I want to compute discrete difference of identity matrix.
The code below use numpy and scipy.
import numpy as np
from scipy.sparse import identity
from scipy.sparse import csc_matrix
x = identity(4).toarray()
y = csc_matrix(np.diff(x, n=2))
print(y)
I would like to improve performance or memory usage.
Since identity matrix produce many zeros, it would reduce memory usage to perform calculation in compressed sparse column(csc) format. However, np.diff() does not accept csc format, so converting between csc and normal format using csc_matrix would slow it down a bit.
Normal format
x = identity(4).toarray()
print(x)
[[1. 0. 0. 0.]
[0. 1. 0. 0.]
[0. 0. 1. 0.]
[0. 0. 0. 1.]]
csc format
x = identity(4)
print(x)
(0, 0) 1.0
(1, 1) 1.0
(2, 2) 1.0
(3, 3) 1.0
Thanks
Here is my hacky solution to get the sparse matrix as you want.
L - the length of the original identity matrix,
n - the parameter of np.diff.
In your question they are:
L = 4
n = 2
My code produces the same y as your code, but without the conversions between csc and normal formats.
Your code:
from scipy.sparse import identity, csc_matrix
x = identity(L).toarray()
y = csc_matrix(np.diff(x, n=n))
My code:
from scipy.linalg import pascal
def get_data(n, L):
nums = pascal(n + 1, kind='lower')[-1].astype(float)
minuses_from = n % 2 + 1
nums[minuses_from : : 2] *= -1
return np.tile(nums, L - n)
data = get_data(n, L)
row_ind = (np.arange(n + 1) + np.arange(L - n).reshape(-1, 1)).flatten()
col_ind = np.repeat(np.arange(L - n), n + 1)
y = csc_matrix((data, (row_ind, col_ind)), shape=(L, L - n))
I have noticed that after applying np.diff to the identity matrix n times, the values of the columns are the binomial coefficients with their signs alternating. This is my variable data.
Then I am just constructing the csc_matrix.
Unfortunately, it does not seem that SciPy provides any tools for this kind of sparse matrix manipulation. Regardless, by cleverly manipulating the indices and data of the entries one can emulate np.diff(x,n) in a straightforward fashion.
Given 2D NumPy array (matrix) of dimension MxN, np.diff() multiplies each column (of column index y) with -1 and adds the next column to it (column index y+1). Difference of order k is just the iterative application of k differences of order 1. A difference of order 0 is just the returns the input matrix.
The method below makes use of this, iterateively eliminating duplicate entries by addition through sum_duplicates(), reducing the number of columns by one, and filtering non-valid indices.
def csc_diff(x, n):
'''Emulates np.diff(x,n) for a sparse matrix by iteratively taking difference of order 1'''
assert isinstance(x, csc_matrix) or (isinstance(x, np.ndarray) & len(x.shape) == 2), "Input matrix must be a 2D np.ndarray or csc_matrix."
assert isinstance(n, int) & n >= 0, "Integer n must be larger or equal to 0."
if n >= x.shape[1]:
return csc_matrix(([], ([], [])), shape=(x.shape[0], 0))
if isinstance(x, np.ndarray):
x = csc_matrix(x)
# set-up of data/indices via column-wise difference
if(n > 0):
for k in range(1,n+1):
# extract data/indices of non-zero entries of (current) sparse matrix
M, N = x.shape
idx, idy = x.nonzero()
dat = x.data
# difference: this row (y) * (-1) + next row (y+1)
idx = np.concatenate((idx, idx))
idy = np.concatenate((idy, idy-1))
dat = np.concatenate(((-1)*dat, dat))
# filter valid indices
validInd = (0<=idy) & (idy<N-1)
# x_diff: csc_matrix emulating np.diff(x,1)'s output'
x_diff = csc_matrix((dat[validInd], (idx[validInd], idy[validInd])), shape=(M, N-1))
x_diff.sum_duplicates()
x = x_diff
return x
Moreover, the method outputs an empty csc_matrix of dimension Mx0 when the difference order is larger or equal to the number of columns of the input matrix. This is why the output is identical, see
csc_diff(x, 2).toarray()
> array([[ 1., 0.],
[-2., 1.],
[ 1., -2.],
[ 0., 1.]])
which is identical to
np.diff(x.toarray(), 2)
> array([[ 1., 0.],
[-2., 1.],
[ 1., -2.],
[ 0., 1.]])
