Mixing
How to find the sum of elements above and below the diagonal of a matrix in python?
I need to find the sum of the elements that are above and below the main diagonal. I have no idea how to condition the algorithm to sum only those numbers. This is the code I have so far, with A being the matrix
A =
N = int(raw_input("Input matrix size: "))
for i in range(0, N):
row =
for j in range(0, N):
row.append(int(raw_input("Input elements: ")))
A.append(row)
sum = 0
for i in range(0, N):
sum += A[i][i]
print sum
sum2 = 0
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
print sum2
I am guessing I should use more for statements.
Thank you
python arrays numpy matrix diagonal
edited Nov 19 '18 at 16:03
rahlf23
5,1522630
asked Nov 19 '18 at 15:44
Martin
62
add a comment |
I need to find the sum of the elements that are above and below the main diagonal. I have no idea how to condition the algorithm to sum only those numbers. This is the code I have so far, with A being the matrix
A =
N = int(raw_input("Input matrix size: "))
for i in range(0, N):
row =
for j in range(0, N):
row.append(int(raw_input("Input elements: ")))
A.append(row)
sum = 0
for i in range(0, N):
sum += A[i][i]
print sum
sum2 = 0
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
print sum2
I am guessing I should use more for statements.
Thank you
python arrays numpy matrix diagonal
edited Nov 19 '18 at 16:03
rahlf23
5,1522630
asked Nov 19 '18 at 15:44
Martin
62
2
Possible duplicate of Python - How to get the sum of the upper triangle of a matrix without using numpy?
– Austin
Nov 19 '18 at 15:47
add a comment |
I need to find the sum of the elements that are above and below the main diagonal. I have no idea how to condition the algorithm to sum only those numbers. This is the code I have so far, with A being the matrix
A =
N = int(raw_input("Input matrix size: "))
for i in range(0, N):
row =
for j in range(0, N):
row.append(int(raw_input("Input elements: ")))
A.append(row)
sum = 0
for i in range(0, N):
sum += A[i][i]
print sum
sum2 = 0
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
print sum2
I am guessing I should use more for statements.
Thank you
python arrays numpy matrix diagonal
edited Nov 19 '18 at 16:03
rahlf23
5,1522630
asked Nov 19 '18 at 15:44
Martin
62
I need to find the sum of the elements that are above and below the main diagonal. I have no idea how to condition the algorithm to sum only those numbers. This is the code I have so far, with A being the matrix
A =
N = int(raw_input("Input matrix size: "))
for i in range(0, N):
row =
for j in range(0, N):
row.append(int(raw_input("Input elements: ")))
A.append(row)
sum = 0
for i in range(0, N):
sum += A[i][i]
print sum
sum2 = 0
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
print sum2
I am guessing I should use more for statements.
Thank you
python arrays numpy matrix diagonal
python arrays numpy matrix diagonal
edited Nov 19 '18 at 16:03
rahlf23
5,1522630
asked Nov 19 '18 at 15:44
Martin
62
edited Nov 19 '18 at 16:03
rahlf23
5,1522630
asked Nov 19 '18 at 15:44
Martin
62
edited Nov 19 '18 at 16:03
rahlf23
5,1522630
edited Nov 19 '18 at 16:03
rahlf23
5,1522630
edited Nov 19 '18 at 16:03
rahlf23
5,1522630
5,1522630
asked Nov 19 '18 at 15:44
Martin
62
asked Nov 19 '18 at 15:44
Martin
62
asked Nov 19 '18 at 15:44
Martin
62
62
2
Possible duplicate of Python - How to get the sum of the upper triangle of a matrix without using numpy?
– Austin
Nov 19 '18 at 15:47
add a comment |
2
Possible duplicate of Python - How to get the sum of the upper triangle of a matrix without using numpy?
– Austin
Nov 19 '18 at 15:47
2
2
Possible duplicate of Python - How to get the sum of the upper triangle of a matrix without using numpy?
– Austin
Nov 19 '18 at 15:47
Possible duplicate of Python - How to get the sum of the upper triangle of a matrix without using numpy?
– Austin
Nov 19 '18 at 15:47
add a comment |
6 Answers
6
active
oldest
votes
You can use np.triu, np.tril and np.trace to compute these sums (your question does not specify whether or not you are allowed to leverage numpy):
import numpy as np
np.random.seed(0)
A = np.random.randint(0,10,size=(5,5))
Gives:
[[5 0 3 3 7]
[9 3 5 2 4]
[7 6 8 8 1]
[6 7 7 8 1]
[5 9 8 9 4]]
Then:
upper_sum = np.triu(A).sum()-np.trace(A)
lower_sum = np.tril(A).sum()-np.trace(A)
Yields:
34
73
answered Nov 19 '18 at 15:59
rahlf23
5,1522630
Neat, have a +1.
