I'm having trouble creating a multivariate normal density with sympy 0.7.6.1.
Here is my code.
from sympy import *
from sympy.stats import *
mu = Matrix([5, 13])
Sigma = Matrix([[2, 0], [0, 2]])
X = Normal('X', mu, Sigma)
y = MatrixSymbol('y', 2, 1)
density(X)(y)
The last line gives me this error:
Power of non-square matrix Matrix([
[ -5],
[-13]]) + y
The problem is simple: the formula to calculate the density is not the one supporting matrices, have a look:
https://github.com/sympy/sympy/blob/sympy-0.7.6.1/sympy/stats/crv_types.py#L1641
In this expression, (x-self.mean) gets squared (i.e. raised to the power of 2), but the square of non-square matrix is not defined.
In short, it looks like multivariate normal distributions are not supported, but you could try a workaround by defining a new distribution:
from sympy.stats.crv_types import rv, SingleContinuousDistribution, _value_check
class MultivariateNormalDistribution(SingleContinuousDistribution):
_argnames = ('mean', 'std')
#staticmethod
def check(mean, std):
_value_check(std > 0, "Standard deviation must be positive")
def pdf(self, x):
return exp(-S.Half * (x - self.mean).T * (self.std.inv()) * (x - self.mean)) / (sqrt(2*pi)**(self.std.shape[0])*self.std.det())
def sample(self):
pass
# define sampling function here
def MultivariateNormal(name, mean, std):
return rv(name, MultivariateNormalDistribution, (mean, std))
Unfortunately, your example still doesn't work, because of missing features in the matrix module (that is, no exponentiation of expressions with MatrixSymbol are supported, yet), but you could get the point density:
In[12]: X = MultivariateNormal('X', mu, Sigma)
In [13]: density(X)(Matrix([0, 0]))
Out[13]:
[ -97/2]
[e ]
[------]
[ 8*pi ]
Or with symbols in the matrix:
In [14]: x1, x2 = symbols('x1, x2')
In [15]: density(X)(Matrix([x1, x2]))
Out[15]:
[ 2 2 ]
[ x1 5*x1 x2 13*x2 97]
[ - --- + ---- - --- + ----- - --]
[ 4 2 4 2 2 ]
[e ]
[--------------------------------]
[ 8*pi ]
Related
im really new for gpu coding i found this Kmeans cupy code my propouse is work with a large data base (n,3) for example to realize about the timing difference on gpu and cpu , i wanna have a huge number of clusters but i am getting a memory management error. Can someone give me the route I should take to research and fix it, i already research but i have not a clear start yet.
import contextlib
import time
import cupy
import matplotlib.pyplot as plt
import numpy
#contextlib.contextmanager
def timer(message):
cupy.cuda.Stream.null.synchronize()
start = time.time()
yield
cupy.cuda.Stream.null.synchronize()
end = time.time()
print('%s: %f sec' % (message, end - start))
var_kernel = cupy.ElementwiseKernel(
'T x0, T x1, T c0, T c1', 'T out',
'out = (x0 - c0) * (x0 - c0) + (x1 - c1) * (x1 - c1)',
'var_kernel'
)
sum_kernel = cupy.ReductionKernel(
'T x, S mask', 'T out',
'mask ? x : 0',
'a + b', 'out = a', '0',
'sum_kernel'
)
count_kernel = cupy.ReductionKernel(
'T mask', 'float32 out',
'mask ? 1.0 : 0.0',
'a + b', 'out = a', '0.0',
'count_kernel'
)
def fit_xp(X, n_clusters, max_iter):
assert X.ndim == 2
# Get NumPy or CuPy module from the supplied array.
xp = cupy.get_array_module(X)
n_samples = len(X)
# Make an array to store the labels indicating which cluster each sample is
# contained.
pred = xp.zeros(n_samples)
# Choose the initial centroid for each cluster.
initial_indexes = xp.random.choice(n_samples, n_clusters, replace=False)
centers = X[initial_indexes]
for _ in range(max_iter):
# Compute the new label for each sample.
distances = xp.linalg.norm(X[:, None, :] - centers[None, :, :], axis=2)
new_pred = xp.argmin(distances, axis=1)
# If the label is not changed for each sample, we suppose the
# algorithm has converged and exit from the loop.
if xp.all(new_pred == pred):
