Numpy: Efficient way to convert indices of a square matrix to its upper triangular indices

up vote
6
down vote

favorite

Question: given a tuple of index, return its order in upper triangular indices. Here is an example:

Suppose we have a square matrix A of shape (3, 3).

A has 6 upper triangular indices, namely, (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).

Now I know an element at index (1, 2), which is a index belongs to the upper triangular part of A. I would like to return 4 (which means it is the 5th element in all upper triangular indices.)

Any ideas on how to do that in general?

Best,
Zhihao

edited 2 hours ago

RafaelC

25k82547

asked 2 hours ago

Zhihao Cui

554

More general, if I have a list of indices, do we have a function to convert them to triu indices together [return a list/array of converted results]?
– Zhihao Cui
2 hours ago

add a comment |

up vote
6
down vote

favorite

Question: given a tuple of index, return its order in upper triangular indices. Here is an example:

Suppose we have a square matrix A of shape (3, 3).

A has 6 upper triangular indices, namely, (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).

Now I know an element at index (1, 2), which is a index belongs to the upper triangular part of A. I would like to return 4 (which means it is the 5th element in all upper triangular indices.)

Any ideas on how to do that in general?

Best,
Zhihao

edited 2 hours ago

RafaelC

25k82547

asked 2 hours ago

Zhihao Cui

554

More general, if I have a list of indices, do we have a function to convert them to triu indices together [return a list/array of converted results]?
– Zhihao Cui
2 hours ago

add a comment |

up vote
6
down vote

favorite

Question: given a tuple of index, return its order in upper triangular indices. Here is an example:

Suppose we have a square matrix A of shape (3, 3).

A has 6 upper triangular indices, namely, (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).

Now I know an element at index (1, 2), which is a index belongs to the upper triangular part of A. I would like to return 4 (which means it is the 5th element in all upper triangular indices.)

Any ideas on how to do that in general?

Best,
Zhihao

edited 2 hours ago

RafaelC

25k82547

asked 2 hours ago

Zhihao Cui

554

Question: given a tuple of index, return its order in upper triangular indices. Here is an example:

Suppose we have a square matrix A of shape (3, 3).

A has 6 upper triangular indices, namely, (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).

Now I know an element at index (1, 2), which is a index belongs to the upper triangular part of A. I would like to return 4 (which means it is the 5th element in all upper triangular indices.)

Any ideas on how to do that in general?

Best,
Zhihao

python numpy matrix

edited 2 hours ago

RafaelC

25k82547

asked 2 hours ago

Zhihao Cui

554

edited 2 hours ago

RafaelC

25k82547

asked 2 hours ago

Zhihao Cui

554

edited 2 hours ago

RafaelC

25k82547

edited 2 hours ago

RafaelC

25k82547

edited 2 hours ago

RafaelC

25k82547

asked 2 hours ago

Zhihao Cui

554

asked 2 hours ago

Zhihao Cui

554

asked 2 hours ago

Zhihao Cui

554

More general, if I have a list of indices, do we have a function to convert them to triu indices together [return a list/array of converted results]?
– Zhihao Cui
2 hours ago

add a comment |

More general, if I have a list of indices, do we have a function to convert them to triu indices together [return a list/array of converted results]?
– Zhihao Cui
2 hours ago

More general, if I have a list of indices, do we have a function to convert them to triu indices together [return a list/array of converted results]?
– Zhihao Cui
2 hours ago

add a comment |

6 Answers
6

active

oldest

votes

up vote
2
down vote

accepted

One can write down the explicit formula:

def utr_idx(N, i, j):
 return (2*N+1-i)*i//2 + j-i

Demo:

>>> N = 127
>>> X = np.transpose(np.triu_indices(N))
>>> utr_idx(N, *X[2123])
2123

answered 1 hour ago

Paul Panzer

28.3k21138

I just realized it lol
– Zhihao Cui
1 hour ago

add a comment |

up vote
2
down vote

IIUC, you can get the indexes using itertools combinations with replacement

>>> ind = tuple(itertools.combinations_with_replacement(range(3),2))
((0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2))

To retrieve the index, just use index method

>>> ind.index((1,2))
4

answered 1 hour ago

RafaelC

25k82547

add a comment |

up vote
2
down vote

You could use np.triu_indices and a dictionary:

import numpy as np

iu1 = np.triu_indices(3)
table = (i, j): c for c, (i, j) in enumerate(zip(*iu1))
print(table[(1, 2)])

Output

answered 1 hour ago

Daniel Mesejo

7,0411821

add a comment |

up vote
2
down vote

For an n×n matrix, the (i, j)-th item of the upper triangle is the i×(2×n-i+1)/2+j-i-th element of the matrix.

