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I'm working on an optimization problem, but to avoid getting into the details, I'm going to provide a simple example of a bug that's been giving me headaches for a few days.
Say I have a 2D numpy array with observed x-y coordinates:
from scipy.optimize import distance
x = np.array([1,2], [2,3], [4,5], [5,6])
I also have a list of x-y coordinates to compare to these points (y):
y = np.array([11,13], [12, 14])
I have a function that takes the sum of manhattan differences between a value of x and all of the values in y:
def find_sum(ref_row, comp_rows):
modeled_counts = []
y = ref_row * len(comp_rows)
res = list(map(distance.cityblock, ref_row, comp_rows))
modeled_counts.append(sum(res))
return sum(modeled_counts)
Essentially, what I would like to do is find the sum of manhattan distances for every item in y with each item in x (so basically for each item in x, find the sum of the Manhattan distances between that (x,y) pair and every (x,y) pair in y).
I've tried this out with the following line of code:
z = list(map(find_sum, x, y))
However, z is of length 2 (like y), and not 4 like x. Is there a way to ensure that z is the result of consecutive one-to-all calculations? That is, I'd like to calculate the sum of all of the manhattan differences between x[0] and every set in y, and so on and so forth, so the length of z should be equal to the length of x.
Is there a simple way to do this without a for loop? My data is rather large (~ 4 million rows), so I'd really appreciate fast solutions. I'm fairly new to Python programming, so any explanations about why the solution works and is fast would be appreciated as well, but definitely isn't required!
Thanks!
This solution implements the distance in numpy, as I think it is a good example of broadcasting, which is a very useful thing to know if you need to use arrays and matrices.
By definition of Manhattan distance, you need to evaluate the sum of the absolute value of difference between each column. However, the first column of x, x[:, 0], has shape (4,) and the first column of y, y[:, 0], has shape (2,), so they are not compatible in the sense of applying subtraction: the broadcasting property says that each shape is compared starting with the trailing dimensions and two dimensions are compatible when they are equal or one of them is 1. Sadly, none of them are true for your columns.
However, you can add a new dimension of value 1 using np.newaxis, so
x[:, 0]
is array([1, 2, 4, 5]), but
x[:, 0, np.newaxis]
is
array([[1],
[2],
[4],
[5]])
and its shape is (4 ,1). Now, a matrix of shape (4, 1) subtracted by an array of shape 2 results in a matrix of shape (4, 2), by numpy's broadcasting treatment:
4 x 1
2
= 4 x 2
You can obtain the differences for each column:
first_column_difference = x[:, 0, np.newaxis] - y[:, 0]
second_column_difference = x[:, 1, np.newaxis] - y[:, 1]
and evaluate the sum of their absolute values:
np.abs(first_column_difference) + np.abs(second_column_difference)
which results in a (4, 2) matrix. Now, you want to sum the values for each row, so that you have 4 values:
np.sum(np.abs(first_column_difference) + np.abs(second_column_difference), axis=1)
which results in array([73, 69, 61, 57]). The rule is simple: the parameter axis will eliminate that dimension from the result, therefore using axis=1 for a (4, 2) matrix generates 4 values -- if you use axis=0, it will generate 2 values.
So, this will solve your problem:
x = np.array([[1, 2], [2, 3], [4, 5], [5, 6]])
y = np.array([[11, 13], [12, 43]])
first_column_difference = x[:, 0, np.newaxis] - y[:, 0]
second_column_difference = x[:, 1, np.newaxis] - y[:, 1]
z = np.abs(first_column_difference) + np.abs(second_column_difference)
print(np.sum(z, axis=1))
You can also skip the intermediate steps for each column and evaluate everything at once (it is a little bit harder to understand, so I prefer the method described above to explain what is happening):
print(np.abs(x[:, np.newaxis] - y).sum(axis=(1, 2)))
It is a general case for an n-dimensional Manhattan distance: if x is (u, n) and y is (v, n), it generates u rows by broadcasting (u, 1, n) by (v, n) = (u, v, n), then applying sum to eliminate the second and third axis.