break
pred = new_pred
# Compute the new centroid for each cluster.
i = xp.arange(n_clusters)
mask = pred == i[:, None]
sums = xp.where(mask[:, :, None], X, 0).sum(axis=1)
counts = xp.count_nonzero(mask, axis=1).reshape((n_clusters, 1))
centers = sums / counts
return centers, pred
def fit_custom(X, n_clusters, max_iter):
assert X.ndim == 2
n_samples = len(X)
pred = cupy.zeros(n_samples,dtype='float32')
initial_indexes = cupy.random.choice(n_samples, n_clusters, replace=False)
centers = X[initial_indexes]
for _ in range(max_iter):
distances = var_kernel(X[:, None, 0], X[:, None, 1],
centers[None, :, 1], centers[None, :, 0])
new_pred = cupy.argmin(distances, axis=1)
if cupy.all(new_pred == pred):
break
pred = new_pred
i = cupy.arange(n_clusters)
mask = pred == i[:, None]
sums = sum_kernel(X, mask[:, :, None], axis=1)
counts = count_kernel(mask, axis=1).reshape((n_clusters, 1))
centers = sums / counts
return centers, pred
def draw(X, n_clusters, centers, pred, output):
# Plot the samples and centroids of the fitted clusters into an image file.
for i in range(n_clusters):
labels = X[pred == i]
plt.scatter(labels[:, 0], labels[:, 1], c=numpy.random.rand(3))
plt.scatter(
centers[:, 0], centers[:, 1], s=120, marker='s', facecolors='y',
edgecolors='k')
plt.savefig(output)
def run_cpu(gpuid, n_clusters, num, max_iter, use_custom_kernel):##, output
samples = numpy.random.randn(num, 3)
X_train = numpy.r_[samples + 1, samples - 1]
with timer(' CPU '):
centers, pred = fit_xp(X_train, n_clusters, max_iter)
def run_gpu(gpuid, n_clusters, num, max_iter, use_custom_kernel):##, output
samples = numpy.random.randn(num, 3)
X_train = numpy.r_[samples + 1, samples - 1]
with cupy.cuda.Device(gpuid):
X_train = cupy.asarray(X_train)
with timer(' GPU '):
if use_custom_kernel:
centers, pred = fit_custom(X_train, n_clusters, max_iter)
else:
centers, pred = fit_xp(X_train, n_clusters, max_iter)
btw i am working in colab pro 25GB(RAM), the code is working with n_clusters=200 and num= 1000000 but if i use bigger numbers the error appear, i am running the code like this:
run_gpu(0,200,1000000,10,True)
This is the error that i have
Any suggestion will be welcome, thanks for your time.
Assuming that CuPy is clever enough not to create explicit copies of the broadcasted input of var_kernel, the output distances has to have a size of 2 * num * num_clusters which are exactly the 6,400,000,000 Bytes it is trying to allocate. You could have a way smaller memory footprint by never actually writing the distances to memory which means fusing the var_kernel with argmin. See this part of the docs.
If I understand the example there correctly, this should work:
#cupy.fuse(kernel_name='argmin_distance')
def argmin_distance(x1, y1, x2, y2):
return cupy.argmin((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2), axis = 1)
The next question would be where the other 13.7GB come from. A big part of them might just be the instances of distances from earlier iterations. I'm not a CuPy expert, but at least in Python/Numpy your use of distances inside the loop would not reuse the same memory, but allocate more memory each time you call the var_kernel. The same problem is visible with pred which is allocated before the loop. If CuPy does things the Numpy way, the solution would be to just put [:] in there like
pred[:] = new_pred
or
distances[:,:,:] = var_kernel(X[:, None, 0], X[:, None, 1],
centers[None, :, 1], centers[None, :, 0])
For this to work, you need to allocate distances before the loop as well. Also this isn't needed anymore when using kernel fusion, so just take it as an example. It may be best to allocate everything beforehand and then use this syntax everywhere in the loop.
I don't know enough about CuPy to answer why fit_xp doesn't have the same problem (or does it?). But my guess would be that garbage collection with CuPy objects works differently there. If garbage collection were "quick enough" in fit_custom it should work even without kernel fusion or reusing already allocated arrays.
Other problems or at least oddities with your code:
Why are you comparing the zeroth coordinate of centers with the first coordinate of X? Wouldn't it make more sense to call
distances = var_kernel(X[:, None, 0], X[:, None, 1],
centers[None, :, 0], centers[None, :, 1])
Why are you creating 3D data when only using the projection on the 2D plane? So why not
samples = numpy.random.randn(num, 2)
Why are you using floats for (the initial version of) pred? The argmin should give an integer type result.