We can also do the math in reverse and obtain the (i, j) element for the k-th element with:

i = ⌊(-√((2n+1)²-8k)+2n+1)/2⌋ and j = k+i-i×(2×n-i+1)/2

So for example:

from math import floor, sqrt

def coor_to_idx(n, i, j):
 return i*(2*n-i+1)//2+j-i

def idx_to_coor(n, k):
 i = floor((-sqrt((2*n+1)*(2*n+1)-8*k)+2*n+1)/2)
 j = k + i - i*(2*n-i+1)//2
 return i, j

For example:

>>> [idx_to_coor(4, i) for i in range(10)]
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)]
>>> [coor_to_idx(4, i, j) for i in range(4) for j in range(i, 4)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Given the numbers are not huge (well if these are huge, calculations are no longer done in constant time), we can thus calculate the k-th coordinate in O(1), for example:

>>> idx_to_coor(1234567, 123456789)
(100, 5139)

which is equivalent to obtaining it through enumeration:

>>> next(islice(((i, j) for i in range(1234567) for j in range(i, 1234567)), 123456789, None))
(100, 5139)

Here converting indices to a coordinate can also have, for large numbers, some rounding errors due to floating point imprecision.

edited 1 hour ago

answered 1 hour ago

Willem Van Onsem

138k16129219

add a comment |

up vote
1
down vote

Similar to @DanielMesejo, you can use np.triu_indices with either argwhere or nonzero:

my_index = (1,2)

>>> np.nonzero((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
(array([4]),)
>>> np.argwhere((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
array([[4]])

Explanation:

np.stack(np.triu_indices(3), axis=1) gives you the indices of your upper triangle in order:

array([[0, 0],
 [0, 1],
 [0, 2],
 [1, 1],
 [1, 2],
 [2, 2]])

So all you have to do is find where it matches [1,2] (which you can do with the == operator and all)

answered 1 hour ago

sacul

25.9k41638

add a comment |

up vote
1
down vote

Constructing upper indices would be costly. We can directly get the corresponding index like so -

def triu_index(N, x, y):
 # Get index corresponding to (x,y) in upper triangular list
 idx = np.r_[0,np.arange(N,1,-1).cumsum()]
 return idx[x]+y-x

Sample run -

In [271]: triu_index(N=3, x=1, y=2)
Out[271]: 4

answered 1 hour ago

Divakar

151k1475166

add a comment |

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53233695%2fnumpy-efficient-way-to-convert-indices-of-a-square-matrix-to-its-upper-triangul%23new-answer', 'question_page');

);

Post as a guest

Name

6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

up vote
2
down vote

accepted

One can write down the explicit formula:

def utr_idx(N, i, j):
 return (2*N+1-i)*i//2 + j-i

Demo:

>>> N = 127
>>> X = np.transpose(np.triu_indices(N))
>>> utr_idx(N, *X[2123])
2123

answered 1 hour ago

Paul Panzer

28.3k21138

I just realized it lol
– Zhihao Cui
1 hour ago

add a comment |

up vote
2
down vote

accepted

One can write down the explicit formula:

def utr_idx(N, i, j):
 return (2*N+1-i)*i//2 + j-i

Demo:

>>> N = 127
>>> X = np.transpose(np.triu_indices(N))
>>> utr_idx(N, *X[2123])
2123

answered 1 hour ago

Paul Panzer

28.3k21138

I just realized it lol
– Zhihao Cui
1 hour ago

add a comment |

up vote
2
down vote

accepted

One can write down the explicit formula:

def utr_idx(N, i, j):
 return (2*N+1-i)*i//2 + j-i

Demo:

>>> N = 127
>>> X = np.transpose(np.triu_indices(N))
>>> utr_idx(N, *X[2123])
2123

answered 1 hour ago

Paul Panzer

28.3k21138

One can write down the explicit formula:

def utr_idx(N, i, j):
 return (2*N+1-i)*i//2 + j-i

Demo:

>>> N = 127
>>> X = np.transpose(np.triu_indices(N))
>>> utr_idx(N, *X[2123])
2123

answered 1 hour ago

Paul Panzer

28.3k21138

answered 1 hour ago

Paul Panzer

28.3k21138

answered 1 hour ago

Paul Panzer

28.3k21138

answered 1 hour ago

Paul Panzer

28.3k21138

I just realized it lol
– Zhihao Cui
1 hour ago

add a comment |

I just realized it lol
– Zhihao Cui
1 hour ago

I just realized it lol
– Zhihao Cui
1 hour ago

add a comment |

up vote
2
down vote

IIUC, you can get the indexes using itertools combinations with replacement

>>> ind = tuple(itertools.combinations_with_replacement(range(3),2))
((0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2))

