How to convert triangle matrix to square in NumPy?

Question

I\'m doing some computations on a full matrix that is redundant (i.e. can be a triangle matrix without losing info). I realized I can compute only the lower portion of the

Answer 1

Since the matrix is symetric, you can do:

m = np.array([1,1,0,1,1,1,0,1,1]).reshape((3,3))

# after some computation you get x
x = np.tril(m)

m_recomposed = x + x.transpose() - np.diag(np.diag(x))

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

#In [152]: np.array_equal(m, m_recomposed)
#Out[152]: True

Answer 2

Assuming A as the input array, few methods are listed below.

Approach #1 : Using np.triu on a transposed version of A -

np.triu(A.T,1) + A

Approach #2 : Avoid np.triu with direct summation between A.T and A and then indexing to set diagonal elements -

out = A.T + A
idx = np.arange(A.shape[0])
out[idx,idx] = A[idx,idx]

Approach #3 : Same as previous one, but compact using in-builts for indexing -

out = A.T + A
np.fill_diagonal(out,np.diag(A))

Approach #4 : Same as previous one, but with boolean indexing to set diagonal elements -

out = A.T + A
mask = np.eye(out.shape[0],dtype=bool)
out[mask] = A[mask]

Approach #5 : Using mask based selection for diagonal elements with np.where -

np.where(np.eye(A.shape[0],dtype=bool),A,A.T+A)

Approach #6 : Using mask based selection for all elements with np.where -

np.where(np.triu(np.ones(A.shape[0],dtype=bool),1),A.T,A)

---

Runtime tests

Functions -

def func1(A):
    return np.triu(A.T,1) + A

def func2(A):
    out = A.T + A
    idx = np.arange(A.shape[0])
    out[idx,idx] = A[idx,idx]
    return out

def func3(A):
    out = A.T + A
    np.fill_diagonal(out,np.diag(A))
    return out

def func4(A):
    out = A.T + A
    mask = np.eye(out.shape[0],dtype=bool)
    out[mask] = A[mask]
    return out

def func5(A):
    return np.where(np.eye(A.shape[0],dtype=bool),A,A.T+A)

def func6(A):
    return np.where(np.triu(np.ones(A.shape[0],dtype=bool),1),A.T,A)

Timings -

In [140]: # Input array
     ...: N = 5000
     ...: A = np.tril(np.random.randint(0,9,(N,N)))
     ...: 

In [141]: %timeit func1(A)
     ...: %timeit func2(A)
     ...: %timeit func3(A)
     ...: %timeit func4(A)
     ...: %timeit func5(A)
     ...: %timeit func6(A)
     ...: 
1 loops, best of 3: 617 ms per loop
1 loops, best of 3: 354 ms per loop
1 loops, best of 3: 354 ms per loop
1 loops, best of 3: 395 ms per loop
1 loops, best of 3: 597 ms per loop
1 loops, best of 3: 440 ms per loop

Looks like the approaches # 2 & #3 are pretty efficient!