In Python's statsmodels.formula.api, the ols functionality automatically includes and estimates an intercept:
results = sm.ols(formula="s ~ x + y + z", data=somedata).fit()
results.params
(* Intercept 0.632646, x -1.258761, y 0.465076, z 0.497991 *)
Because I'm using it in a linear probability model, is there any way to fix the intercept to 0.5?
You can reproduce this behavior in 2 steps:
Subtract the predefined_intercept from your targets
Fit OLS without intercept: include "-1" in your formula
Minimal example:
from statsmodels.formula.api import ols
import pandas as pd
import numpy as np
n_samples = 100
predefined_intercept = 0.5
somedata = pd.DataFrame(np.random.random((n_samples, 3)), columns = ['x', 'y', 'z'])
somedata['s'] = somedata['x'] - 2 * somedata['y'] + 5 * somedata['z'] - predefined_intercept
results = ols(formula="s ~ x + y + z - 1", data=somedata).fit()
print(results.params)
Output:
x 0.671561
y -2.315076
z 4.759542
See an official example notebook on formulas for detailed explanations and more.
How can we efficiently calculate pairwise cosine distances in a matrix using TensorFlow? Given an MxN matrix, the result should be an MxM matrix, where the element at position [i][j] is the cosine distance between i-th and j-th rows/vectors in the input matrix.
This can be done with Scikit-Learn fairly easily as follows:
from sklearn.metrics.pairwise import pairwise_distances
pairwise_distances(input_matrix, metric='cosine')
Is there an equivalent method in TensorFlow?
There is an answer for getting a single cosine distance here: https://stackoverflow.com/a/46057597/288875 . This is based on tf.losses.cosine_distance .
Here is a solution which does this for matrices:
import tensorflow as tf
import numpy as np
with tf.Session() as sess:
M = 3
# input
input = tf.placeholder(tf.float32, shape = (M, M))
# normalize each row
normalized = tf.nn.l2_normalize(input, dim = 1)
# multiply row i with row j using transpose
# element wise product
prod = tf.matmul(normalized, normalized,
adjoint_b = True # transpose second matrix
)
dist = 1 - prod
input_matrix = np.array(
[[ 1, 1, 1 ],
[ 0, 1, 1 ],
[ 0, 0, 1 ],
],
dtype = 'float32')
print "input_matrix:"
print input_matrix
from sklearn.metrics.pairwise import pairwise_distances
print "sklearn:"
print pairwise_distances(input_matrix, metric='cosine')
print "tensorflow:"
print sess.run(dist, feed_dict = { input : input_matrix })
which gives me:
input_matrix:
[[ 1. 1. 1.]
[ 0. 1. 1.]
[ 0. 0. 1.]]
sklearn:
[[ 0. 0.18350345 0.42264974]
[ 0.18350345 0. 0.29289323]
[ 0.42264974 0.29289323 0. ]]
tensorflow:
[[ 5.96046448e-08 1.83503449e-01 4.22649741e-01]
[ 1.83503449e-01 5.96046448e-08 2.92893231e-01]
[ 4.22649741e-01 2.92893231e-01 0.00000000e+00]]
Note that this solution may not be the optimal one as it calculates all entries of the (symmetric) result matrix, i.e. does almost twice of the calculations. This is likely not a problem for small matrices, for large matrices a combination of loops may be faster.
Note also that this does not have a minibatch dimension so works for a single matrix only.
