I am working on my own implementation of the weighted knn algorithm.
To simplify the logic, let's represent this as a predict method, which takes three parameters:
indices - matrix of nearest j neighbors from the training sample for object i (i=1...n, n objects in total). [i, j] - index of object from the training sample.
For example, for 4 objects and 3 neighbors:
indices = np.asarray([[0, 3, 1],
[0, 3, 1],
[1, 2, 0],
[5, 4, 3]])
distances - matrix of distances from j nearest neighbors from the training sample to object i. (i=1...n, n objects in total). For example, for 4 objects and 3 neighbors:
distances = np.asarray([[ 4.12310563, 7.07106781, 7.54983444],
[ 4.89897949, 6.70820393, 8.24621125],
[ 0., 1.73205081, 3.46410162],
[1094.09368886, 1102.55022561, 1109.62245832]])
labels - vector with true labels of classes for each object j of training sample. For example:
labels = np.asarray([0, 0, 0, 1, 1, 2])
Thus, the function signature is:
def predict(indices, distances, labels):
....
# return [np.bincount(x).argmax() for x in labels[indices]]
return predict
In the commentary you can see the code that returns the prediction for the "non-weighted" knn-method, which does not use distances. Can you please show, how predictions can be calculated with using the distance matrix? I found the algorithm, but now I'm completely stumped becase I don't know how to realize it with numpy.
Thank you!
This should work:
# compute inverses of distances
# suppress division by 0 warning,
# replace np.inf with a very large number
with np.errstate(divide='ignore'):
dinv = np.nan_to_num(1 / distances)
# an array with distinct class labels
distinct_labels = np.array(list(set(labels)))
# an array with labels of neighbors
neigh_labels = labels[indices]
# compute the weighted score for each potential label
weighted_scores = ((neigh_labels[:, :, np.newaxis] == distinct_labels) * dinv[:, :, np.newaxis]).sum(axis=1)
# choose the label with the highest score
predictions = distinct_labels[weighted_scores.argmax(axis=1)]
I want to make a torch tensor or numpy array efficiently of a matrix that is a shifting window of 1s.
So for example the matrix below would be a window=3.
The diagonal element has 3 1s to it's right, and 3 1s to it's left but it doesn't wrap round like a circulant matrix, so row 1 just has 4 1s.
Has anyone got any ideas, this is to be used as a mask.
Pytorch provides the tensor.diagonal method, which gives you access to any diagonal of a tensor. To assign a value to the resulting view of your tensor, you can use tensor.copy_. That would give you something like :
def circulant(n, window):
circulant_t = torch.zeros(n,n)
# [0, 1, 2, ..., window, -1, -2, ..., window]
offsets = [0] + [i for i in range(window)] + [-i for i in range(window)]
for offset in offsets:
#size of the 1-tensor depends on the length of the diagonal
circulant_t.diagonal(offset=offset).copy_(torch.ones(n-abs(offset)))
return circulant_t
I'm trying to apply world transformation to a numpy matrix. However, I can't seem to find a numpy way to perform a 4x4 matrix multiplication by Nx4 vector where N is the number of vertices.
I have both tried Nx4x4#Nx4 and 4x4#Nx4 multiplications. Sure, I could do this element wise but I'm hoping there's a smarter way to do this.
vertices = np.ones([VERTEX_COUNT, 4])
vertices[:, 0:3] = vertex_map[element.path_vertices]
matrix = np.full([VERTEX_COUNT, 4, 4], np.reshape(element.matrix, [4, 4]))
transformed = matrix # vertices # dimension mismatch
# i would rather not do this
# matrix = np.reshape(element.matrix, [4, 4])
# transformed = np.array([matrix # vertex for vertex in vertices])
Getting the inverse of a diagonal matrix is very simple and does not require complex methods. Does scipy.linalg.inv check whether the matrix is diagonal before it applies more complex methods or do I need to check this myself?