This identity holds for other difference orders, too
(csc_diff(x, 0).toarray() == np.diff(x.toarray(), 0)).all()
>True
(csc_diff(x, 3).toarray() == np.diff(x.toarray(), 3)).all()
>True
(csc_diff(x, 13).toarray() == np.diff(x.toarray(), 13)).all()
>True
Related
I have a list of m n x n matrices and a list of m real values (alphas). The values of n and m can be quite large. I am trying to calculate the weighted sum of the matrices with weights in alphas.
I was wondering if there is a numpy function (or any other library) that can do this faster than the manual for-loop method.
I have included my current function below.
def calculate_matrix_sums(mats, alphas):
"""
Calculate the weighted sum of matrices in mats with weights alpha
"""
k_mults = [np.multiply(mats[i], alphas[i]) for i in range(len(alphas))]
k_sums1 = np.matrix(k_mults[0]) + np.matrix(k_mults[1])
for i in range(2, len(k_mults)):
k_sums1 = k_sums1 + np.asmatrix(k_mults[i])
k_sums2 = np.asarray(k_sums1).astype(float)
k_sums2 = k_sums2.reshape(len(mats[0]), len(mats[0]))
return k_sums2
and a sample code:
matrices = np.asarray([np.array([[1., 0.77841638, 0.53239253, 0.9444068, 0.93024477],
[0.77841638, 1., 0.7221497, 0.5805838, 0.68501944],
[0.53239253, 0.7221497, 1., 0.36986265, 0.62792847],
[0.9444068, 0.5805838, 0.36986265, 1., 0.88303226],
[0.93024477, 0.68501944, 0.62792847, 0.88303226, 1.]]),
np.array([[1., 0.45650032, 0.13898701, 0.83605729, 0.79743304],
[0.45650032, 1., 0.36094014, 0.18229867, 0.30596445],
[0.13898701, 0.36094014, 1., 0.04443844, 0.23300302],
[0.83605729, 0.18229867, 0.04443844, 1., 0.67745532],
[0.79743304, 0.30596445, 0.23300302, 0.67745532, 1.]])])
alpha_vals = [0.47547796, 0.52452204]
print(calculate_matrix_sums(matrices, alpha_vals))
Any suggestions are appreciated.
You can reshape alpha_vals so that it broadcasts across the first axis of matrices correctly:
(np.array(alpha_vals)[:, None, None] * matrices).sum(axis=0)
Alternatively, you can adjust the strides of matrices, so that the last dimension corresponds to alpha_vals:
(np.moveaxis(matrices, 0, -1) * alpha_vals).sum(axis=-1)
You can also use np.einsum for this sort of thing (probably the most elegant solution):
np.einsum('ijk,i->jk', matrices, alpha_vals)
If you want a matrix result, it's simply:
def calculate_matrix_sums(mats, alphas):
"""
Calculate the weighted sum of matrices in mats with weights alpha
"""
return np.sum( m * a for m,a in zip(mats,alphas))
I am trying to solve the Isoperimetric problem (7.14) from Additional Exercises for Convex Optimization by Stephen Boyd using CVXPY. The problem formulation is:
The code of constraints is given below:
constraints = [ y[1] == 0,
y[N] == 0,
y[F] == yfixed[F],
cp.abs(( y[i+2] - 2 * y[i+1] + y[i]) / h**2) <= C for i in np.arange(1,199) ] #not using the first constraint here
The constraints have for loops, and when I tried to formulate the problem according to the CVXPY documentation, I got the following error
Invalid syntax
cp.abs(( y[i+2] - 2 * y[i+1] + y[i]) / h**2) <= C for i in np.arange(1,199) ]
^
How to use the loops in CVXPY constraints?
