht, c, t, whf, wxf, bf, whi, wxi, bi, whl, wxl, bl, who, wxo, bo): h = ht[0:1] for i in range(t - 1, -1, -1): c, h = cell(xt[i:i + 1], h, c, whf, wxf, bf, whi, wxi, bi, whl, wxl, bl, who, wxo, bo) ht[i] = h return ht # LSTM-bi 1 hf = LSTMf(xt, ht.copy(), c, t, whff, wxff, bff, whif, wxif, bif, whlf, wxlf, blf, whof, wxof, bof) hb = LSTMb(xt, ht.copy(), c, t, whfb, wxfb, bfb, whib, wxib, bib, whlb, wxlb, blb, whob, wxob, bob) xt = np.concatenate((hf, hb), axis=1) return xt The input data and the rest of parameters can be artificially generated with the following code: t = 500 # input's number of points u = 64 # layer's number of units xt = np.zeros((t, 1), dtype=np.float32) # input ht = np.zeros((t, u), dtype=np.float32) ou = np.zeros((1, u), dtype=np.float32) uu = np.zeros((u, u), dtype=np.float32) weights = {'wxif':ou,'wxff':ou,'wxlf':ou,'wxof':ou,'whif':uu,'whff':uu,'whlf':uu,'whof':uu,'bif':ou,'bff':ou,'blf':ou,'bof':ou, 'wxib':ou,'wxfb':ou,'wxlb':ou,'wxob':ou,'whib':uu,'whfb':uu,'whlb':uu,'whob':uu,'bib':ou,'bfb':ou,'blb':ou,'bob':ou} yt = predict(xt, ht, ou, **weights) # Call example I have timed it (1) like this, (2) with Numba, and (3) with Cython: import numpy as np from predict import predict from predict_numba import predict_numba from predict_cython import predict_cython import timeit n = 100 print(timeit.Timer(lambda: predict(xt, ht, ou, u, t, **weights)).timeit(n)/n) # 0.05198 s predict_numba(xt, ht, ou, u, t, **weights) # Dummy slow numba call print(timeit.Timer(lambda: predict_numba(xt, ht, ou, u, t, **weights)).timeit(n)/n) # 0.01149 s print(timeit.Timer(lambda: predict_cython(xt, ht, ou, u, t, **weights)).timeit(n)/n) # 0.13345 s I would like to make this prediction faster than 0.03 s. Numba is fast enough but I cannot have a very slow first call (more than 30 s for the three layers) Cython is very slow; I'm not sure if this is the reason, but following the advice here (Cython: matrix multiplication) I did not type most parameters since the operation '#' does not support memory views. Originally I was using Keras with CPU or GPU, but NumPy is faster than either. I have also heard of TorchScript which might be applicable. What can I do to make the prediction faster? __ Context: This function predicts the R-peaks in an ECG window, and is meant to be called as frequently as possible, to predict the R-peaks of an ECG being acquired in real-time. PS. In case you want to make sense of the calculations, this description of how an LSTM cell works might be of use: https://imgur.com/UFrd9oa
What is wrong with my cython implementation of erosion operation of mathematical morphology
I have produced a naive implementation of "erosion". The performance is not relevant since I just trying to understand the algorithm. However, the output of my implementation does not match the one I get from scipy.ndimage. What is wrong with my implementation ? Here is my implementation with a small test case: import numpy as np from PIL import Image # a small image to play with a cross structuring element imgmat = np.array([ [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,1,0,0,1,1,1,1,0,0,0,1,1,0,0,1,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,1,0,0,1,1,1,1,0,0,0,1,1,1,0,1,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,1,0,0,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,1,0,0,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,1,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,1,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,1,0,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], [0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,1,1,1,1,0,0,1,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,1,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0], ]) imgmat2 = np.where(imgmat == 0, 0, 255).astype(np.uint8) imarr = Image.fromarray(imgmat2).resize((100, 200)) imarr = np.array(imgrrr) imarr = np.where(imarr == 0, 0, 1) se_mat3 = np.array([ [0,1,0], [1,1,1], [0,1,0] ]) se_mat31 = np.where(se_mat3 == 1, 0, 1) The imarr is . My implementation of erosion: %%cython -a import numpy as np cimport numpy as cnp cdef erosionC(cnp.ndarray[cnp.int_t, ndim=2] img, cnp.ndarray[cnp.int_t, ndim=2] B, cnp.ndarray[cnp.int_t, ndim=2] X): """ X: image coordinates struct_element_mat: black and white image, black region is considered as the shape of structuring element This operation checks whether (B *includes* X) = $B \subset X$ as per defined in Serra (Jean), « Introduction to mathematical morphology », Computer Vision, Graphics, and Image Processing, vol. 35, nᵒ 3 (septembre 1986). URL : https://linkinghub.elsevier.com/retrieve/pii/0734189X86900022.. doi: 10.1016/0734-189X(86)90002-2 Consulted le 6 août 2020, p. 283‑305. """ cdef cnp.ndarray[cnp.int_t, ndim=1] a, x, bx cdef cnp.ndarray[cnp.int_t, ndim=2] Bx, B_frame, Xcp, b cdef bint check a = B[0] # get an anchor point from the structuring element coordinates B_frame = B - a # express the se element coordinates in with respect to anchor point Xcp = X.copy() b = img.copy() for x in X: # X contains the foreground coordinates in the image Bx = B_frame + x # translate relative coordinates with respect to foreground coordinates considering it as the anchor point check = True # this is erosion so if any of the se coordinates is not in foreground coordinates we consider it a miss for bx in Bx: # Bx contains all the translated coordinates of se if bx not in Xcp: check = False if check: b[x[0], x[1]] = 1 # if there is a hit else: b[x[0], x[1]] = 0 # if there is no hit return b def erosion(img: np.ndarray, struct_el_mat: np.ndarray, foregroundValue = 0): B = np.argwhere(struct_el_mat == 0) X = np.argwhere(img == foregroundValue) nimg = erosionC(img, B, X) return np.where(nimg == 1, 255, 0) The calling code for both is: from scipy import ndimage as nd err = nd.binary_erosion(imarr, se_mat3) imerrCustom = erosion(imarr, se_mat31, foregroundValue=1) err produces imerrCustom produces
In the end, I am still not sure about it, but after having read several papers more, I assume that my interpretation of X as foreground coordinates was an error. It should have probably been the entire image that is being iterated. As I have stated I am not sure if this interpretation is correct as well. But I made a new implementation which iterates over the image, and it gives a more plausible result. I am sharing it in here, hoping that it might help someone: %%cython -a import numpy as np cimport numpy as cnp cdef dilation_c(cnp.ndarray[cnp.uint8_t, ndim=2] X, cnp.ndarray[cnp.uint8_t, ndim=2] SE): """ X: boolean image SE: structuring element matrix origin: coordinate of the origin of the structuring element This operation checks whether (B *hits* X) = $B \cap X \not = \emptyset$ as per defined in Serra (Jean), « Introduction to mathematical morphology », Computer Vision, Graphics, and Image Processing, vol. 35, nᵒ 3 (septembre 1986). URL : https://linkinghub.elsevier.com/retrieve/pii/0734189X86900022.. doi: 10.1016/0734-189X(86)90002-2 Consulted le 6 août 2020, p. 283‑305. The algorithm adapts DILDIRECT of Najman (Laurent) et Talbot (Hugues), Mathematical morphology: from theory to applications, 2013. ISBN : 9781118600788, p. 329 to the formula given in Jähne (Bernd), Digital image processing, 6th rev. and ext. ed, Berlin ; New York, 2005. TA1637 .J34 2005. ISBN : 978-3-540-24035-8. """ cdef cnp.ndarray[cnp.uint8_t, ndim=2] O cdef list elst cdef int r, c, X_rows, X_cols, SE_rows, SE_cols, se_r, se_c cdef cnp.ndarray[cnp.int_t, ndim=1] bp cdef list conds cdef bint check, b, p, cond O = np.zeros_like(X) X_rows, X_cols = X.shape[:2] SE_rows, SE_cols = SE.shape[:2] # a boolean convolution for r in range(0, X_rows-SE_rows): for c in range(0, X_cols - SE_cols): conds = [] for se_r in range(SE_rows): for se_c in range(SE_cols): b = <bint>SE[se_r, se_c] p = <bint>X[se_r+r, se_c+c] conds.append(b and p) O[r,c] = <cnp.uint8_t>any(conds) return O def dilation_erosion( img: np.ndarray, struct_el_mat: np.ndarray, foregroundValue: int = 1, isErosion: bool = False): """ img: image matrix struct_el: NxN mesh grid of the structuring element whose center is SE's origin structuring element is encoded as 1 foregroundValue: value to be considered as foreground in the image """ B = struct_el_mat.astype(np.uint8) if isErosion: X = np.where(img == foregroundValue, 0, 1).astype(np.uint8) else: X = np.where(img == foregroundValue, 1, 0).astype(np.uint8) nimg = dilation_c(X, B) foreground, background = (255, 0) if foregroundValue == 1 else (0, 1) if isErosion: return np.where(nimg == 1, background, foreground).astype(np.uint8) else: return np.where(nimg == 1, foreground, background).astype(np.uint8) # return nimg
In Tensorflow, is there a built in function to compute states over time given a transition matrix?