Here is how you can do it using numpy broadcast with simplified explanation
Adjust Shape For Broadcasting
import numpy as np
start_points = np.array([[1,2], [2,3], [4,5], [5,6]])
dest_points = np.array([[11,13], [12, 14]])
## using np.newaxis as index add a new dimension at that position
## : give all the elements on that dimension
start_points = start_points[np.newaxis, :, :]
dest_points = dest_points[:, np.newaxis, :]
## Now lets check he shape of the point arrays
print('start_points.shape: ', start_points.shape) # (1, 4, 2)
print('dest_points.shape', dest_points.shape) # (2, 1, 2)
Lets try to understand
last element of shape represent x and y of a point, size 2
we can think of start_points as having 1 row and 4 columns of points
we can think of dest_points as having 2 rows and 1 columns of points
We can think start_points and dest_points as matrix or a table of points of size (1X4) and (2X1)
We clearly see that size are not compatible. What will happen if we perform arithmatic
operation between them? Here is where a smart part of numpy comes, called broadcast.
It will repeat rows of start_points to match that of dest_point making matrix of (2X4)
It will repeat columns of dest_point to match that of start_points making matrix of (2X4)
Result is arithmetic operation between every pair of elements on start_points and dest_points
Calculate the distance
diff_x_y = start_points - dest_points
print(diff_x_y.shape) # (2, 4, 2)
abs_diff_x_y = np.abs(start_points - dest_points)
man_distance = np.sum(abs_diff_x_y, axis=2)
print('man_distance:\n', man_distance)
sum_distance = np.sum(man_distance, axis=0)
print('sum_distance:\n', sum_distance)
Oneliner
start_points = np.array([[1,2], [2,3], [4,5], [5,6]])
dest_points = np.array([[11,13], [12, 14]])
np.sum(np.abs(start_points[np.newaxis, :, :] - dest_points[:, np.newaxis, :]), axis=(0,2))
Here is more detail explanation of broadcasting if you want to understand it more
With so many rows you can make substantial savings by using a smart algorithm. Let us for simplicity assume there is just one dimension; once we have established the algorithm, getting back to the general case is a simple matter of summing over coordinates.
The naive algorithm is O(mn) where m,n are the sizes of sets X,Y. Our algorithm is O((m+n)log(m+n)) so it scales much better.
We first have to sort the union of X and Y by coordinate and then form the cumsum over Y. Next, we find for each x in X the number YbefX of y in Y to its left and use it to look up the corresponding cumsum item YbefXval. The summed distances to all y to the left of x are YbefX times coordinate of x minus YbefXval, the distances to all y to the right are sum of all y coordinates minus YbefXval minus n - YbefX times coordinate of x.
Where does the saving come from? Sorting coordinates enables us to recycle the summations we have done before, instead of starting each time from scratch. This uses the fact that up to a sign we always sum the same y coordinates and going from left to right the signs flip one by one.
Code:
import numpy as np
from scipy.spatial.distance import cdist
from timeit import timeit
def pp(X,Y):
(m,k),(n,k) = X.shape,Y.shape
XY = np.concatenate([X.T,Y.T],1)
idx = XY.argsort(1)
Xmsk = idx<m
Ymsk = ~Xmsk
Xidx = np.arange(k)[:,None],idx[Xmsk].reshape(k,m)
Yidx = np.arange(k)[:,None],idx[Ymsk].reshape(k,n)
YbefX = Ymsk.cumsum(1)[Xmsk].reshape(k,m)
YbefXval = XY[Yidx].cumsum(1)[np.arange(k)[:,None],YbefX-1]
YbefXval[YbefX==0] = 0
XY[Xidx] = ((2*YbefX-n)*XY[Xidx]) - 2*YbefXval + Y.sum(0)[:,None]
return XY[:,:m].sum(0)
def summed_cdist(X,Y):
return cdist(X,Y,"minkowski",p=1).sum(1)
# demo
m,n,k = 1000,500,10
X,Y = np.random.randn(m,k),np.random.randn(n,k)
print("same result:",np.allclose(pp(X,Y),summed_cdist(X,Y)))
print("sort :",timeit(lambda:pp(X,Y),number=1000),"ms")
print("scipy cdist:",timeit(lambda:summed_cdist(X,Y),number=100)*10,"ms")
Sample run, comparing smart algo "sort" to naive algo implemented using cdist library function:
same result: True
sort : 1.4447695480193943 ms
scipy cdist: 36.41934019047767 ms
I am using scikit-learn preprocessing scaling for sparse matrices.