To retrieve the index, just use index method

>>> ind.index((1,2))
4

answered 1 hour ago

RafaelC

25k82547

add a comment |

up vote
2
down vote

IIUC, you can get the indexes using itertools combinations with replacement

>>> ind = tuple(itertools.combinations_with_replacement(range(3),2))
((0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2))

To retrieve the index, just use index method

>>> ind.index((1,2))
4

answered 1 hour ago

RafaelC

25k82547

add a comment |

up vote
2
down vote

IIUC, you can get the indexes using itertools combinations with replacement

>>> ind = tuple(itertools.combinations_with_replacement(range(3),2))
((0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2))

To retrieve the index, just use index method

>>> ind.index((1,2))
4

answered 1 hour ago

RafaelC

25k82547

IIUC, you can get the indexes using itertools combinations with replacement

>>> ind = tuple(itertools.combinations_with_replacement(range(3),2))
((0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2))

To retrieve the index, just use index method

>>> ind.index((1,2))
4

answered 1 hour ago

RafaelC

25k82547

answered 1 hour ago

RafaelC

25k82547

answered 1 hour ago

RafaelC

25k82547

answered 1 hour ago

RafaelC

25k82547

add a comment |

up vote
2
down vote

You could use np.triu_indices and a dictionary:

import numpy as np

iu1 = np.triu_indices(3)
table = (i, j): c for c, (i, j) in enumerate(zip(*iu1))
print(table[(1, 2)])

Output

answered 1 hour ago

Daniel Mesejo

7,0411821

add a comment |

up vote
2
down vote

You could use np.triu_indices and a dictionary:

import numpy as np

iu1 = np.triu_indices(3)
table = (i, j): c for c, (i, j) in enumerate(zip(*iu1))
print(table[(1, 2)])

Output

answered 1 hour ago

Daniel Mesejo

7,0411821

add a comment |

up vote
2
down vote

You could use np.triu_indices and a dictionary:

import numpy as np

iu1 = np.triu_indices(3)
table = (i, j): c for c, (i, j) in enumerate(zip(*iu1))
print(table[(1, 2)])

Output

answered 1 hour ago

Daniel Mesejo

7,0411821

You could use np.triu_indices and a dictionary:

import numpy as np

iu1 = np.triu_indices(3)
table = (i, j): c for c, (i, j) in enumerate(zip(*iu1))
print(table[(1, 2)])

Output

answered 1 hour ago

Daniel Mesejo

7,0411821

answered 1 hour ago

Daniel Mesejo

7,0411821

answered 1 hour ago

Daniel Mesejo

7,0411821

answered 1 hour ago

Daniel Mesejo

7,0411821

add a comment |

up vote
2
down vote

For an n×n matrix, the (i, j)-th item of the upper triangle is the i×(2×n-i+1)/2+j-i-th element of the matrix.

We can also do the math in reverse and obtain the (i, j) element for the k-th element with:

i = ⌊(-√((2n+1)²-8k)+2n+1)/2⌋ and j = k+i-i×(2×n-i+1)/2

So for example:

from math import floor, sqrt

def coor_to_idx(n, i, j):
 return i*(2*n-i+1)//2+j-i

def idx_to_coor(n, k):
 i = floor((-sqrt((2*n+1)*(2*n+1)-8*k)+2*n+1)/2)
 j = k + i - i*(2*n-i+1)//2
 return i, j

For example:

>>> [idx_to_coor(4, i) for i in range(10)]
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)]
>>> [coor_to_idx(4, i, j) for i in range(4) for j in range(i, 4)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Given the numbers are not huge (well if these are huge, calculations are no longer done in constant time), we can thus calculate the k-th coordinate in O(1), for example:

>>> idx_to_coor(1234567, 123456789)
(100, 5139)

which is equivalent to obtaining it through enumeration:

>>> next(islice(((i, j) for i in range(1234567) for j in range(i, 1234567)), 123456789, None))
(100, 5139)

Here converting indices to a coordinate can also have, for large numbers, some rounding errors due to floating point imprecision.

edited 1 hour ago

answered 1 hour ago

Willem Van Onsem

138k16129219

add a comment |

up vote
2
down vote

For an n×n matrix, the (i, j)-th item of the upper triangle is the i×(2×n-i+1)/2+j-i-th element of the matrix.