– b-fg
Nov 19 '18 at 16:01
2
You can usetriu(A, 1)andtril(A, -1)to exclude the (main) diagonal.
– Paul Panzer
Nov 19 '18 at 19:43
add a comment |
Your first loop will compute the sum of the diagonal
for i in range(0, N):
sum += A[i][i]
This second loop will do the job, and compute the sum of every elements above the diagonal !
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
So, the tricks is: computing the sum for each i and j, with j > i.
Applying the same tricks for elements below diagonal is computing the sum for each i and j, with j < i.
for i in range(0, N):
for j in range(0, i):
sum3 += A[i][j]
answered Nov 19 '18 at 15:52
NiziL
4,1661731
add a comment |
Let's assume you have a 3x3 matrix.
[[a11, a12, a13],
[a21, a22, a23],
[a31, a32, a33]]
Do you find any similarity in the indices of the upper triangular and lower triangular parts? (Hover on the below text part to know the answer).
The first index of the upper triangle is lower in value as compared to the second index. Similarly, the first index is smaller than the second index in the lower triangle. Moreover, for diagonal elements, the indices are same!
Now, assuming you have written the above code yourself, I hope that you can do this one yourself, for now you know the logic. It will be two loops (one to traverse the rows, and one to traverse the columns), and some if statements.
answered Nov 19 '18 at 15:53
MaJoR
430111
If there's any doubts regarding how to code this bit, you can have a look at other answers. ^.^
– MaJoR
Nov 19 '18 at 15:54
add a comment |
You can acomplish this with numpy.triu_indices. I have commented below each step to guide you through it. Basically you get the upper right indices with numpy.triu_indices and loop over them to get the elements. You sum all of the elements except the ones in the diagonal.
import numpy as np
m = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
triu = np.triu_indices(m.ndim+1) # Get upper right indices of m
triu_idx = list(zip(triu[0],triu[1])) # List of tuples for the upper right indices
cum_sum = 0 # Initialize sum
for x in triu_idx: # Loop over the upper right indices
if x[0] != x[1]: # Exclude diagonal elements
cum_sum += m[x] # Add to sum
print(cum_sum) # Output 11
Given the matrix
[[1 2 3]
[4 5 6]
[7 8 9]]
It outputs 11.
answered Nov 19 '18 at 15:59
b-fg
1,96411422
This is a bit long-winded in terms of simply finding the sum, see me answer.
– rahlf23
Nov 19 '18 at 16:00
Yeap, looks better!
– b-fg
Nov 19 '18 at 16:01
add a comment |
Here is an example case demonstrating how to find the sum in both cases, using nested loop:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
totalSum = 0
for i in range(1,len(matrix)):
for j in range(i):
totalSum += matrix[i][j]
print("Below sum: ", totalSum)
totalSum = 0
for i in range(len(matrix)):
for j in range(i+1,len(matrix)):
totalSum += matrix[i][j]
print("Above sum: ", totalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Below sum: 18
Above sum: 18
You can also use these one-liners:
Below diagonal:
totalSum = sum([matrix[i][j] for i in range(1,len(matrix)) for j in range(i)])
Above diagonal:
totalSum = sum([matrix[i][j] for i in range(len(matrix)) for j in range(i+1,len(matrix))])
If you want to sum all numbers above and below the main diagonal you can perform an index check:
totalSum = 0
for i in range(len(matrix)):
for j in range(len(matrix)):
if not i==j:
totalSum += matrix[i][j]
print("Sum: ", totalSum)
Yet, another way of finding that sum (although not recommended) is to find the total sum of the matrix and the sum of the main diagonal and then perform a subtraction to find the final sum:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
matrixSum = sum([sum(elem for elem in row) for row in matrix])
diagonalSum = sum([matrix[i][i] for i in range(len(matrix))])
finalSum = matrixSum - diagonalSum
print("Matrix sum: ", matrixSum)
print("Diagonal sum: ", diagonalSum)
print("Final sum: ", finalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Matrix sum: 48
Diagonal sum: 12
Final sum: 36
Note: Mind the syntax in the print statements as you're using Python 2 and my answer is in Python 3.
answered Nov 19 '18 at 15:50
Vasilis G.
