Fast diagonal matrix creation from two lists
2
Given 2 lists I would like to create a Diagonal matrix.
One list will fill the diagonal-constant and the other will fill the matrix.
For example:
fast_matrix([1,2], [6,7,8])
Should output 2 matrices:
[2] # unused
[1, 6, 7, 8]
[6, 1, 7, 8]
[6, 7, 1, 8]
[6, 7, 8, 1]
and
[1] # unused
[2, 6, 7, 8]
[6, 2, 7, 8]
[6, 7, 2, 8]
[6, 7, 8, 2]
My code makes 10000 transformations in 2.5secs in my pc.
from pprint import pprint
import timeit
def not_so_fast_matrix(A, B):
rt_obj =
for i,_ in enumerate(A):
for z in range(len(B) + 1):
new_from = A.copy()
new_from.remove(A[i])
new_list = B.copy()
new_list.insert(z, A[i])
rt_obj.append({'remain': new_from, 'to_list': new_list})
return rt_obj
# pprint(not_so_fast_matrix([1,2], [6,7,8]))
A = ([1,2,3,4,5,6,7,8,9,10])
B = ([60,70,80,90,100,200,300])
t = timeit.Timer(lambda: not_so_fast_toeplitz(A, B))
print("not_so_fast_matrix took: {:.3f}secs for 10000 iterations".format(t.timeit(number=10000)))
I was wondering if it can go faster using another approach.
Circulant from scipy.linalg looks like what I'm after but without rolling:from scipy.linalg import circulant
print(circulant([1, 8,7,6])) # <- Should be inverted
outputs:
[[1 6 7 8]
[8 1 6 7]
[7 8 1 6]
[6 7 8 1]]
Elements were shifted (pushed) right.
python python-3.x algorithm linear-algebra
asked Nov 21 '18 at 19:48
Panos Kal.Panos Kal.
8,27984361
|
show 5 more comments
2
Given 2 lists I would like to create a Diagonal matrix.
One list will fill the diagonal-constant and the other will fill the matrix.
For example:
fast_matrix([1,2], [6,7,8])
Should output 2 matrices:
[2] # unused
[1, 6, 7, 8]
[6, 1, 7, 8]
[6, 7, 1, 8]
[6, 7, 8, 1]
and
[1] # unused
[2, 6, 7, 8]
[6, 2, 7, 8]
[6, 7, 2, 8]
[6, 7, 8, 2]
My code makes 10000 transformations in 2.5secs in my pc.
from pprint import pprint
import timeit
def not_so_fast_matrix(A, B):
rt_obj =
for i,_ in enumerate(A):
for z in range(len(B) + 1):
new_from = A.copy()
new_from.remove(A[i])
new_list = B.copy()
new_list.insert(z, A[i])
rt_obj.append({'remain': new_from, 'to_list': new_list})
return rt_obj
# pprint(not_so_fast_matrix([1,2], [6,7,8]))
A = ([1,2,3,4,5,6,7,8,9,10])
B = ([60,70,80,90,100,200,300])
t = timeit.Timer(lambda: not_so_fast_toeplitz(A, B))
print("not_so_fast_matrix took: {:.3f}secs for 10000 iterations".format(t.timeit(number=10000)))
I was wondering if it can go faster using another approach.
Circulant from scipy.linalg looks like what I'm after but without rolling:from scipy.linalg import circulant
print(circulant([1, 8,7,6])) # <- Should be inverted
outputs:
[[1 6 7 8]
[8 1 6 7]
[7 8 1 6]
[6 7 8 1]]
Elements were shifted (pushed) right.
python python-3.x algorithm linear-algebra
asked Nov 21 '18 at 19:48
Panos Kal.Panos Kal.
8,27984361
Are you aware ofscipy.linalg.toeplitz?
– Warren Weckesser
Nov 21 '18 at 20:24
Yes, but for whatever reason it was slower
– Panos Kal.
Nov 21 '18 at 20:25
Have you tried it using SciPy version 1.1.0? Some changes were made in that release that improved the performance oftoeplitz.