As you can see the Github code of scipy.linalg.inv, function inv first calls
getrf, getri, getri_lwork = get_lapack_funcs(('getrf', 'getri','getri_lwork'),
Then function getrf does it job to give the LU decomposition and so on. Now we have to investigate how getrf function gives the LU decomposition. Because if it checks if it's diagonal before to process the input matrix, then no need to check it yourself.
Function getrf is obtained by calling _get_funcs but I can't go further from there (_get_funcs is called with the following arguments _get_funcs(names, arrays, dtype, "LAPACK", _flapack, _clapack, "flapack", "clapack", _lapack_alias)).
I suggest that you run an experiment with a large diagonal matrix to compare the time given to spit the output with linalg and an inversion by hand.
Update (by question author):
import numpy as np
from scipy.linalg import inv
a = np.diag(np.random.random(19999))
b = a.copy()
np.fill_diagonal(a, 1/a.diagonal())
c = inv(b)
does not even require a time measuring tool: It it very obvious that inv is much slower... (that is surprisingly disappointing).
Please check: scipy.linalg.inv
If you put scipy.linalg.inv in try except if it raises LinAlgError when matrix a is singular. The determinant for singular matrix it zero.
try:
# your code that will (maybe) throw scipy.linalg.inv(your matrix)
except np.linalg.LinAlgError as err:
# It shows your matrix is singular
# Its determinant of a matrix is equal to zero
# The matrix does not have an inverse.
# You can conclude if the matrix is diagonal or not
If the determinant of a matrix is equal to zero:
The matrix is less than full rank. The matrix is singular. The matrix
does not have an inverse.
As manually:
def is_diagonal(matrix):
#create a dummy matrix
dummy_matrix = np.ones(matrix.shape, dtype=np.uint8)
# Fill the diagonal of dummy matrix with 0.
np.fill_diagonal(dummy_matrix, 0)
return np.count_nonzero(np.multiply(dummy_matrix, matrix)) == 0
diagonal_matrix = np.array([[3, 0, 0],
[0, 7, 0],
[0, 0, 4]])
print is_diagonal(diagonal_matrix)
>>> True
random_matrix = np.array([[3, 8, 0],
[1, 7, 8],
[5, 0, 4]])
print is_diagonal(random_matrix)
>>> False
scipy.sparse.dia_matrix.diagonal returns the k-th diagonal of the matrix.
from scipy.sparse import csr_matrix
A = csr_matrix([[1, 2, 0], [0, 0, 3], [4, 0, 5]])
A.diagonal()
array([1, 0, 5])
A.diagonal(k=1)
array([2, 3])
Also, from scipy.linalg import block_diag creates diagonal matrix if input arrays are square, therefore if they are not square, it can not create diagonal matrix.
Please consider in Jupyter you can find out time complexity. %timeit yourfunctionname
EDIT
I realized that I did not check my mwe very well and as such asked something of the wrong question. The main problem is when the numpy array is passed in as a 2d array instead of 1d (or even when a python list is passed in as 1d instead of 2d). So if we have
x = np.array([[1], [2], [3]])
then obviously if you try to index this then you will get arrays out (if you use item you do not). this same thing also applies to standard python lists.
Sorry about the confusion.
Original
I am trying to form a new numpy array from something that may be a numpy array or may be a standard python list.
for example
import numpy as np
x = [2, 3, 1]
y = np.array([[0, -x[2], x[1]], [x[2], 0, -x[0]], [-x[1], x[0], 0]])
Now I would like to form a function such that I can make y easily.
def skew(vector):
"""
this function returns a numpy array with the skew symmetric cross product matrix for vector.
the skew symmetric cross product matrix is defined such that
np.cross(a, b) = np.dot(skew(a), b)
:param vector: An array like vector to create the skew symmetric cross product matrix for
:return: A numpy array of the skew symmetric cross product vector
"""
return np.array([[0, -vector[2], vector[1]],
[vector[2], 0, -vector[0]],
[-vector[1], vector[0], 0]])
This works great and I can now write (assuming the above function is included)
import numpy as np
x=[2, 3, 1]
y = skew(x)
However, I would also like to be able to call skew on existing 1d or 2d numpy arrays. For instance
import numpy as np
x = np.array([2, 3, 1])
y = skew(x)
Unfortunately, doing this returns a numpy array where the elements are also numpy arrays, not python floats as I would like them to be.