Elegant solution (output is the same as from scikit-learn pairwise_distances function):
def compute_cosine_distances(a, b):
# x shape is n_a * dim
# y shape is n_b * dim
# results shape is n_a * n_b
normalize_a = tf.nn.l2_normalize(a,1)
normalize_b = tf.nn.l2_normalize(b,1)
distance = 1 - tf.matmul(normalize_a, normalize_b, transpose_b=True)
return distance
test
input_matrix = np.array([[1, 1, 1],
[0, 1, 1],
[0, 0, 1]], dtype = 'float32')
compute_cosine_distances(input_matrix, input_matrix)
output:
<tf.Tensor: id=442, shape=(3, 3), dtype=float32, numpy=
array([[5.9604645e-08, 1.8350345e-01, 4.2264974e-01],
[1.8350345e-01, 5.9604645e-08, 2.9289323e-01],
[4.2264974e-01, 2.9289323e-01, 0.0000000e+00]], dtype=float32)>
I'm trying to create a symbolic matrix (S) of general size (let's say LxL), and I want to set each element of the matrix as a function of the indices, i.e.:
S[m,n] = (u+i/2*(n-m))/(u-i/2*(n-m)) * (u+i/2*(n+m))/(u-i/2*(n+m))
I tried running this in sympy, and I got
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-11-a456d47e99e7> in <module>()
2 S_l = MatrixSymbol('S_l',2*l+1,2*l+1)
3 S_k = MatrixSymbol('S_k',2*k+1,2*k+1)
----> 4 S_l[m,n] = (u+i/2*(n-m))/(u-i/2*(n-m)) * (u+i/2*(n+m))/(u-i/2*(n+m))
TypeError: 'MatrixSymbol' object does not support item assignment
Searching through Stack Exchange I found this question from last year:
Sympy - Dense Display of Matrices of Arbitrary Size
Which is unanswered and not exactly the same. Is it the same issue, or am I just trying to do an impossible thing in sympy (or computers in general)?
I know this is ancient, but I came across the same issue and figured I'd share a solution that works for me. You'll need to use a FunctionMatrix object instead of a MatrixSymbol. For background, I'm using SymPy 1.6.1 on Python 3.5.2.
Here's an example. Using the code below, I've setup some iteration symbols and the function f(i,j) I'd like to use for the elements of my matrix u.
# Import SymPy for symbolic computations
import sympy as sym
# Index variables
i,j = sym.symbols('i j', integer=True);
N = sym.Symbol('N', real=True, integer=True, zero=False, positive=True);
# The function we'll use for our matrix
def f(i,j):
# Some arbitrary function...
return i + j;
# Define a function matrix where elements of the matrix
# are a function of the indices
U = sym.FunctionMatrix(N, N, sym.Lambda((i,j), f(i,j)));
Now, let's try using the elements in the matrix by summing them all up...
U_sum = sym.Sum(u[i,j], (i, 0, N), (j, 0, N));
U_sum
>>>
N N
___ ___
╲ ╲
╲ ╲
╱ ╱ (i + j)
╱ ╱
‾‾‾ ‾‾‾
j = 0 i = 0
Then, let's tell SymPy to calculate the summation
our_sum.doit().simplify()
>>> N * ( N**2 + 2*N + 1 )
This certainly can be done. The docs offer some examples. Here's one
>>> Matrix(3, 4, lambda i,j: 1 - (i+j) % 2)
Matrix([
[1, 0, 1, 0],
[0, 1, 0, 1],
[1, 0, 1, 0]])
I was looking for example code showing how to compute a singular value decomposition of a 2x2 matrix that can contain complex values.
For example, this would be useful for "repairing" user-entered matrices to be unitary. You just take u, s, v = svd(m) then omit the s part from the product: repaired = u * v.
Here's some python code that does the trick. It basically just extracts the complex parts then delegates to the solution from this answer for real 2x2 matrices.
I've written the code in python, using numpy. This is a bit ironic, because if you have numpy you should just use np.linalg.svd. Clearly this is intended as example code suitable for learning or translating into other languages in a pinch.
I'm also not an expert on numerical stability, so... buyer beware.
import numpy as np
import math
# Note: in practice in python just use np.linalg.svd instead
def singular_value_decomposition_complex_2x2(m):
"""
Returns a singular value decomposition of the given 2x2 complex numpy
matrix.
:param m: A 2x2 numpy matrix with complex values.
:returns: A tuple (U, S, V) where U*S*V ~= m, where U and V are complex
2x2 unitary matrices, and where S is a 2x2 diagonal matrix with
non-negative real values.
"""
# Make top row non-imaginary and non-negative by column phasing.
# m2 = m p = | > > |
# | ?+?i ?+?i |
p = phase_cancel_matrix(m[0, 0], m[0, 1])
m2 = m * p
# Cancel top-right value by rotation.
# m3 = m p r = | ?+?i 0 |
# | ?+?i ?+?i |
r = rotation_matrix(math.atan2(m2[0, 1].real, m2[0, 0].real))
m3 = m2 * r
# Make bottom row non-imaginary and non-negative by column phasing.
# m4 = m p r q = | ?+?i 0 |
# | > > |
q = phase_cancel_matrix(m3[1, 0], m3[1, 1])
m4 = m3 * q
# Cancel imaginary part of top left value by row phasing.
# m5 = t m p r q = | > 0 |
# | > > |
t = phase_cancel_matrix(m4[0, 0], 1)
m5 = t * m4
# All values are now real (also the top-right is zero), so delegate to a
# singular value decomposition that works for real matrices.
# t m p r q = u s v
u, s, v = singular_value_decomposition_real_2x2(np.real(m5))
# m = (t* u) s (v q* r* p*)
return adjoint(t) * u, s, v * adjoint(q) * adjoint(r) * adjoint(p)
def singular_value_decomposition_real_2x2(m):
"""
Returns a singular value decomposition of the given 2x2 real numpy matrix.