– Warren Weckesser
Nov 21 '18 at 20:30
1
You can take a look at the source code on github. Your function returns a list of dictionaries, whilescipy.linalg.toeplitzreturns a numpy array, sotoeplitzwon't be a direct replacement for your function.
– Warren Weckesser
Nov 21 '18 at 20:37
1
Hint: transposed and flattened matrix data is structured. Creation can be done in reverse direction. Something like repeat() and reshape() while taking a care about diagonal and one missing element.
– Ante
Nov 22 '18 at 21:35
|
show 5 more comments
2
2
2
Given 2 lists I would like to create a Diagonal matrix.
One list will fill the diagonal-constant and the other will fill the matrix.
For example:
fast_matrix([1,2], [6,7,8])
Should output 2 matrices:
[2] # unused
[1, 6, 7, 8]
[6, 1, 7, 8]
[6, 7, 1, 8]
[6, 7, 8, 1]
and
[1] # unused
[2, 6, 7, 8]
[6, 2, 7, 8]
[6, 7, 2, 8]
[6, 7, 8, 2]
My code makes 10000 transformations in 2.5secs in my pc.
from pprint import pprint
import timeit
def not_so_fast_matrix(A, B):
rt_obj =
for i,_ in enumerate(A):
for z in range(len(B) + 1):
new_from = A.copy()
new_from.remove(A[i])
new_list = B.copy()
new_list.insert(z, A[i])
rt_obj.append({'remain': new_from, 'to_list': new_list})
return rt_obj
# pprint(not_so_fast_matrix([1,2], [6,7,8]))
A = ([1,2,3,4,5,6,7,8,9,10])
B = ([60,70,80,90,100,200,300])
t = timeit.Timer(lambda: not_so_fast_toeplitz(A, B))
print("not_so_fast_matrix took: {:.3f}secs for 10000 iterations".format(t.timeit(number=10000)))
I was wondering if it can go faster using another approach.
Circulant from scipy.linalg looks like what I'm after but without rolling:from scipy.linalg import circulant
print(circulant([1, 8,7,6])) # <- Should be inverted
outputs:
[[1 6 7 8]
[8 1 6 7]
[7 8 1 6]
[6 7 8 1]]
Elements were shifted (pushed) right.
python python-3.x algorithm linear-algebra
asked Nov 21 '18 at 19:48
Panos Kal.Panos Kal.
8,27984361
Given 2 lists I would like to create a Diagonal matrix.
One list will fill the diagonal-constant and the other will fill the matrix.
For example:
fast_matrix([1,2], [6,7,8])
Should output 2 matrices:
[2] # unused
[1, 6, 7, 8]
[6, 1, 7, 8]
[6, 7, 1, 8]
[6, 7, 8, 1]
and
[1] # unused
[2, 6, 7, 8]
[6, 2, 7, 8]
[6, 7, 2, 8]
[6, 7, 8, 2]
My code makes 10000 transformations in 2.5secs in my pc.
from pprint import pprint
import timeit
def not_so_fast_matrix(A, B):
rt_obj =
for i,_ in enumerate(A):
for z in range(len(B) + 1):
new_from = A.copy()
new_from.remove(A[i])
new_list = B.copy()
new_list.insert(z, A[i])
rt_obj.append({'remain': new_from, 'to_list': new_list})
return rt_obj
# pprint(not_so_fast_matrix([1,2], [6,7,8]))
A = ([1,2,3,4,5,6,7,8,9,10])
B = ([60,70,80,90,100,200,300])
t = timeit.Timer(lambda: not_so_fast_toeplitz(A, B))
print("not_so_fast_matrix took: {:.3f}secs for 10000 iterations".format(t.timeit(number=10000)))
I was wondering if it can go faster using another approach.