You need to express the constraints in terms of matrix-vector equalities and inequalities which follow the DCP protocol for cvxpy.
To elaborate, I can see three kinds of constraints in this problem: (I am going to assume 0-based indexing for programming convenience for the rest of the answer.)
Take y as the N+1 dimensional optimization variable.
Fixed point equality constraints: These basically set some indices of the y vector to a set of given values. Note that the zero boundary conditions y[0] == 0 and y[N] == 0 also fall under this.
Perimeter constraint: This is to be computed using successive differences. And finally we set something like the sum of the square roots of 1 plus the squares of the differences to be less than L. This part is probably the most involved one to write following the cvxpy protocols.
Curvature constraints: This also involves a calculation similar to successive differences like above but this is much easier to write as you'll see this is simply a matrix-vector multiplication type constraint like the first one.
Now let's write the constraints in code.
Necessary imports:
import numpy as np
import cvxpy as cp
import matplotlib.pyplot as plt
from scipy.linalg import circulant
1. Equality constraints:
These basically pick some indices from y and set those to given values. This can be implemented as follows:
def equality_constraints(N, F, vals):
'''
Sets some indices (F) in the y vector to given values. Also sets
y[0] == 0 and y[N] == 0.
Returns E (a matrix) and e (a vector) such that E # y == e.
'''
E = np.zeros((F.shape[0]+2, N+1)) # '+2' for selecting y[0] and y[N] also
e = np.zeros(F.shape[0]+2)
E[0, 0] = 1
E[-1, -1] = 1
if F.shape[0]:
E[1:-1:1][np.arange(F.shape[0]), F] = 1
e[1:-1:1] = vals
return E, e
E is a binary matrix of shape (F.shape[0] + 2) x (N+1) and it has exactly one column set to 1 in each row, giving an index for the (N+1) dimensional vector y and e contains the value for that index of y.
2. Perimeter constraint:
For this, we need successive differences of the form y[i+1] - y[i] for i = 0, 1, 2, . . . , N-1. Note that we can similarly construct a vector having this N successive differences as its elements. We can perform the square root and other operations on this vector easily using vectorized computations. Here we are constructing an N x (N+1) matrix M, which when multiplied by y will give the N differences.
def length_matrix(N):
'''
Returns L with [-1, 1, 0, . . . , 0] as first row and sliding it
to the right to get the following rows.
'''
val = np.array([-1, 1])
offsets = np.array([0, 1])
col0 = np.zeros(N+1)
col0[offsets] = val
M = circulant(col0).T[:-(len(val) - 1)]
return M
The matrix M will be a circulant matrix. I simply transposed it and removed the last row to get the desired matrix. You can see this post to know how to create one such matrix. M looks like this:
array([[-1., 1., 0., ..., 0., 0., 0.],
[ 0., -1., 1., ..., 0., 0., 0.],
[ 0., 0., -1., ..., 0., 0., 0.],
...,
[ 0., 0., 0., ..., 1., 0., 0.],
[ 0., 0., 0., ..., -1., 1., 0.],
[ 0., 0., 0., ..., 0., -1., 1.]])
3. Curvature constraints:
Exactly same matrix calculation like the last one. Just repeat and slide [1, -2, 1] along the rows!
def curvature_constraints(N, C, h):
'''
Returns D and C_vec to be used as D # y <= C and D # y >= -C
as curvature constraints.
'''
val = np.array([1, -2, 1])
offsets = np.array([0, 1, 2])
col0 = np.zeros(N+1)
col0[offsets] = val
D = circulant(col0).T[:-(len(val) - 1)]
D /= h**2
C_vec = np.ones(D.shape[0]) * C
return D, C_vec
I am dividing by h**2 in the matrix itself.
Example:
I have taken this example from the site of the book itself. The data is also available here.