Is there an easy way to form a new numpy array like I have done from something that is either a python list or a numpy array and have the result be just a standard numpy array with floats in each element?
Now obviously one solution is to check to see if the input is a numpy array or not:
def skew(vector):
"""
this function returns a numpy array with the skew symmetric cross product matrix for vector.
the skew symmetric cross product matrix is defined such that
np.cross(a, b) = np.dot(skew(a), b)
:param vector: An array like vector to create the skew symmetric cross product matrix for
:return: A numpy array of the skew symmetric cross product vector
"""
if isinstance(vector, np.ndarray):
return np.array([[0, -vector.item(2), vector.item(1)],
[vector.item(2), 0, -vector.item(0)],
[-vector.item(1), vector.item(0), 0]])
else:
return np.array([[0, -vector[2], vector[1]],
[vector[2], 0, -vector[0]],
[-vector[1], vector[0], 0]])
however, it gets very tedious having to write these instance checks all over the place.
Another solution would be to cast everything to an array first and then just use the array call
def skew(vector):
"""
this function returns a numpy array with the skew symmetric cross product matrix for vector.
the skew symmetric cross product matrix is defined such that
np.cross(a, b) = np.dot(skew(a), b)
:param vector: An array like vector to create the skew symmetric cross product matrix for
:return: A numpy array of the skew symmetric cross product vector
"""
vector = np.array(vector)
return np.array([[0, -vector.item(2), vector.item(1)],
[vector.item(2), 0, -vector.item(0)],
[-vector.item(1), vector.item(0), 0]])
but I feel like this is inefficient as it requires creating a new copy of vector (in this case not a big deal since vector is small but this is just a simple example).
My question is, is there a different way to do this outside of what I've discussed or am I stuck using one of these methods?
Arrays are iterable. You can write in your skew function:
def skew(x):
return np.array([[0, -x[2], x[1]],
[x[2], 0, -x[0]],
[-x[1], x[0], 0]])
x = [1,2,3]
y = np.array([1,2,3])
>>> skew(y)
array([[ 0, -3, 2],
[ 3, 0, -1],
[-2, 1, 0]])
>>> skew(x)
array([[ 0, -3, 2],
[ 3, 0, -1],
[-2, 1, 0]])
In any case your methods ended with 1st dimension elements being numpy arrays containing floats. You'll need in any case a call on the 2nd dimension to get the floats inside.
Regarding what you told me in the comments, you may add an if condition for 2d arrays:
def skew(x):
if (isinstance(x,ndarray) and len(x.shape)>=2):
return np.array([[0, -x[2][0], x[1][0]],
[x[2][0], 0, -x[0][0]],
[-x[1][0], x[0][0], 0]])
else:
return np.array([[0, -x[2], x[1]],
[x[2], 0, -x[0]],
[-x[1], x[0], 0]])
You can implement the last idea efficiently using numpy.asarray():
vector = np.asarray(vector)
Then, if vector is already a NumPy array, no copying occurs.
You can keep the first version of your function and convert the numpy array to list:
def skew(vector):
if isinstance(vector, np.ndarray):
vector = vector.tolist()
return np.array([[0, -vector[2], vector[1]],
[vector[2], 0, -vector[0]],
[-vector[1], vector[0], 0]])
In [58]: skew([2, 3, 1])
Out[58]:
array([[ 0, -1, 3],
[ 1, 0, -2],
[-3, 2, 0]])
In [59]: skew(np.array([2, 3, 1]))
Out[59]:
array([[ 0, -1, 3],
[ 1, 0, -2],
[-3, 2, 0]])
This is not an optimal solution but is a very easy one.
You can just convert the vector into list by default.
def skew(vector):
vector = list(vector)
return np.array([[0, -vector[2], vector[1]],
[vector[2], 0, -vector[0]],
[-vector[1], vector[0], 0]])