:param m: A 2x2 numpy matrix with real values.
:returns: A tuple (U, S, V) where U*S*V ~= m, where U and V are 2x2
rotation matrices, and where S is a 2x2 diagonal matrix with
non-negative real values.
"""
a = m[0, 0]
b = m[0, 1]
c = m[1, 0]
d = m[1, 1]
t = a + d
x = b + c
y = b - c
z = a - d
theta_0 = math.atan2(x, t) / 2.0
theta_d = math.atan2(y, z) / 2.0
s_0 = math.sqrt(t**2 + x**2) / 2.0
s_d = math.sqrt(z**2 + y**2) / 2.0
return \
rotation_matrix(theta_0 - theta_d), \
np.mat([[s_0 + s_d, 0], [0, s_0 - s_d]]), \
rotation_matrix(theta_0 + theta_d)
def adjoint(m):
"""
Returns the adjoint, i.e. the conjugate transpose, of the given matrix.
When the matrix is unitary, the adjoint is also its inverse.
:param m: A numpy matrix to transpose and conjugate.
:return: A numpy matrix.
"""
return m.conjugate().transpose()
def rotation_matrix(theta):
"""
Returns a 2x2 unitary matrix corresponding to a 2d rotation by the given angle.
:param theta: The angle, in radians, that the matrix should rotate by.
:return: A 2x2 orthogonal matrix.
"""
c, s = math.cos(theta), math.sin(theta)
return np.mat([[c, -s],
[s, c]])
def phase_cancel_complex(c):
"""
Returns a unit complex number p that cancels the phase of the given complex
number c. That is, c * p will be real and non-negative (approximately).
:param c: A complex number.
:return: A complex number on the complex unit circle.
"""
m = abs(c)
# For small values, where the division is in danger of exploding small
# errors, use trig functions instead.
if m < 0.0001:
theta = math.atan2(c.imag, c.real)
return math.cos(theta) - math.sin(theta) * 1j
return (c / float(m)).conjugate()
def phase_cancel_matrix(p, q):
"""
Returns a 2x2 unitary matrix M such that M cancels out the phases in the
column {{p}, {q}} so that the result of M * {{p}, {q}} should be a vector
with non-negative real values.
:param p: A complex number.
:param q: A complex number.
:return: A 2x2 diagonal unitary matrix.
"""
return np.mat([[phase_cancel_complex(p), 0],
[0, phase_cancel_complex(q)]])
I tested the above code by fuzzing it with matrices filled with random values in [-10, 10] + [-10, 10]i, and checking that the decomposed factors had the right properties (i.e. unitary, diagonal, real, as appropriate) and that their product was (approximately) equal to the input.
But here's a simple smoke test:
m = np.mat([[5, 10], [1j, -1]])
u, s, v = singular_value_decomposition_complex_2x2(m)
np.set_printoptions(precision=5, suppress=True)
print "M:\n", m
print "U*S*V:\n", u*s*v
print "U:\n", u
print "S:\n", s
print "V:\n", v
print "M ~= U*S*V:", np.all(np.abs(m - u*s*v) < 0.1**14)
Which outputs the following. You can confirm that the factored S matches the svd from wolfram alpha, although of course the U and V can be (and are) different.
M:
[[ 5.+0.j 10.+0.j]
[ 0.+1.j -1.+0.j]]
U*S*V:
[[ 5.+0.j 10.+0.j]
[ 0.+1.j -1.-0.j]]
U:
[[-0.89081-0.44541j 0.08031+0.04016j]
[ 0.08979+0.j 0.99596+0.j ]]
S:
[[ 11.22533 0. ]
[ 0. 0.99599]]
V:
[[-0.39679+0.20639j -0.80157+0.39679j]
[ 0.40319+0.79837j -0.19359-0.40319j]]
M ~= U*S*V: True