3,3702722
add a comment |
Here is a fast method using scipy.spatial.distance.squareform for the triangles and np.einsum for the diagonal:
>>> import numpy as np
>>> from scipy.spatial.distance import squareform
>>>
>>> x = np.arange(36).reshape(6, 6)
>>>
>>> sum_ut = squareform(x, checks=False).sum()
>>> sum_dg = np.einsum('ii', x)
>>> sum_lt = squareform(x.T, checks=False).sum()
Timings:
>>> timeit(lambda: squareform(x, checks=False).sum())
6.272806407185271
>>> timeit(lambda: np.einsum('ii', x))
1.3961836302187294
>>> timeit(lambda: squareform(x.T, checks=False).sum())
6.6827554509509355
For comparison:
>>> timeit(lambda: np.triu(x, 1).sum())
13.85556498519145
>>> timeit(lambda: np.trace(x))
3.081781509099528
>>> timeit(lambda: np.tril(x, -1).sum())
13.659938262077048
answered Nov 19 '18 at 19:38
Paul Panzer
29.9k21240
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use np.triu, np.tril and np.trace to compute these sums (your question does not specify whether or not you are allowed to leverage numpy):
import numpy as np
np.random.seed(0)
A = np.random.randint(0,10,size=(5,5))
Gives:
[[5 0 3 3 7]
[9 3 5 2 4]
[7 6 8 8 1]
[6 7 7 8 1]
[5 9 8 9 4]]
Then:
upper_sum = np.triu(A).sum()-np.trace(A)
lower_sum = np.tril(A).sum()-np.trace(A)
Yields:
34
73
answered Nov 19 '18 at 15:59
rahlf23
5,1522630
Neat, have a +1.
– b-fg
Nov 19 '18 at 16:01
2
You can usetriu(A, 1)andtril(A, -1)to exclude the (main) diagonal.
– Paul Panzer
Nov 19 '18 at 19:43
add a comment |
You can use np.triu, np.tril and np.trace to compute these sums (your question does not specify whether or not you are allowed to leverage numpy):
import numpy as np
np.random.seed(0)
A = np.random.randint(0,10,size=(5,5))
Gives:
[[5 0 3 3 7]
[9 3 5 2 4]
[7 6 8 8 1]
[6 7 7 8 1]
[5 9 8 9 4]]
Then:
upper_sum = np.triu(A).sum()-np.trace(A)
lower_sum = np.tril(A).sum()-np.trace(A)
Yields:
34
73
answered Nov 19 '18 at 15:59
rahlf23
5,1522630
Neat, have a +1.
– b-fg
Nov 19 '18 at 16:01
2
You can usetriu(A, 1)andtril(A, -1)to exclude the (main) diagonal.
– Paul Panzer
Nov 19 '18 at 19:43
add a comment |
You can use np.triu, np.tril and np.trace to compute these sums (your question does not specify whether or not you are allowed to leverage numpy):
import numpy as np
np.random.seed(0)
A = np.random.randint(0,10,size=(5,5))
Gives:
[[5 0 3 3 7]
[9 3 5 2 4]
[7 6 8 8 1]
[6 7 7 8 1]
[5 9 8 9 4]]
Then:
upper_sum = np.triu(A).sum()-np.trace(A)
lower_sum = np.tril(A).sum()-np.trace(A)
Yields:
34
73
answered Nov 19 '18 at 15:59
rahlf23
5,1522630
You can use np.triu, np.tril and np.trace to compute these sums (your question does not specify whether or not you are allowed to leverage numpy):
import numpy as np
np.random.seed(0)
A = np.random.randint(0,10,size=(5,5))
Gives:
[[5 0 3 3 7]
[9 3 5 2 4]
[7 6 8 8 1]
[6 7 7 8 1]
[5 9 8 9 4]]
Then:
upper_sum = np.triu(A).sum()-np.trace(A)
lower_sum = np.tril(A).sum()-np.trace(A)
Yields:
34
73
answered Nov 19 '18 at 15:59
rahlf23
5,1522630
answered Nov 19 '18 at 15:59
rahlf23
5,1522630
answered Nov 19 '18 at 15:59
rahlf23
5,1522630
answered Nov 19 '18 at 15:59
rahlf23
5,1522630
5,1522630
Neat, have a +1.
– b-fg
Nov 19 '18 at 16:01
2
You can usetriu(A, 1)andtril(A, -1)to exclude the (main) diagonal.
– Paul Panzer
Nov 19 '18 at 19:43
add a comment |
Neat, have a +1.
– b-fg
Nov 19 '18 at 16:01
2
You can usetriu(A, 1)andtril(A, -1)to exclude the (main) diagonal.
– Paul Panzer
Nov 19 '18 at 19:43
Neat, have a +1.