My goal is to "scale" each feature-column by taking the logarithm-base the column maximum value. My wording may be inexact. I try to explain.
Say feature-column has values: 0, 8, 2:
Max value = 8
Log-8 of feature value 0 should be 0.0 = math.log(0+1, 8+1) (the +1 is to cope with zeros; so yes, we are actually taking log-base 9)
Log-8 of feature value 8 should be 1.0 = math.log(8+1, 8+1)
Log-8 of feature value 2 should be 0.5 = math.log(2+1, 8+1)
Yes, I can easily apply any arbitrary function-based transformer with FunctionTransformer, but I want the base of the log change (based on) each column (in particular, the maximum value). That is, I want to do something like the MaxAbsScaler, only taking logarithms.
I see that MaxAbsScaler gets first a vector (scale) of the maximum values of each column (code) and then multiples the original matrix times 1 / scale in code.
However, I don't know what to do if I want to take the logarithms-based on the scale vector. Is it even possible to transform the logarithm operation to a multiplication (?) or do I have other possibilities of scipy sparse operations that are efficient?
I hope my intent is clear (and possible).
Logarithm of x in base b is the same as log(x)/log(b), where logs are natural. So, the process you describe amounts to first applying log(x+1) transformation to everything, and then scaling by max absolute value. Conveniently, log(x+1) is a built-in function, log1p. Example:
from sklearn.preprocessing import FunctionTransformer, maxabs_scale
from scipy.sparse import csc_matrix
import numpy as np
logtran = FunctionTransformer(np.log1p, accept_sparse=True)
X = csc_matrix([[ 1., 0, 8], [ 2., 0, 0], [ 0, 1., 2]])
Y = maxabs_scale(logtran.transform(X))
Output (sparse matrix Y):
(0, 0) 0.630929753571
(1, 0) 1.0
(2, 1) 1.0
(0, 2) 1.0
(2, 2) 0.5
How to identify the linearly independent rows from a matrix? For instance,
The 4th rows is independent.
First, your 3rd row is linearly dependent with 1t and 2nd row. However, your 1st and 4th column are linearly dependent.
Two methods you could use:
Eigenvalue
If one eigenvalue of the matrix is zero, its corresponding eigenvector is linearly dependent. The documentation eig states the returned eigenvalues are repeated according to their multiplicity and not necessarily ordered. However, assuming the eigenvalues correspond to your row vectors, one method would be:
import numpy as np
matrix = np.array(
[
[0, 1 ,0 ,0],
[0, 0, 1, 0],
[0, 1, 1, 0],
[1, 0, 0, 1]
])
lambdas, V = np.linalg.eig(matrix.T)
# The linearly dependent row vectors
print matrix[lambdas == 0,:]
Cauchy-Schwarz inequality
To test linear dependence of vectors and figure out which ones, you could use the Cauchy-Schwarz inequality. Basically, if the inner product of the vectors is equal to the product of the norm of the vectors, the vectors are linearly dependent. Here is an example for the columns:
import numpy as np
matrix = np.array(
[
[0, 1 ,0 ,0],
[0, 0, 1, 0],
[0, 1, 1, 0],
[1, 0, 0, 1]
])
print np.linalg.det(matrix)
for i in range(matrix.shape[0]):
for j in range(matrix.shape[0]):
if i != j:
inner_product = np.inner(
matrix[:,i],
matrix[:,j]
)
norm_i = np.linalg.norm(matrix[:,i])
norm_j = np.linalg.norm(matrix[:,j])
print 'I: ', matrix[:,i]
print 'J: ', matrix[:,j]
print 'Prod: ', inner_product
print 'Norm i: ', norm_i
print 'Norm j: ', norm_j
if np.abs(inner_product - norm_j * norm_i) < 1E-5:
print 'Dependent'
else:
print 'Independent'
To test the rows is a similar approach.