We can also do the math in reverse and obtain the (i, j) element for the k-th element with:

i = ⌊(-√((2n+1)²-8k)+2n+1)/2⌋ and j = k+i-i×(2×n-i+1)/2

So for example:

from math import floor, sqrt

def coor_to_idx(n, i, j):
 return i*(2*n-i+1)//2+j-i

def idx_to_coor(n, k):
 i = floor((-sqrt((2*n+1)*(2*n+1)-8*k)+2*n+1)/2)
 j = k + i - i*(2*n-i+1)//2
 return i, j

For example:

>>> [idx_to_coor(4, i) for i in range(10)]
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)]
>>> [coor_to_idx(4, i, j) for i in range(4) for j in range(i, 4)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Given the numbers are not huge (well if these are huge, calculations are no longer done in constant time), we can thus calculate the k-th coordinate in O(1), for example:

>>> idx_to_coor(1234567, 123456789)
(100, 5139)

which is equivalent to obtaining it through enumeration:

>>> next(islice(((i, j) for i in range(1234567) for j in range(i, 1234567)), 123456789, None))
(100, 5139)

Here converting indices to a coordinate can also have, for large numbers, some rounding errors due to floating point imprecision.

edited 1 hour ago

answered 1 hour ago

Willem Van Onsem

138k16129219

add a comment |

up vote
2
down vote

For an n×n matrix, the (i, j)-th item of the upper triangle is the i×(2×n-i+1)/2+j-i-th element of the matrix.

We can also do the math in reverse and obtain the (i, j) element for the k-th element with:

i = ⌊(-√((2n+1)²-8k)+2n+1)/2⌋ and j = k+i-i×(2×n-i+1)/2

So for example:

from math import floor, sqrt

def coor_to_idx(n, i, j):
 return i*(2*n-i+1)//2+j-i

def idx_to_coor(n, k):
 i = floor((-sqrt((2*n+1)*(2*n+1)-8*k)+2*n+1)/2)
 j = k + i - i*(2*n-i+1)//2
 return i, j

For example:

>>> [idx_to_coor(4, i) for i in range(10)]
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)]
>>> [coor_to_idx(4, i, j) for i in range(4) for j in range(i, 4)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Given the numbers are not huge (well if these are huge, calculations are no longer done in constant time), we can thus calculate the k-th coordinate in O(1), for example:

>>> idx_to_coor(1234567, 123456789)
(100, 5139)

which is equivalent to obtaining it through enumeration:

>>> next(islice(((i, j) for i in range(1234567) for j in range(i, 1234567)), 123456789, None))
(100, 5139)

Here converting indices to a coordinate can also have, for large numbers, some rounding errors due to floating point imprecision.

edited 1 hour ago

answered 1 hour ago

Willem Van Onsem

138k16129219

For an n×n matrix, the (i, j)-th item of the upper triangle is the i×(2×n-i+1)/2+j-i-th element of the matrix.

We can also do the math in reverse and obtain the (i, j) element for the k-th element with:

i = ⌊(-√((2n+1)²-8k)+2n+1)/2⌋ and j = k+i-i×(2×n-i+1)/2

So for example:

from math import floor, sqrt

def coor_to_idx(n, i, j):
 return i*(2*n-i+1)//2+j-i

def idx_to_coor(n, k):
 i = floor((-sqrt((2*n+1)*(2*n+1)-8*k)+2*n+1)/2)
 j = k + i - i*(2*n-i+1)//2
 return i, j

For example:

>>> [idx_to_coor(4, i) for i in range(10)]
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)]
>>> [coor_to_idx(4, i, j) for i in range(4) for j in range(i, 4)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Given the numbers are not huge (well if these are huge, calculations are no longer done in constant time), we can thus calculate the k-th coordinate in O(1), for example:

>>> idx_to_coor(1234567, 123456789)
(100, 5139)

which is equivalent to obtaining it through enumeration:

>>> next(islice(((i, j) for i in range(1234567) for j in range(i, 1234567)), 123456789, None))
(100, 5139)

Here converting indices to a coordinate can also have, for large numbers, some rounding errors due to floating point imprecision.

edited 1 hour ago

answered 1 hour ago

Willem Van Onsem

138k16129219

edited 1 hour ago

answered 1 hour ago

Willem Van Onsem

138k16129219

answered 1 hour ago

Willem Van Onsem

138k16129219

answered 1 hour ago

Willem Van Onsem

138k16129219

add a comment |

up vote
1
down vote

Similar to @DanielMesejo, you can use np.triu_indices with either argwhere or nonzero:

my_index = (1,2)