– b-fg
Nov 19 '18 at 16:01
Neat, have a +1.
– b-fg
Nov 19 '18 at 16:01
2
2
You can use
triu(A, 1) and tril(A, -1) to exclude the (main) diagonal.– Paul Panzer
Nov 19 '18 at 19:43
You can use
triu(A, 1) and tril(A, -1) to exclude the (main) diagonal.– Paul Panzer
Nov 19 '18 at 19:43
add a comment |
Your first loop will compute the sum of the diagonal
for i in range(0, N):
sum += A[i][i]
This second loop will do the job, and compute the sum of every elements above the diagonal !
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
So, the tricks is: computing the sum for each i and j, with j > i.
Applying the same tricks for elements below diagonal is computing the sum for each i and j, with j < i.
for i in range(0, N):
for j in range(0, i):
sum3 += A[i][j]
answered Nov 19 '18 at 15:52
NiziL
4,1661731
add a comment |
Your first loop will compute the sum of the diagonal
for i in range(0, N):
sum += A[i][i]
This second loop will do the job, and compute the sum of every elements above the diagonal !
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
So, the tricks is: computing the sum for each i and j, with j > i.
Applying the same tricks for elements below diagonal is computing the sum for each i and j, with j < i.
for i in range(0, N):
for j in range(0, i):
sum3 += A[i][j]
answered Nov 19 '18 at 15:52
NiziL
4,1661731
add a comment |
Your first loop will compute the sum of the diagonal
for i in range(0, N):
sum += A[i][i]
This second loop will do the job, and compute the sum of every elements above the diagonal !
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
So, the tricks is: computing the sum for each i and j, with j > i.
Applying the same tricks for elements below diagonal is computing the sum for each i and j, with j < i.
for i in range(0, N):
for j in range(0, i):
sum3 += A[i][j]
answered Nov 19 '18 at 15:52
NiziL
4,1661731
Your first loop will compute the sum of the diagonal
for i in range(0, N):
sum += A[i][i]
This second loop will do the job, and compute the sum of every elements above the diagonal !
for i in range(0, N):
for j in range(i+1, N):
sum2 += A[i][j]
So, the tricks is: computing the sum for each i and j, with j > i.
Applying the same tricks for elements below diagonal is computing the sum for each i and j, with j < i.
for i in range(0, N):
for j in range(0, i):
sum3 += A[i][j]
answered Nov 19 '18 at 15:52
NiziL
4,1661731
answered Nov 19 '18 at 15:52
NiziL
4,1661731
answered Nov 19 '18 at 15:52
NiziL
4,1661731
answered Nov 19 '18 at 15:52
NiziL
4,1661731
4,1661731
add a comment |
add a comment |
Let's assume you have a 3x3 matrix.
[[a11, a12, a13],
[a21, a22, a23],
[a31, a32, a33]]
Do you find any similarity in the indices of the upper triangular and lower triangular parts? (Hover on the below text part to know the answer).
The first index of the upper triangle is lower in value as compared to the second index. Similarly, the first index is smaller than the second index in the lower triangle. Moreover, for diagonal elements, the indices are same!
Now, assuming you have written the above code yourself, I hope that you can do this one yourself, for now you know the logic. It will be two loops (one to traverse the rows, and one to traverse the columns), and some if statements.
answered Nov 19 '18 at 15:53
MaJoR
430111
If there's any doubts regarding how to code this bit, you can have a look at other answers. ^.^
– MaJoR
Nov 19 '18 at 15:54
add a comment |
Let's assume you have a 3x3 matrix.
[[a11, a12, a13],
[a21, a22, a23],
[a31, a32, a33]]
Do you find any similarity in the indices of the upper triangular and lower triangular parts? (Hover on the below text part to know the answer).
The first index of the upper triangle is lower in value as compared to the second index. Similarly, the first index is smaller than the second index in the lower triangle. Moreover, for diagonal elements, the indices are same!
Now, assuming you have written the above code yourself, I hope that you can do this one yourself, for now you know the logic. It will be two loops (one to traverse the rows, and one to traverse the columns), and some if statements.
answered Nov 19 '18 at 15:53
MaJoR
430111
If there's any doubts regarding how to code this bit, you can have a look at other answers. ^.^
– MaJoR
Nov 19 '18 at 15:54
add a comment |
Let's assume you have a 3x3 matrix.
[[a11, a12, a13],
[a21, a22, a23],
[a31, a32, a33]]
Do you find any similarity in the indices of the upper triangular and lower triangular parts? (Hover on the below text part to know the answer).