I have a system given by this recursive relationship: xt = At xt-1 + bt. I wish to compute xt for all t, with At, bt and x0 given. Is there are built-in function for that? If I use a loop it would be extremely slow. Thanks!
There is sort of a way. Let's say you have your A matrices in a 3D tensor with shape (T, N, N), where T is the total number of time steps and N is the size of your vector. Similarly, B values are in a 2D tensor (T, N). The first step in the computation would be: x1 = A[0] # x0 + B[0] Where # represents matrix product. But you can convert this into a single matrix product. Suppose we add a value 1 at the end of x0, and we call that x0p (for prime): x0p = tf.concat([x, [1]], axis=0) And now we build a new 3D tensor Ap with shape (T, N+1, N+1), such that for each A[i] we concatenate B[i] as a new column, and then we add a row with N zeros and a single one at the end: AwithB = tf.concat([tf.concat([A, tf.expand_dims(B, 2)], axis=2)], axis=1) AnewRow = tf.concat([tf.zeros((T, 1, N), A.dtype), tf.ones((T, 1, 1), A.dtype)], axis=2) Ap = tf.concat([AwithB, AnewRow], axis=1) As it turns out, you can now say: x1p = Ap[0] # x0p And therefore: x2p = Ap[1] # x1p = Ap[1] # Ap[0] # x0p So we just need to compute all the matrix product of all matrices in Ap across the first dimension. Unfortunately, there does not seem to be a direct operation to compute that with TensorFlow, but you can do it relatively fast with tf.scan: Ap_prod = tf.scan(tf.matmul, Ap)[-1] And with that you just have to do: xtp = Ap_prod # x0p Here is a proof of concept (the code is tweaked to support single examples and batches, either in the A and B values or in the x) import tensorflow as tf def compute_state(a, b, x): s = tf.shape(a) t = s[-3] n = s[-1] # Add final 1 to x xp = tf.concat([x, tf.ones_like(x[..., :1])], axis=-1) # Add B column to A a_b = tf.concat([tf.concat([a, tf.expand_dims(b, axis=-1)], axis=-1)], axis=-2) # Make new final row for A a_row = tf.concat([tf.zeros_like(a[..., :1, :]), tf.ones_like(a[..., :1, :1])], axis=-1) # Add new row to A ap = tf.concat([a_b, a_row], axis=-2) # Compute matrix product reduction ap_prod = tf.scan(tf.matmul, ap)[..., -1, :, :] # Compute final result outp = tf.linalg.matvec(ap_prod, xp) return outp[..., :-1] #Test tf.random.set_seed(0) a = tf.random.uniform((10, 5, 5), -1, 1) b = tf.random.uniform((10, 5), -1, 1) x = tf.random.uniform((5,), -1, 1) y = compute_state(a, b, x) # Also works with batches of (a, b) or x a = tf.random.uniform((100, 10, 5, 5), -1, 1) b = tf.random.uniform((100, 10, 5), -1, 1) x = tf.random.uniform((100, 5), -1, 1) y = compute_state(a, b, x)
matmul function for vector with tensor multiplication in tensorflow
In general when we multiply a vector v of dimension 1*n with a tensor T of dimension m*n*k, we expect to get a matrix/tensor of dimension m*k/m*1*k. This means that our tensor has m slices of matrices with dimension n*k, and v is multiplied to each matrix and the resulting vectors are stacked together. In order to do this multiplication in tensorflow, I came up with the following formulation. I am just wondering if there is any built-in function that does this standard multiplication straightforward? T = tf.Variable(tf.random_normal((m,n,k)), name="tensor") v = tf.Variable(tf.random_normal((1,n)), name="vector") c = tf.stack([v,v]) # m times, here set m=2 output = tf.matmul(c,T)