Circulant from scipy.linalg looks like what I'm after but without rolling:from scipy.linalg import circulant
print(circulant([1, 8,7,6])) # <- Should be inverted
outputs:
[[1 6 7 8]
[8 1 6 7]
[7 8 1 6]
[6 7 8 1]]
Elements were shifted (pushed) right.
python python-3.x algorithm linear-algebra
python python-3.x algorithm linear-algebra
asked Nov 21 '18 at 19:48
Panos Kal.Panos Kal.
8,27984361
asked Nov 21 '18 at 19:48
Panos Kal.Panos Kal.
8,27984361
edited Nov 22 '18 at 16:45
Panos Kal.
asked Nov 21 '18 at 19:48
Panos Kal.Panos Kal.
8,27984361
asked Nov 21 '18 at 19:48
Panos Kal.Panos Kal.
8,27984361
asked Nov 21 '18 at 19:48
Panos Kal.Panos Kal.
8,27984361
8,27984361
Are you aware ofscipy.linalg.toeplitz?
– Warren Weckesser
Nov 21 '18 at 20:24
Yes, but for whatever reason it was slower
– Panos Kal.
Nov 21 '18 at 20:25
Have you tried it using SciPy version 1.1.0? Some changes were made in that release that improved the performance oftoeplitz.
– Warren Weckesser
Nov 21 '18 at 20:30
1
You can take a look at the source code on github. Your function returns a list of dictionaries, whilescipy.linalg.toeplitzreturns a numpy array, sotoeplitzwon't be a direct replacement for your function.
– Warren Weckesser
Nov 21 '18 at 20:37
1
Hint: transposed and flattened matrix data is structured. Creation can be done in reverse direction. Something like repeat() and reshape() while taking a care about diagonal and one missing element.
– Ante
Nov 22 '18 at 21:35
|
show 5 more comments
Are you aware ofscipy.linalg.toeplitz?
– Warren Weckesser
Nov 21 '18 at 20:24
Yes, but for whatever reason it was slower
– Panos Kal.
Nov 21 '18 at 20:25
Have you tried it using SciPy version 1.1.0? Some changes were made in that release that improved the performance oftoeplitz.
– Warren Weckesser
Nov 21 '18 at 20:30
1
You can take a look at the source code on github. Your function returns a list of dictionaries, whilescipy.linalg.toeplitzreturns a numpy array, sotoeplitzwon't be a direct replacement for your function.
– Warren Weckesser
Nov 21 '18 at 20:37
1
Hint: transposed and flattened matrix data is structured. Creation can be done in reverse direction. Something like repeat() and reshape() while taking a care about diagonal and one missing element.
– Ante
Nov 22 '18 at 21:35
Are you aware of
scipy.linalg.toeplitz?– Warren Weckesser
Nov 21 '18 at 20:24
Are you aware of
scipy.linalg.toeplitz?– Warren Weckesser
Nov 21 '18 at 20:24
Yes, but for whatever reason it was slower
– Panos Kal.
Nov 21 '18 at 20:25
Yes, but for whatever reason it was slower
– Panos Kal.
Nov 21 '18 at 20:25
Have you tried it using SciPy version 1.1.0? Some changes were made in that release that improved the performance of
toeplitz.– Warren Weckesser
Nov 21 '18 at 20:30
Have you tried it using SciPy version 1.1.0? Some changes were made in that release that improved the performance of
toeplitz.– Warren Weckesser
Nov 21 '18 at 20:30
1
1
You can take a look at the source code on github. Your function returns a list of dictionaries, while
scipy.linalg.toeplitz returns a numpy array, so toeplitz won't be a direct replacement for your function.– Warren Weckesser
Nov 21 '18 at 20:37
You can take a look at the source code on github. Your function returns a list of dictionaries, while
scipy.linalg.toeplitz returns a numpy array, so toeplitz won't be a direct replacement for your function.– Warren Weckesser
Nov 21 '18 at 20:37
1
1
Hint: transposed and flattened matrix data is structured. Creation can be done in reverse direction. Something like repeat() and reshape() while taking a care about diagonal and one missing element.
– Ante
Nov 22 '18 at 21:35
Hint: transposed and flattened matrix data is structured. Creation can be done in reverse direction. Something like repeat() and reshape() while taking a care about diagonal and one missing element.