The first index of the upper triangle is lower in value as compared to the second index. Similarly, the first index is smaller than the second index in the lower triangle. Moreover, for diagonal elements, the indices are same!
Now, assuming you have written the above code yourself, I hope that you can do this one yourself, for now you know the logic. It will be two loops (one to traverse the rows, and one to traverse the columns), and some if statements.
answered Nov 19 '18 at 15:53
MaJoR
430111
Let's assume you have a 3x3 matrix.
[[a11, a12, a13],
[a21, a22, a23],
[a31, a32, a33]]
Do you find any similarity in the indices of the upper triangular and lower triangular parts? (Hover on the below text part to know the answer).
The first index of the upper triangle is lower in value as compared to the second index. Similarly, the first index is smaller than the second index in the lower triangle. Moreover, for diagonal elements, the indices are same!
Now, assuming you have written the above code yourself, I hope that you can do this one yourself, for now you know the logic. It will be two loops (one to traverse the rows, and one to traverse the columns), and some if statements.
answered Nov 19 '18 at 15:53
MaJoR
430111
answered Nov 19 '18 at 15:53
MaJoR
430111
answered Nov 19 '18 at 15:53
MaJoR
430111
answered Nov 19 '18 at 15:53
MaJoR
430111
430111
If there's any doubts regarding how to code this bit, you can have a look at other answers. ^.^
– MaJoR
Nov 19 '18 at 15:54
add a comment |
If there's any doubts regarding how to code this bit, you can have a look at other answers. ^.^
– MaJoR
Nov 19 '18 at 15:54
If there's any doubts regarding how to code this bit, you can have a look at other answers. ^.^
– MaJoR
Nov 19 '18 at 15:54
If there's any doubts regarding how to code this bit, you can have a look at other answers. ^.^
– MaJoR
Nov 19 '18 at 15:54
add a comment |
You can acomplish this with numpy.triu_indices. I have commented below each step to guide you through it. Basically you get the upper right indices with numpy.triu_indices and loop over them to get the elements. You sum all of the elements except the ones in the diagonal.
import numpy as np
m = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
triu = np.triu_indices(m.ndim+1) # Get upper right indices of m
triu_idx = list(zip(triu[0],triu[1])) # List of tuples for the upper right indices
cum_sum = 0 # Initialize sum
for x in triu_idx: # Loop over the upper right indices
if x[0] != x[1]: # Exclude diagonal elements
cum_sum += m[x] # Add to sum
print(cum_sum) # Output 11
Given the matrix
[[1 2 3]
[4 5 6]
[7 8 9]]
It outputs 11.
answered Nov 19 '18 at 15:59
b-fg
1,96411422
This is a bit long-winded in terms of simply finding the sum, see me answer.
– rahlf23
Nov 19 '18 at 16:00
Yeap, looks better!
– b-fg
Nov 19 '18 at 16:01
add a comment |
You can acomplish this with numpy.triu_indices. I have commented below each step to guide you through it. Basically you get the upper right indices with numpy.triu_indices and loop over them to get the elements. You sum all of the elements except the ones in the diagonal.
import numpy as np
m = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
triu = np.triu_indices(m.ndim+1) # Get upper right indices of m
triu_idx = list(zip(triu[0],triu[1])) # List of tuples for the upper right indices
cum_sum = 0 # Initialize sum
for x in triu_idx: # Loop over the upper right indices
if x[0] != x[1]: # Exclude diagonal elements
cum_sum += m[x] # Add to sum
print(cum_sum) # Output 11
Given the matrix
[[1 2 3]
[4 5 6]
[7 8 9]]
It outputs 11.
answered Nov 19 '18 at 15:59
b-fg
1,96411422
This is a bit long-winded in terms of simply finding the sum, see me answer.
– rahlf23
Nov 19 '18 at 16:00
Yeap, looks better!
– b-fg
Nov 19 '18 at 16:01
add a comment |
You can acomplish this with numpy.triu_indices. I have commented below each step to guide you through it. Basically you get the upper right indices with numpy.triu_indices and loop over them to get the elements. You sum all of the elements except the ones in the diagonal.
import numpy as np
m = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
triu = np.triu_indices(m.ndim+1) # Get upper right indices of m
triu_idx = list(zip(triu[0],triu[1])) # List of tuples for the upper right indices
cum_sum = 0 # Initialize sum
for x in triu_idx: # Loop over the upper right indices
if x[0] != x[1]: # Exclude diagonal elements
cum_sum += m[x] # Add to sum
print(cum_sum) # Output 11
Given the matrix
[[1 2 3]
[4 5 6]
[7 8 9]]
It outputs 11.