L = 1.5
a = 1
C = 15
N = 200
h = a/N
F = np.array([20,40,140,180]) # fixed points
vals = np.array([0.1, 0.15, 0.15, 0.2])
# Declare an array for plotting purposes
yfixed = np.zeros(N+1)
yfixed[F] = vals
x = np.linspace(0, a, N+1)
Problem Formulation and Solution:
I am leaving it for you to understand how I have assembled the matrices in formulating the constraints, especially the perimeter one. This is not difficult, but might require you some practice depending on how comfortable you are with vectorization. The DCP page is a very good place to start.
y = cp.Variable(N+1)
E, e = equality_constraints(N, F, vals)
M = length_matrix(N)
D, d = curvature_constraints(N, C, h)
constraints = [
E # y == e,
h * cp.sum(cp.norm(cp.vstack([(M # y)/h, np.ones(N)]), p = 2, axis = 0)) <= L,
D # y <= d,
D # y >= -d
]
objective_function = h * cp.sum(y)
objective = cp.Maximize(objective_function)
problem = cp.Problem(objective, constraints)
problem.solve()
plt.plot(0, 0, 'ko')
plt.plot(a, 0, 'ko')
for i in F:
plt.plot(x[i], yfixed[i], 'bo')
plt.plot(x, y.value) # y.value gives the value of the cp Variable
plt.savefig('curve.png')
I got the answer as 0.1594237500556726 for the above example and the curve looks like this:
I have checked this solution with few other contrived test cases to verify correctness. However, there might be other more efficient solutions formulating this problem differently or there might even be some unexpected or embarrassing errors here! Feel free to let me know in case there is some error or you find anything difficult to understand in the answer.
Try to split in two:
constraints = [ y[1] == 0,
y[N] == 0,
y[F] == yfixed[F] ] +
[ cp.abs(( y[i+2] - 2 * y[i+1] + y[i]) / h**2) <= C for i in np.arange(1,199) ]
This should be a simple question, either I am missing information, or I have mis-coded this.
I am trying to implement Mahalanabois distance in python which I am following from the formula in python.
My code is as follows:
a = np.array([[1, 3, 5]])
b = np.array([[4, 5, 6]])
X = np.empty((0,3), float)
X = np.vstack([X, [2,3,4]])
X = np.vstack([X, a])
X = np.vstack([X, b])
n = ((a-b).T)*(np.cov(X)**-1)*(a-b)
dist = np.sqrt(n)
dist returns a 3x3 array but should I not be expecting a single number representing the distance?
dist = array([[ 1.5 , 1.73205081, 1.22474487],
[ 1.73205081 , 2. , 1.41421356],
[ 1.22474487 , 1.41421356, 1. ]])
Wikipedia does not suggest (to me) that it should return a matrix. Googling implementations of mahalanbois distance in python I have not found something to compare it to.
From wiki page you could see, that a and b are vectors but in your case they are arrays. So you need reverse transposing. And also there should be matrix multiplication. In numpy * means element-wise multiplication, for matrix you should use np.dot function or .dot method of the np.array. For your case answer is:
n = (a-b).dot((np.cov(X)**-1).dot((a-b).T))
dist = np.sqrt(n)
In [54]: n
Out[54]: array([[ 25.]])
In [55]: dist
Out[55]: array([[ 5.]])
EDIT
As #roadrunner66 noticed you should use inverse matrix instead of inverse matrix of element. Usually np.linalg.inv works for that cases but for that you've got Singular Error and you need to use np.linalg.pinv:
n = (a-b).dot((np.linalg.pinv(np.cov(X))).dot((a-b).T))
dist = np.sqrt(n)
In [90]: n
Out[90]: array([[ 1.77777778]])
In [91]: dist
Out[91]: array([[ 1.33333333]])
m,n=input()
a=[0]*n
for i in range(0,m):
a[i]=[0]*m
for i in range(0,m):
for j in range(0,n):
a[i][j]=input()
print a
Consider the above piece of code written in Python 2.7.4 to accept a 2-D array and then print it. This code functions well but it should accept any 2-D array means for example the values of m and could be 3,2 respectively let's say but it only accepts a square matrix. We cannot create a rectangular matrix because it gives the error: index out of range if the values of m and n are not equal. Is there any way to create a rectangular matrix just like we can do in C/C++ easily?