Then you could extend this to test all combinations of vectors, but I imagine this solution scale badly with size.
With sympy you can find the linear independant rows using: sympy.Matrix.rref:
>>> import sympy
>>> import numpy as np
>>> mat = np.array([[0,1,0,0],[0,0,1,0],[0,1,1,0],[1,0,0,1]]) # your matrix
>>> _, inds = sympy.Matrix(mat).T.rref() # to check the rows you need to transpose!
>>> inds
[0, 1, 3]
Which basically tells you the rows 0, 1 and 3 are linear independant while row 2 isn't (it's a linear combination of row 0 and 1).
Then you could remove these rows with slicing:
>>> mat[inds]
array([[0, 1, 0, 0],
[0, 0, 1, 0],
[1, 0, 0, 1]])
This also works well for rectangular (not only for quadratic) matrices.
I edited the code for Cauchy-Schwartz inequality which scales better with dimension: the inputs are the matrix and its dimension, while the output is a new rectangular matrix which contains along its rows the linearly independent columns of the starting matrix. This works in the assumption that the first column in never null, but can be readily generalized in order to implement this case too. Another thing that I observed is that 1e-5 seems to be a "sloppy" threshold, since some particular pathologic vectors were found to be linearly dependent in that case: 1e-4 doesn't give me the same problems. I hope this could be of some help: it was pretty difficult for me to find a really working routine to extract li vectors, and so I'm willing to share mine. If you find some bug, please report them!!
from numpy import dot, zeros
from numpy.linalg import matrix_rank, norm
def find_li_vectors(dim, R):
r = matrix_rank(R)
index = zeros( r ) #this will save the positions of the li columns in the matrix
counter = 0
index[0] = 0 #without loss of generality we pick the first column as linearly independent
j = 0 #therefore the second index is simply 0
for i in range(R.shape[0]): #loop over the columns
if i != j: #if the two columns are not the same
inner_product = dot( R[:,i], R[:,j] ) #compute the scalar product
norm_i = norm(R[:,i]) #compute norms
norm_j = norm(R[:,j])
#inner product and the product of the norms are equal only if the two vectors are parallel
#therefore we are looking for the ones which exhibit a difference which is bigger than a threshold
if absolute(inner_product - norm_j * norm_i) > 1e-4:
counter += 1 #counter is incremented
index[counter] = i #index is saved
j = i #j is refreshed
#do not forget to refresh j: otherwise you would compute only the vectors li with the first column!!
R_independent = zeros((r, dim))
i = 0
#now save everything in a new matrix
while( i < r ):
R_independent[i,:] = R[index[i],:]
i += 1
return R_independent
I know this was asked a while ago, but here is a very simple (although probably inefficient) solution. Given an array, the following finds a set of linearly independent vectors by progressively adding a vector and testing if the rank has increased:
from numpy.linalg import matrix_rank
def LI_vecs(dim,M):
LI=[M[0]]
for i in range(dim):
tmp=[]
for r in LI:
tmp.append(r)
tmp.append(M[i]) #set tmp=LI+[M[i]]
if matrix_rank(tmp)>len(LI): #test if M[i] is linearly independent from all (row) vectors in LI
LI.append(M[i]) #note that matrix_rank does not need to take in a square matrix
return LI #return set of linearly independent (row) vectors
#Example
mat=[[1,2,3,4],[4,5,6,7],[5,7,9,11],[2,4,6,8]]
LI_vecs(4,mat)
I interpret the problem as finding rows that are linearly independent from other rows.
That is equivalent to finding rows that are linearly dependent on other rows.
Gaussian elimination and treat numbers smaller than a threshold as zeros can do that. It is faster than finding eigenvalues of a matrix, testing all combinations of rows with Cauchy-Schwarz inequality, or singular value decomposition.
