CODEWITHSUNDEEP
pythonperformancenumpy
cgreen
cgreen

Reputation: 361

Copy flat list of upper triangle entries to full matrix?

I have the upper triangle entries (including diagonal) of a symmetric matrix in a flat list (concatenated rows), and I want to use them to fill in the full matrix, including the lower triangle. What's the fastest way to do this?

Here is my current approach. Seems like a lot of work for such a simple operation.

import numpy as np
def utri2mat(utri,ntotal):
    iu1 = np.triu_indices(ntotal)
    ret = np.zeros([ntotal,ntotal])
    ret[iu1] = utri
    ret = ret + ret.transpose() - np.diag(ret.diagonal())
    return ret

Upvotes: 2

Views: 1767

Answers (4)

seralouk
seralouk

Reputation: 33197

Assuming that you have a vector containing the upper triangular values of a symmetric matrix (n x n) then you can re-build the full matrix as follows: 

import numpy as np

# dimension of the full matrix
n = 80

# artificial upper triangle entries n(n-1) / 2 if matrix is symmetric
entries = np.array(range((80*79) / 2))

full_matrix = np.zeros((n,n))
inds = np.triu_indices_from(full_matrix, k = 1)
full_matrix[inds] = entries
full_matrix[(inds[1], inds[0])] = entries

print(full_matrix)

Verify:

np.allclose(full_matrix, full_matrix.T)

Upvotes: 0

hpaulj
hpaulj

Reputation: 231665

Here's my nomination for a faster, and possibly better, way to make a symmetric matrix from flat values:

def make_sym(val, n):
    # uses boolean mask
    # uses the same lower tri as np.triu
    mask = ~np.tri(5,k=-1,dtype=bool)
    out = np.zeros((n,n),dtype=val.dtype)
    out[mask] = val
    out.T[mask] = val
    return out

testing:

In [939]: val=np.arange(1,16)
In [940]: make_sym(val, 5)
Out[940]: 
array([[ 1,  2,  3,  4,  5],
       [ 2,  6,  7,  8,  9],
       [ 3,  7, 10, 11, 12],
       [ 4,  8, 11, 13, 14],
       [ 5,  9, 12, 14, 15]])

Like the other answers it uses out.T[] to assign the lower triangle.

Warren's answer uses np.triu_indices, which are the where values. This type of indexing is a bit slower than boolean masking.

But as I noted the np.triu that Divakar uses does not return a boolean mask in earlier numpy versions (e.g. 1.9). This is what prompted me to dig into the issue.

In 1.10 this function was rewritten as:

mask = np.tri(*m.shape[-2:], k=k-1, dtype=bool)
return np.where(mask, np.zeros(1, m.dtype), m)

I gain a bit of speed by replacing the where with ~mask. Same result, but just cutting out an intermediate step.

Upvotes: 1

Divakar
Divakar

Reputation: 221684

Inspired by this solution, you can use boolean indexing to set elements and as such might be pretty efficient. Here's one way to implement it -

def mask_based_utri2mat(utri,ntotal):
    # Setup output array
    out = np.empty((ntotal,ntotal))

    # Create upper triang. mask
    mask = np.triu(np.ones((ntotal,ntotal),dtype=bool))

    # Set upper triang. elements with mask
    out[mask] = utri

    # Set lower triang. elements with transposed mask
    out.T[mask] = utri
    return out

Runtime tests -

In [52]: # Inputs
    ...: ntotal = 100
    ...: utri = np.random.rand(np.triu_indices(ntotal)[0].size)
    ...: 

In [53]: np.allclose(mask_based_utri2mat(utri,ntotal),utri2mat(utri,ntotal))
Out[53]: True

In [54]: %timeit utri2mat(utri,ntotal)
1000 loops, best of 3: 270 µs per loop

In [55]: %timeit mask_based_utri2mat(utri,ntotal)
10000 loops, best of 3: 127 µs per loop

In [56]: # Inputs
    ...: ntotal = 1000
    ...: utri = np.random.rand(np.triu_indices(ntotal)[0].size)
    ...: 

In [57]: np.allclose(mask_based_utri2mat(utri,ntotal),utri2mat(utri,ntotal))
Out[57]: True

In [58]: %timeit utri2mat(utri,ntotal)
10 loops, best of 3: 53.9 ms per loop

In [59]: %timeit mask_based_utri2mat(utri,ntotal)
100 loops, best of 3: 15.1 ms per loop

Upvotes: 1

Warren Weckesser
Warren Weckesser

Reputation: 114956

This version is a slight variation of yours:

import numpy as np

def utri2mat(utri):
    n = int(-1 + np.sqrt(1 + 8*len(utri))) // 2
    iu1 = np.triu_indices(n)
    ret = np.empty((n, n))
    ret[iu1] = utri
    ret.T[iu1] = utri
    return ret

I replaced

    ret = ret + ret.transpose() - np.diag(ret.diagonal())

with an assignment of utri directly to the transpose of ret:

    ret.T[iu1] = utri

I also removed the argument ntotal and instead computed what n has to be based on the length of utri.

Upvotes: 1

Related Questions