Numpy is a great module for fast linear algebra operations. You can create a rectangular array with Numpy which is essentially a matrix. (Numpy also has matrix functions too but they are a little more tedious to work with).
As an example, create a 3x4 array as follows
import numpy as np
input = np.zeros((3, 4)) #This creates a 3x4 array. It is good practice to initialize your array with 0's
input[0][3] = 5 #Fill out your array. The 0,0 index is the top left corner
In [42]:input
Out[42]:
array([[ 0., 0., 0., 5.],
[ 0., 0., 0., 0.],
[ 0., 0., 0., 0.]])
Proceed to fill out the rest of your rectangular matrix as per normal.
You can do something like this:
rows = input()
cols = input()
arr = [[None] * cols for _ in range(rows)]
for i in range(rows):
for j in range(cols):
print(arr)
arr[i][j] = input()
print(arr)
You should see this similar question: How to initialize a two-dimensional array in Python?
As far as getting your exact code to work, you just have m and n switched on lines 2 & 4:
m = input() # rows
n = input() # cols
a = [0] * m # Initialize rows
for i in range(0, m): # For each row (must be n rows)
a[i] = [0] * n # Initialize a column
for i in range(0, m): # For each row
for j in range(0, n): # For each column
a[i][j] = input() # Input a value
print a
I have very large matrix, so dont want to sum by going through each row and column.
a = [[1,2,3],[3,4,5],[5,6,7]]
def neighbors(i,j,a):
return [a[i][j-1], a[i][(j+1)%len(a[0])], a[i-1][j], a[(i+1)%len(a)][j]]
[[np.mean(neighbors(i,j,a)) for j in range(len(a[0]))] for i in range(len(a))]
This code works well for 3x3 or small range of matrix, but for large matrix like 2k x 2k this is not feasible. Also this does not work if any of the value in matrix is missing or it's like na
This code works well for 3x3 or small range of matrix, but for large matrix like 2k x 2k this is not feasible. Also this does not work if any of the value in matrix is missing or it's like na. If any of the neighbor values is na then skip that neighbour in getting the average
Shot #1
This assumes you are looking to get sliding windowed average values in an input array with a window of 3 x 3 and considering only the north-west-east-south neighborhood elements.
For such a case, signal.convolve2d with an appropriate kernel could be used. At the end, you need to divide those summations by the number of ones in kernel, i.e. kernel.sum() as only those contributed to the summations. Here's the implementation -
import numpy as np
from scipy import signal
# Inputs
a = [[1,2,3],[3,4,5],[5,6,7],[4,8,9]]
# Convert to numpy array
arr = np.asarray(a,float)
# Define kernel for convolution
kernel = np.array([[0,1,0],
[1,0,1],
[0,1,0]])
# Perform 2D convolution with input data and kernel
out = signal.convolve2d(arr, kernel, boundary='wrap', mode='same')/kernel.sum()
Shot #2
This makes the same assumptions as in shot #1, except that we are looking to find average values in a neighborhood of only zero elements with the intention to replace them with those average values.
Approach #1: Here's one way to do it using a manual selective convolution approach -
import numpy as np
# Convert to numpy array
arr = np.asarray(a,float)
# Pad around the input array to take care of boundary conditions
arr_pad = np.lib.pad(arr, (1,1), 'wrap')
R,C = np.where(arr==0) # Row, column indices for zero elements in input array
N = arr_pad.shape[1] # Number of rows in input array
offset = np.array([-N, -1, 1, N])
idx = np.ravel_multi_index((R+1,C+1),arr_pad.shape)[:,None] + offset
arr_out = arr.copy()
arr_out[R,C] = arr_pad.ravel()[idx].sum(1)/4
Sample input, output -
In [587]: arr
Out[587]:
array([[ 4., 0., 3., 3., 3., 1., 3.],
[ 2., 4., 0., 0., 4., 2., 1.],
[ 0., 1., 1., 0., 1., 4., 3.],
[ 0., 3., 0., 2., 3., 0., 1.]])