See:
https://math.stackexchange.com/questions/1297437/using-gauss-elimination-to-check-for-linear-dependence
Problem with floating point numbers:
http://numpy-discussion.10968.n7.nabble.com/Reduced-row-echelon-form-td16486.html
With regards to the following discussion:
Find dependent rows/columns of a matrix using Matlab?
from sympy import *
A = Matrix([[1,1,1],[2,2,2],[1,7,5]])
print(A.nullspace())
It is obvious that the first and second row are multiplication of each other.
If we execute the above code we get [-1/3, -2/3, 1]. The indices of the zero elements in the null space show independence. But why is the third element here not zero? If we multiply the A matrix with the null space, we get a zero column vector. So what's wrong?
The answer which we are looking for is the null space of the transpose of A.
B = A.T
print(B.nullspace())
Now we get the [-2, 1, 0], which shows that the third row is independent.
Two important notes here:
Consider whether we want to check the row dependencies or the column
dependencies.
Notice that the null space of a matrix is not equal to the null
space of the transpose of that matrix unless it is symmetric.
You can basically find the vectors spanning the columnspace of the matrix by using SymPy library's columnspace() method of Matrix object. Automatically, they are the linearly independent columns of the matrix.
import sympy as sp
import numpy as np
M = sp.Matrix([[0, 1, 0, 0],
[0, 0, 1, 0],
[1, 0, 0, 1]])
for i in M.columnspace():
print(np.array(i))
print()
# The output is following.
# [[0]
# [0]
# [1]]
# [[1]
# [0]
# [0]]
# [[0]
# [1]
# [0]]
I have a 3D image with dimensions rows x cols x deps. For every voxel in the image, I have computed a 3x3 real symmetric matrix. They are stored in the array D, which therefore has shape (rows, cols, deps, 6).
D stores the 6 unique elements of the 3x3 symmetric matrix for every voxel in my image. I need to find the Moore-Penrose pseudo inverse of all row*cols*deps matrices simultaneously/in vectorized code (looping through every image voxel and inverting is far too slow in Python).
Some of these 3x3 symmetric matrices are non-singular, and I can find their inverses, in vectorized code, using the analytical formula for the true inverse of a non-singular 3x3 symmetric matrix, and I've done that.
However, for those matrices that ARE singular (and there are sure to be some) I need the Moore-Penrose pseudo inverse. I could derive an analytical formula for the MP of a real, singular, symmetric 3x3 matrix, but it's a really nasty/lengthy formula, and would therefore involve a VERY large number of (element-wise) matrix arithmetic and quite a bit of confusing looking code.
Hence, I would like to know if there is a way to simultaneously find the MP pseudo inverse for all these matrices at once numerically. Is there a way to do this?
Gratefully,
GF
NumPy 1.8 included linear algebra gufuncs, which do exactly what you are after. While np.linalg.pinv is not gufunc-ed, np.linalg.svd is, and behind the scenes that is the function that gets called. So you can define your own gupinv function, based on the source code of the original function, as follows:
def gu_pinv(a, rcond=1e-15):
a = np.asarray(a)
swap = np.arange(a.ndim)
swap[[-2, -1]] = swap[[-1, -2]]
u, s, v = np.linalg.svd(a)
cutoff = np.maximum.reduce(s, axis=-1, keepdims=True) * rcond
mask = s > cutoff
s[mask] = 1. / s[mask]
s[~mask] = 0
return np.einsum('...uv,...vw->...uw',
np.transpose(v, swap) * s[..., None, :],
np.transpose(u, swap))
And you can now do things like:
a = np.random.rand(50, 40, 30, 6)
b = np.empty(a.shape[:-1] + (3, 3), dtype=a.dtype)
# Expand the unique items into a full symmetrical matrix
b[..., 0, :] = a[..., :3]
b[..., 1:, 0] = a[..., 1:3]
b[..., 1, 1:] = a[..., 3:5]
b[..., 2, 1:] = a[..., 4:]
# make matrix at [1, 2, 3] singular
b[1, 2, 3, 2] = b[1, 2, 3, 0] + b[1, 2, 3, 1]
# Find all the pseudo-inverses
pi = gu_pinv(b)
And of course the results are correct, both for singular and non-singular matrices:
>>> np.allclose(pi[0, 0, 0], np.linalg.pinv(b[0, 0, 0]))
True
>>> np.allclose(pi[1, 2, 3], np.linalg.pinv(b[1, 2, 3]))
True
And for this example, with 50 * 40 * 30 = 60,000 pseudo-inverses calculated:
In [2]: %timeit pi = gu_pinv(b)
1 loops, best of 3: 422 ms per loop
Which is really not that bad, although it is noticeably (4x) slower than simply calling np.linalg.inv, but this of course fails to properly handle the singular arrays:
In [8]: %timeit np.linalg.inv(b)
10 loops, best of 3: 98.8 ms per loop
EDIT: See #Jaime's answer. Only the discussion in the comments to this answer is useful now, and only for the specific problem at hand.