answered Nov 19 '18 at 15:59
b-fg
1,96411422
You can acomplish this with numpy.triu_indices. I have commented below each step to guide you through it. Basically you get the upper right indices with numpy.triu_indices and loop over them to get the elements. You sum all of the elements except the ones in the diagonal.
import numpy as np
m = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
triu = np.triu_indices(m.ndim+1) # Get upper right indices of m
triu_idx = list(zip(triu[0],triu[1])) # List of tuples for the upper right indices
cum_sum = 0 # Initialize sum
for x in triu_idx: # Loop over the upper right indices
if x[0] != x[1]: # Exclude diagonal elements
cum_sum += m[x] # Add to sum
print(cum_sum) # Output 11
Given the matrix
[[1 2 3]
[4 5 6]
[7 8 9]]
It outputs 11.
answered Nov 19 '18 at 15:59
b-fg
1,96411422
answered Nov 19 '18 at 15:59
b-fg
1,96411422
answered Nov 19 '18 at 15:59
b-fg
1,96411422
answered Nov 19 '18 at 15:59
b-fg
1,96411422
1,96411422
This is a bit long-winded in terms of simply finding the sum, see me answer.
– rahlf23
Nov 19 '18 at 16:00
Yeap, looks better!
– b-fg
Nov 19 '18 at 16:01
add a comment |
This is a bit long-winded in terms of simply finding the sum, see me answer.
– rahlf23
Nov 19 '18 at 16:00
Yeap, looks better!
– b-fg
Nov 19 '18 at 16:01
This is a bit long-winded in terms of simply finding the sum, see me answer.
– rahlf23
Nov 19 '18 at 16:00
This is a bit long-winded in terms of simply finding the sum, see me answer.
– rahlf23
Nov 19 '18 at 16:00
Yeap, looks better!
– b-fg
Nov 19 '18 at 16:01
Yeap, looks better!
– b-fg
Nov 19 '18 at 16:01
add a comment |
Here is an example case demonstrating how to find the sum in both cases, using nested loop:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
totalSum = 0
for i in range(1,len(matrix)):
for j in range(i):
totalSum += matrix[i][j]
print("Below sum: ", totalSum)
totalSum = 0
for i in range(len(matrix)):
for j in range(i+1,len(matrix)):
totalSum += matrix[i][j]
print("Above sum: ", totalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Below sum: 18
Above sum: 18
You can also use these one-liners:
Below diagonal:
totalSum = sum([matrix[i][j] for i in range(1,len(matrix)) for j in range(i)])
Above diagonal:
totalSum = sum([matrix[i][j] for i in range(len(matrix)) for j in range(i+1,len(matrix))])
If you want to sum all numbers above and below the main diagonal you can perform an index check:
totalSum = 0
for i in range(len(matrix)):
for j in range(len(matrix)):
if not i==j:
totalSum += matrix[i][j]
print("Sum: ", totalSum)
Yet, another way of finding that sum (although not recommended) is to find the total sum of the matrix and the sum of the main diagonal and then perform a subtraction to find the final sum:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
matrixSum = sum([sum(elem for elem in row) for row in matrix])
diagonalSum = sum([matrix[i][i] for i in range(len(matrix))])
finalSum = matrixSum - diagonalSum
print("Matrix sum: ", matrixSum)
print("Diagonal sum: ", diagonalSum)
print("Final sum: ", finalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Matrix sum: 48
Diagonal sum: 12
Final sum: 36
Note: Mind the syntax in the print statements as you're using Python 2 and my answer is in Python 3.
answered Nov 19 '18 at 15:50
Vasilis G.
3,3702722
add a comment |
Here is an example case demonstrating how to find the sum in both cases, using nested loop:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
totalSum = 0
for i in range(1,len(matrix)):
for j in range(i):
totalSum += matrix[i][j]
print("Below sum: ", totalSum)
totalSum = 0
for i in range(len(matrix)):
for j in range(i+1,len(matrix)):
totalSum += matrix[i][j]
print("Above sum: ", totalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Below sum: 18
Above sum: 18
You can also use these one-liners:
Below diagonal:
totalSum = sum([matrix[i][j] for i in range(1,len(matrix)) for j in range(i)])
Above diagonal:
totalSum = sum([matrix[i][j] for i in range(len(matrix)) for j in range(i+1,len(matrix))])
If you want to sum all numbers above and below the main diagonal you can perform an index check:
totalSum = 0
for i in range(len(matrix)):
for j in range(len(matrix)):
if not i==j:
totalSum += matrix[i][j]
print("Sum: ", totalSum)
Yet, another way of finding that sum (although not recommended) is to find the total sum of the matrix and the sum of the main diagonal and then perform a subtraction to find the final sum:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
matrixSum = sum([sum(elem for elem in row) for row in matrix])
diagonalSum = sum([matrix[i][i] for i in range(len(matrix))])
finalSum = matrixSum - diagonalSum
print("Matrix sum: ", matrixSum)
print("Diagonal sum: ", diagonalSum)
print("Final sum: ", finalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Matrix sum: 48
Diagonal sum: 12
Final sum: 36
Note: Mind the syntax in the print statements as you're using Python 2 and my answer is in Python 3.