In [588]: arr_out
Out[588]:
array([[ 4. , 3.5 , 3. , 3. , 3. , 1. , 3. ],
[ 2. , 4. , 2. , 1.75, 4. , 2. , 1. ],
[ 1.5 , 1. , 1. , 1. , 1. , 4. , 3. ],
[ 2. , 3. , 2.25, 2. , 3. , 2.25, 1. ]])
To take care of the boundary conditions, there are other options for padding. Look at numpy.pad for more info.
Approach #2: This would be a modified version of convolution based approach listed earlier in Shot #1. This is same as that earlier approach, except that at the end, we selectively replace
the zero elements with the convolution output. Here's the code -
import numpy as np
from scipy import signal
# Inputs
a = [[1,2,3],[3,4,5],[5,6,7],[4,8,9]]
# Convert to numpy array
arr = np.asarray(a,float)
# Define kernel for convolution
kernel = np.array([[0,1,0],
[1,0,1],
[0,1,0]])
# Perform 2D convolution with input data and kernel
conv_out = signal.convolve2d(arr, kernel, boundary='wrap', mode='same')/kernel.sum()
# Initialize output array as a copy of input array
arr_out = arr.copy()
# Setup a mask of zero elements in input array and
# replace those in output array with the convolution output
mask = arr==0
arr_out[mask] = conv_out[mask]
Remarks: Approach #1 would be the preferred way when you have fewer number of zero elements in input array, otherwise go with Approach #2.
This is an appendix to comments under #Divakar's answer (rather than an independent answer).
Out of curiosity I tried different 'pseudo' convolutions against the scipy convolution. The fastest one was the % (modulus) wrapping one, which surprised me: obviously numpy does something clever with its indexing, though obviously not having to pad will save time.
fn3 -> 9.5ms, fn1 -> 21ms, fn2 -> 232ms
import timeit
setup = """
import numpy as np
from scipy import signal
N = 1000
M = 750
P = 5 # i.e. small number -> bigger proportion of zeros
a = np.random.randint(0, P, M * N).reshape(M, N)
arr = np.asarray(a,float)"""
fn1 = """
arr_pad = np.lib.pad(arr, (1,1), 'wrap')
R,C = np.where(arr==0)
N = arr_pad.shape[1]
offset = np.array([-N, -1, 1, N])
idx = np.ravel_multi_index((R+1,C+1),arr_pad.shape)[:,None] + offset
arr[R,C] = arr_pad.ravel()[idx].sum(1)/4"""
fn2 = """
kernel = np.array([[0,1,0],
[1,0,1],
[0,1,0]])
conv_out = signal.convolve2d(arr, kernel, boundary='wrap', mode='same')/kernel.sum()
mask = arr == 0.0
arr[mask] = conv_out[mask]"""
fn3 = """
R,C = np.where(arr == 0.0)
arr[R, C] = (arr[(R-1)%M,C] + arr[R,(C-1)%N] + arr[R,(C+1)%N] + arr[(R+1)%M,C]) / 4.0
"""
print(timeit.timeit(fn1, setup, number = 100))
print(timeit.timeit(fn2, setup, number = 100))
print(timeit.timeit(fn3, setup, number = 100))
Using numpy and scipy.ndimage, you can apply a "footprint" that defines where you look for the neighbours of each element and apply a function to those neighbours:
import numpy as np
import scipy.ndimage as ndimage
# Getting neighbours horizontally and vertically,
# not diagonally
footprint = np.array([[0,1,0],
[1,0,1],
[0,1,0]])
a = [[1,2,3],[3,4,5],[5,6,7]]
# Need to make sure that dtype is float or the
# mean won't be calculated correctly
a_array = np.array(a, dtype=float)
# Can specify that you want neighbour selection to
# wrap around at the borders
ndimage.generic_filter(a_array, np.mean,
footprint=footprint, mode='wrap')
Out[36]:
array([[ 3.25, 3.5 , 3.75],
[ 3.75, 4. , 4.25],
[ 4.25, 4.5 , 4.75]])