– Ante
Nov 22 '18 at 21:35
|
show 5 more comments
1 Answer
1
active
oldest
votes
1
Transposed and flatten matrix has a structure where elements from B are repeated. This approach uses that property to create blueprint of matrix with wrong values on diagonal, and than fill diagonal with correct value.
import numpy
def create_matrices(A, B):
# Create blueprint of result matrix
n = len(B) + 1
b = numpy.empty((n * n,), dtype=numpy.int32)
b[:-1] = numpy.repeat(B, n + 1)
b = b.reshape((n, n)).T # <- added transposition
# Change diagonal elements
for a in A:
m = b.copy()
numpy.fill_diagonal(m, a)
print(m)
create_matrices([1, 2], [6, 7, 8])
answered Nov 23 '18 at 10:25
AnteAnte
4,38251942
Thanks Ante. There is a problem with the output though. It returns [[1 6 6 6] [6 1 7 7] [7 7 1 8] [8 8 8 1]] when it should return [[1 6 7 8] [6 1 7 8] [6 7 1 8] [6 7 8 1]]
– Panos Kal.
Nov 23 '18 at 10:36
@PanosKal. I forget to add transposition. Check the line b = b.reshape...
– Ante
Nov 23 '18 at 10:39
add a comment |
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1 Answer
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1
Transposed and flatten matrix has a structure where elements from B are repeated. This approach uses that property to create blueprint of matrix with wrong values on diagonal, and than fill diagonal with correct value.
import numpy
def create_matrices(A, B):
# Create blueprint of result matrix
n = len(B) + 1
b = numpy.empty((n * n,), dtype=numpy.int32)
b[:-1] = numpy.repeat(B, n + 1)
b = b.reshape((n, n)).T # <- added transposition
# Change diagonal elements
for a in A:
m = b.copy()
numpy.fill_diagonal(m, a)
print(m)
create_matrices([1, 2], [6, 7, 8])
answered Nov 23 '18 at 10:25
AnteAnte
4,38251942
Thanks Ante. There is a problem with the output though. It returns [[1 6 6 6] [6 1 7 7] [7 7 1 8] [8 8 8 1]] when it should return [[1 6 7 8] [6 1 7 8] [6 7 1 8] [6 7 8 1]]
– Panos Kal.
Nov 23 '18 at 10:36
@PanosKal. I forget to add transposition. Check the line b = b.reshape...
– Ante
Nov 23 '18 at 10:39
add a comment |
1
Transposed and flatten matrix has a structure where elements from B are repeated. This approach uses that property to create blueprint of matrix with wrong values on diagonal, and than fill diagonal with correct value.
import numpy
def create_matrices(A, B):
# Create blueprint of result matrix
n = len(B) + 1
b = numpy.empty((n * n,), dtype=numpy.int32)
b[:-1] = numpy.repeat(B, n + 1)
b = b.reshape((n, n)).T # <- added transposition
# Change diagonal elements
for a in A:
m = b.copy()
numpy.fill_diagonal(m, a)
print(m)
create_matrices([1, 2], [6, 7, 8])
answered Nov 23 '18 at 10:25
AnteAnte
4,38251942
Thanks Ante. There is a problem with the output though. It returns [[1 6 6 6] [6 1 7 7] [7 7 1 8] [8 8 8 1]] when it should return [[1 6 7 8] [6 1 7 8] [6 7 1 8] [6 7 8 1]]
– Panos Kal.
Nov 23 '18 at 10:36
@PanosKal. I forget to add transposition. Check the line b = b.reshape...