You can do it with: tf.reduce_sum(tf.expand_dims(v,2)*T,1) Code: m, n, k = 2, 3, 4 T = tf.Variable(tf.random_normal((m,n,k)), name="tensor") v = tf.Variable(tf.random_normal((1,n)), name="vector") c = tf.stack([v,v]) # m times, here set m=2 out1 = tf.matmul(c,T) out2 = tf.reduce_sum(tf.expand_dims(v,2)*T,1) with tf.Session() as sess: sess.run(tf.global_variables_initializer()) n_out1 = sess.run(out1) n_out2 = sess.run(out2) #both n_out1 and n_out2 matches
Not sure if there is a better way, but it sounds like you could use tf.map_fn like this: output = tf.map_fn(lambda x: tf.matmul(v, x), T)
Evaluating the squared term of a gaussian kernel for having a covariance matrix for multi-dimensional inputs [duplicate]
I have the following code. It is taking forever in Python. There must be a way to translate this calculation into a broadcast... def euclidean_square(a,b): squares = np.zeros((a.shape[0],b.shape[0])) for i in range(squares.shape[0]): for j in range(squares.shape[1]): diff = a[i,:] - b[j,:] sqr = diff**2.0 squares[i,j] = np.sum(sqr) return squares
You can use np.einsum after calculating the differences in a broadcasted way, like so - ab = a[:,None,:] - b out = np.einsum('ijk,ijk->ij',ab,ab) Or use scipy's cdist with its optional metric argument set as 'sqeuclidean' to give us the squared euclidean distances as needed for our problem, like so - from scipy.spatial.distance import cdist out = cdist(a,b,'sqeuclidean')
I collected the different methods proposed here, and in two other questions, and measured the speed of the different methods: import numpy as np import scipy.spatial import sklearn.metrics def dist_direct(x, y): d = np.expand_dims(x, -2) - y return np.sum(np.square(d), axis=-1) def dist_einsum(x, y): d = np.expand_dims(x, -2) - y return np.einsum('ijk,ijk->ij', d, d) def dist_scipy(x, y): return scipy.spatial.distance.cdist(x, y, "sqeuclidean") def dist_sklearn(x, y): return sklearn.metrics.pairwise.pairwise_distances(x, y, "sqeuclidean") def dist_layers(x, y): res = np.zeros((x.shape[0], y.shape[0])) for i in range(x.shape[1]): res += np.subtract.outer(x[:, i], y[:, i])**2 return res # inspired by the excellent https://github.com/droyed/eucl_dist def dist_ext1(x, y): nx, p = x.shape x_ext = np.empty((nx, 3*p)) x_ext[:, :p] = 1 x_ext[:, p:2*p] = x x_ext[:, 2*p:] = np.square(x) ny = y.shape[0] y_ext = np.empty((3*p, ny)) y_ext[:p] = np.square(y).T y_ext[p:2*p] = -2*y.T y_ext[2*p:] = 1 return x_ext.dot(y_ext) # https://stackoverflow.com/a/47877630/648741 def dist_ext2(x, y): return np.einsum('ij,ij->i', x, x)[:,None] + np.einsum('ij,ij->i', y, y) - 2 * x.dot(y.T) I use timeit to compare the speed of the different methods. For the comparison, I use vectors of length 10, with 100 vectors in the first group, and 1000 vectors in the second group. import timeit p = 10 x = np.random.standard_normal((100, p)) y = np.random.standard_normal((1000, p)) for method in dir(): if not method.startswith("dist_"): continue t = timeit.timeit(f"{method}(x, y)", number=1000, globals=globals()) print(f"{method:12} {t:5.2f}ms") On my laptop, the results are as follows: dist_direct 5.07ms dist_einsum 3.43ms dist_ext1 0.20ms <-- fastest dist_ext2 0.35ms dist_layers 2.82ms dist_scipy 0.60ms dist_sklearn 0.67ms While the two methods dist_ext1 and dist_ext2, both based on the idea of writing (x-y)**2 as x**2 - 2*x*y + y**2, are very fast, there is a downside: When the distance between x and y is very small, due to cancellation error the numerical result can sometimes be (very slightly) negative.
Another solution besides using cdist is the following difference_squared = np.zeros((a.shape[0], b.shape[0])) for dimension_iterator in range(a.shape[1]): difference_squared = difference_squared + np.subtract.outer(a[:, dimension_iterator], b[:, dimension_iterator])**2.