answered Nov 19 '18 at 15:50
Vasilis G.
3,3702722
add a comment |
Here is an example case demonstrating how to find the sum in both cases, using nested loop:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
totalSum = 0
for i in range(1,len(matrix)):
for j in range(i):
totalSum += matrix[i][j]
print("Below sum: ", totalSum)
totalSum = 0
for i in range(len(matrix)):
for j in range(i+1,len(matrix)):
totalSum += matrix[i][j]
print("Above sum: ", totalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Below sum: 18
Above sum: 18
You can also use these one-liners:
Below diagonal:
totalSum = sum([matrix[i][j] for i in range(1,len(matrix)) for j in range(i)])
Above diagonal:
totalSum = sum([matrix[i][j] for i in range(len(matrix)) for j in range(i+1,len(matrix))])
If you want to sum all numbers above and below the main diagonal you can perform an index check:
totalSum = 0
for i in range(len(matrix)):
for j in range(len(matrix)):
if not i==j:
totalSum += matrix[i][j]
print("Sum: ", totalSum)
Yet, another way of finding that sum (although not recommended) is to find the total sum of the matrix and the sum of the main diagonal and then perform a subtraction to find the final sum:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
matrixSum = sum([sum(elem for elem in row) for row in matrix])
diagonalSum = sum([matrix[i][i] for i in range(len(matrix))])
finalSum = matrixSum - diagonalSum
print("Matrix sum: ", matrixSum)
print("Diagonal sum: ", diagonalSum)
print("Final sum: ", finalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Matrix sum: 48
Diagonal sum: 12
Final sum: 36
Note: Mind the syntax in the print statements as you're using Python 2 and my answer is in Python 3.
answered Nov 19 '18 at 15:50
Vasilis G.
3,3702722
Here is an example case demonstrating how to find the sum in both cases, using nested loop:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
totalSum = 0
for i in range(1,len(matrix)):
for j in range(i):
totalSum += matrix[i][j]
print("Below sum: ", totalSum)
totalSum = 0
for i in range(len(matrix)):
for j in range(i+1,len(matrix)):
totalSum += matrix[i][j]
print("Above sum: ", totalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Below sum: 18
Above sum: 18
You can also use these one-liners:
Below diagonal:
totalSum = sum([matrix[i][j] for i in range(1,len(matrix)) for j in range(i)])
Above diagonal:
totalSum = sum([matrix[i][j] for i in range(len(matrix)) for j in range(i+1,len(matrix))])
If you want to sum all numbers above and below the main diagonal you can perform an index check:
totalSum = 0
for i in range(len(matrix)):
for j in range(len(matrix)):
if not i==j:
totalSum += matrix[i][j]
print("Sum: ", totalSum)
Yet, another way of finding that sum (although not recommended) is to find the total sum of the matrix and the sum of the main diagonal and then perform a subtraction to find the final sum:
matrix = [[i+j for j in range(4)] for i in range(4)]
for row in matrix:
print(" ".join(list(map(str,row))))
matrixSum = sum([sum(elem for elem in row) for row in matrix])
diagonalSum = sum([matrix[i][i] for i in range(len(matrix))])
finalSum = matrixSum - diagonalSum
print("Matrix sum: ", matrixSum)
print("Diagonal sum: ", diagonalSum)
print("Final sum: ", finalSum)
Output:
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
Matrix sum: 48
Diagonal sum: 12
Final sum: 36
Note: Mind the syntax in the print statements as you're using Python 2 and my answer is in Python 3.
answered Nov 19 '18 at 15:50
Vasilis G.
3,3702722
edited Nov 19 '18 at 16:05
answered Nov 19 '18 at 15:50
Vasilis G.
3,3702722
answered Nov 19 '18 at 15:50
Vasilis G.
3,3702722
answered Nov 19 '18 at 15:50
Vasilis G.