– Ante
Nov 23 '18 at 10:39
add a comment |
1
1
1
Transposed and flatten matrix has a structure where elements from B are repeated. This approach uses that property to create blueprint of matrix with wrong values on diagonal, and than fill diagonal with correct value.
import numpy
def create_matrices(A, B):
# Create blueprint of result matrix
n = len(B) + 1
b = numpy.empty((n * n,), dtype=numpy.int32)
b[:-1] = numpy.repeat(B, n + 1)
b = b.reshape((n, n)).T # <- added transposition
# Change diagonal elements
for a in A:
m = b.copy()
numpy.fill_diagonal(m, a)
print(m)
create_matrices([1, 2], [6, 7, 8])
answered Nov 23 '18 at 10:25
AnteAnte
4,38251942
Transposed and flatten matrix has a structure where elements from B are repeated. This approach uses that property to create blueprint of matrix with wrong values on diagonal, and than fill diagonal with correct value.
import numpy
def create_matrices(A, B):
# Create blueprint of result matrix
n = len(B) + 1
b = numpy.empty((n * n,), dtype=numpy.int32)
b[:-1] = numpy.repeat(B, n + 1)
b = b.reshape((n, n)).T # <- added transposition
# Change diagonal elements
for a in A:
m = b.copy()
numpy.fill_diagonal(m, a)
print(m)
create_matrices([1, 2], [6, 7, 8])
answered Nov 23 '18 at 10:25
AnteAnte
4,38251942
edited Nov 23 '18 at 10:39
answered Nov 23 '18 at 10:25
AnteAnte
4,38251942
answered Nov 23 '18 at 10:25
AnteAnte
4,38251942
answered Nov 23 '18 at 10:25
AnteAnte
4,38251942
4,38251942
Thanks Ante. There is a problem with the output though. It returns [[1 6 6 6] [6 1 7 7] [7 7 1 8] [8 8 8 1]] when it should return [[1 6 7 8] [6 1 7 8] [6 7 1 8] [6 7 8 1]]
– Panos Kal.
Nov 23 '18 at 10:36
@PanosKal. I forget to add transposition. Check the line b = b.reshape...
– Ante
Nov 23 '18 at 10:39
add a comment |
Thanks Ante. There is a problem with the output though. It returns [[1 6 6 6] [6 1 7 7] [7 7 1 8] [8 8 8 1]] when it should return [[1 6 7 8] [6 1 7 8] [6 7 1 8] [6 7 8 1]]
– Panos Kal.
Nov 23 '18 at 10:36
@PanosKal. I forget to add transposition. Check the line b = b.reshape...
– Ante
Nov 23 '18 at 10:39
Thanks Ante. There is a problem with the output though. It returns [[1 6 6 6] [6 1 7 7] [7 7 1 8] [8 8 8 1]] when it should return [[1 6 7 8] [6 1 7 8] [6 7 1 8] [6 7 8 1]]
– Panos Kal.
Nov 23 '18 at 10:36
Thanks Ante. There is a problem with the output though. It returns [[1 6 6 6] [6 1 7 7] [7 7 1 8] [8 8 8 1]] when it should return [[1 6 7 8] [6 1 7 8] [6 7 1 8] [6 7 8 1]]
– Panos Kal.
Nov 23 '18 at 10:36
@PanosKal. I forget to add transposition. Check the line b = b.reshape...
– Ante
Nov 23 '18 at 10:39
@PanosKal. I forget to add transposition. Check the line b = b.reshape...
– Ante
Nov 23 '18 at 10:39
add a comment |
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Are you aware of
scipy.linalg.toeplitz?– Warren Weckesser
Nov 21 '18 at 20:24
Yes, but for whatever reason it was slower
– Panos Kal.
Nov 21 '18 at 20:25
Have you tried it using SciPy version 1.1.0? Some changes were made in that release that improved the performance of
toeplitz.– Warren Weckesser
Nov 21 '18 at 20:30
1
You can take a look at the source code on github. Your function returns a list of dictionaries, while
scipy.linalg.toeplitzreturns a numpy array, sotoeplitzwon't be a direct replacement for your function.– Warren Weckesser
Nov 21 '18 at 20:37
1
Hint: transposed and flattened matrix data is structured. Creation can be done in reverse direction. Something like repeat() and reshape() while taking a care about diagonal and one missing element.
– Ante
Nov 22 '18 at 21:35