You can do this matrix by matrix, using scipy, that provides pinv (link) to calculate the Moore-Penrose pseudo inverse. An example follows:
from scipy.linalg import det,eig,pinv
from numpy.random import randint
#generate a random singular matrix M first
while True:
M = randint(0,10,9).reshape(3,3)
if det(M)==0:
break
M = M.astype(float)
#this is the method you need
MPpseudoinverse = pinv(M)
This does not exploit the fact that the matrix is symmetric though. You may also want to try the version of pinv exposed by numpy, that is supposedely faster, and different. See this post.
Having a nxn (6x6 in the example below) matrix filled only with 0 and 1:
old_matrix=[[0,0,0,1,1,0],
[1,1,1,1,0,0],
[0,0,1,0,0,0],
[1,0,0,0,0,1],
[0,1,1,1,1,0],
[1,0,0,1,1,0]]
I want to resize it in a particular way. Taking (2x2) sub-matrice and checking if there are more ones or zeros. This means the new matrix will be (3x3) If there are more 1 than 0 un the sub-matrice a 1 value will be assigned in the new matrix. Otherwise, (if there are less or equal) its new value will be 0.
new_matrix=[[0,1,0],
[0,0,0],
[0,1,0]]
I've tried to achieve this by using lots of whiles. However it doesn seem to work. Here's what I got so far:
def convert_track(a):
#converts original map to a 8x8 tile Track
NEW_TRACK=[]
w=0 #matrix width
h=0 #matrix heigth
t_w=0 #submatrix width
t_h=0 #submatrix heigth
BLACK=0 #number of ones in submatrix
WHITE=0 #number of zeros in submatrix
while h<=6:
while w<=6:
l=[]
while t_h<=2 and h<=6:
t_w=0
while t_w<=2 and w<=6:
if a[h][w]==1:
BLACK+=1
else:
WHITE+=1
t_w+=1
w+=1
h+=1
t_h+=1
t_w=0
t_h+=1
if BLACK<=WHITE:
l.append(0)
else:
l.append(1)
BLACK=0
WHITE=0
t_h=0
NEW_TRACK.append(l)
return NEW_TRACK
Raises the error list index out of range or returns the list
[[0]]
is there an easier way to achieve this? What am i doing wrong?
If you are willing/able to use NumPy you can do something like this. If you're working with anything like the data you've shown it's well worth your time to learn as operations like these can be done very efficiently and with very little code.
import numpy as np
from scipy.signal import convolve2d
old_matrix=[[0,0,0,1,1,0],
[1,1,1,1,0,0],
[0,0,1,0,0,0],
[1,0,0,0,0,1],
[0,1,1,1,1,0],
[1,0,0,1,1,0]]
a = np.array(old_matrix)
k = np.ones((2,2))
# compute sums at each submatrix
local_sums = convolve2d(a, k, mode='valid')
# restrict to sums corresponding to non-overlapping
# sub-matrices with local_sums[::2, ::2] and check if
# there are more 1 than 0 elements
result = local_sums[::2, ::2] > 2
# convert back to Python list if needed
new_matrix = result.astype(np.int).tolist()
Result:
>>> result.astype(np.int).tolist()
[[0, 1, 0], [0, 0, 0], [0, 1, 0]]
Here I've used convolve2d to compute the sums at each submatrix. From what I can tell you are only interested in non-overlapping sub-matrices, so the part local_sums[::2, ::2] chops out only the sums corresponding to those.