3,3702722
3,3702722
add a comment |
add a comment |
Here is a fast method using scipy.spatial.distance.squareform for the triangles and np.einsum for the diagonal:
>>> import numpy as np
>>> from scipy.spatial.distance import squareform
>>>
>>> x = np.arange(36).reshape(6, 6)
>>>
>>> sum_ut = squareform(x, checks=False).sum()
>>> sum_dg = np.einsum('ii', x)
>>> sum_lt = squareform(x.T, checks=False).sum()
Timings:
>>> timeit(lambda: squareform(x, checks=False).sum())
6.272806407185271
>>> timeit(lambda: np.einsum('ii', x))
1.3961836302187294
>>> timeit(lambda: squareform(x.T, checks=False).sum())
6.6827554509509355
For comparison:
>>> timeit(lambda: np.triu(x, 1).sum())
13.85556498519145
>>> timeit(lambda: np.trace(x))
3.081781509099528
>>> timeit(lambda: np.tril(x, -1).sum())
13.659938262077048
answered Nov 19 '18 at 19:38
Paul Panzer
29.9k21240
add a comment |
Here is a fast method using scipy.spatial.distance.squareform for the triangles and np.einsum for the diagonal:
>>> import numpy as np
>>> from scipy.spatial.distance import squareform
>>>
>>> x = np.arange(36).reshape(6, 6)
>>>
>>> sum_ut = squareform(x, checks=False).sum()
>>> sum_dg = np.einsum('ii', x)
>>> sum_lt = squareform(x.T, checks=False).sum()
Timings:
>>> timeit(lambda: squareform(x, checks=False).sum())
6.272806407185271
>>> timeit(lambda: np.einsum('ii', x))
1.3961836302187294
>>> timeit(lambda: squareform(x.T, checks=False).sum())
6.6827554509509355
For comparison:
>>> timeit(lambda: np.triu(x, 1).sum())
13.85556498519145
>>> timeit(lambda: np.trace(x))
3.081781509099528
>>> timeit(lambda: np.tril(x, -1).sum())
13.659938262077048
answered Nov 19 '18 at 19:38
Paul Panzer
29.9k21240
add a comment |
Here is a fast method using scipy.spatial.distance.squareform for the triangles and np.einsum for the diagonal:
>>> import numpy as np
>>> from scipy.spatial.distance import squareform
>>>
>>> x = np.arange(36).reshape(6, 6)
>>>
>>> sum_ut = squareform(x, checks=False).sum()
>>> sum_dg = np.einsum('ii', x)
>>> sum_lt = squareform(x.T, checks=False).sum()
Timings:
>>> timeit(lambda: squareform(x, checks=False).sum())
6.272806407185271
>>> timeit(lambda: np.einsum('ii', x))
1.3961836302187294
>>> timeit(lambda: squareform(x.T, checks=False).sum())
6.6827554509509355
For comparison:
>>> timeit(lambda: np.triu(x, 1).sum())
13.85556498519145
>>> timeit(lambda: np.trace(x))
3.081781509099528
>>> timeit(lambda: np.tril(x, -1).sum())
13.659938262077048
answered Nov 19 '18 at 19:38
Paul Panzer
29.9k21240
Here is a fast method using scipy.spatial.distance.squareform for the triangles and np.einsum for the diagonal:
>>> import numpy as np
>>> from scipy.spatial.distance import squareform
>>>
>>> x = np.arange(36).reshape(6, 6)
>>>
>>> sum_ut = squareform(x, checks=False).sum()
>>> sum_dg = np.einsum('ii', x)
>>> sum_lt = squareform(x.T, checks=False).sum()
Timings:
>>> timeit(lambda: squareform(x, checks=False).sum())
6.272806407185271
>>> timeit(lambda: np.einsum('ii', x))
1.3961836302187294
>>> timeit(lambda: squareform(x.T, checks=False).sum())
6.6827554509509355
For comparison:
>>> timeit(lambda: np.triu(x, 1).sum())
13.85556498519145
>>> timeit(lambda: np.trace(x))
3.081781509099528
>>> timeit(lambda: np.tril(x, -1).sum())
13.659938262077048
answered Nov 19 '18 at 19:38
Paul Panzer
29.9k21240
answered Nov 19 '18 at 19:38
Paul Panzer
29.9k21240
answered Nov 19 '18 at 19:38
Paul Panzer
29.9k21240
answered Nov 19 '18 at 19:38
Paul Panzer
29.9k21240
29.9k21240
add a comment |
add a comment |
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Possible duplicate of Python - How to get the sum of the upper triangle of a matrix without using numpy?
– Austin
Nov 19 '18 at 15:47