>>> np.nonzero((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
(array([4]),)
>>> np.argwhere((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
array([[4]])

Explanation:

np.stack(np.triu_indices(3), axis=1) gives you the indices of your upper triangle in order:

array([[0, 0],
 [0, 1],
 [0, 2],
 [1, 1],
 [1, 2],
 [2, 2]])

So all you have to do is find where it matches [1,2] (which you can do with the == operator and all)

answered 1 hour ago

sacul

25.9k41638

add a comment |

up vote
1
down vote

Similar to @DanielMesejo, you can use np.triu_indices with either argwhere or nonzero:

my_index = (1,2)

>>> np.nonzero((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
(array([4]),)
>>> np.argwhere((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
array([[4]])

Explanation:

np.stack(np.triu_indices(3), axis=1) gives you the indices of your upper triangle in order:

array([[0, 0],
 [0, 1],
 [0, 2],
 [1, 1],
 [1, 2],
 [2, 2]])

So all you have to do is find where it matches [1,2] (which you can do with the == operator and all)

answered 1 hour ago

sacul

25.9k41638

add a comment |

up vote
1
down vote

Similar to @DanielMesejo, you can use np.triu_indices with either argwhere or nonzero:

my_index = (1,2)

>>> np.nonzero((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
(array([4]),)
>>> np.argwhere((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
array([[4]])

Explanation:

np.stack(np.triu_indices(3), axis=1) gives you the indices of your upper triangle in order:

array([[0, 0],
 [0, 1],
 [0, 2],
 [1, 1],
 [1, 2],
 [2, 2]])

So all you have to do is find where it matches [1,2] (which you can do with the == operator and all)

answered 1 hour ago

sacul

25.9k41638

Similar to @DanielMesejo, you can use np.triu_indices with either argwhere or nonzero:

my_index = (1,2)

>>> np.nonzero((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
(array([4]),)
>>> np.argwhere((np.stack(np.triu_indices(3), axis=1) == my_index).all(1))
array([[4]])

Explanation:

np.stack(np.triu_indices(3), axis=1) gives you the indices of your upper triangle in order:

array([[0, 0],
 [0, 1],
 [0, 2],
 [1, 1],
 [1, 2],
 [2, 2]])

So all you have to do is find where it matches [1,2] (which you can do with the == operator and all)

answered 1 hour ago

sacul

25.9k41638

answered 1 hour ago

sacul

25.9k41638

answered 1 hour ago

sacul

25.9k41638

answered 1 hour ago

sacul

25.9k41638

add a comment |

up vote
1
down vote

Constructing upper indices would be costly. We can directly get the corresponding index like so -

def triu_index(N, x, y):
 # Get index corresponding to (x,y) in upper triangular list
 idx = np.r_[0,np.arange(N,1,-1).cumsum()]
 return idx[x]+y-x

Sample run -

In [271]: triu_index(N=3, x=1, y=2)
Out[271]: 4

answered 1 hour ago

Divakar

151k1475166

add a comment |

up vote
1
down vote

Constructing upper indices would be costly. We can directly get the corresponding index like so -

def triu_index(N, x, y):
 # Get index corresponding to (x,y) in upper triangular list
 idx = np.r_[0,np.arange(N,1,-1).cumsum()]
 return idx[x]+y-x

Sample run -

In [271]: triu_index(N=3, x=1, y=2)
Out[271]: 4

answered 1 hour ago

Divakar

151k1475166

add a comment |

up vote
1
down vote

Constructing upper indices would be costly. We can directly get the corresponding index like so -

def triu_index(N, x, y):
 # Get index corresponding to (x,y) in upper triangular list
 idx = np.r_[0,np.arange(N,1,-1).cumsum()]
 return idx[x]+y-x

Sample run -

In [271]: triu_index(N=3, x=1, y=2)
Out[271]: 4

answered 1 hour ago

Divakar

151k1475166

Constructing upper indices would be costly. We can directly get the corresponding index like so -

def triu_index(N, x, y):
 # Get index corresponding to (x,y) in upper triangular list
 idx = np.r_[0,np.arange(N,1,-1).cumsum()]
 return idx[x]+y-x

Sample run -

In [271]: triu_index(N=3, x=1, y=2)
Out[271]: 4

answered 1 hour ago

Divakar

151k1475166

answered 1 hour ago

Divakar

151k1475166

answered 1 hour ago

Divakar

151k1475166

answered 1 hour ago

Divakar

151k1475166

add a comment |

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

Search This Blog

Iyfjky