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 triangle for faster results. How can I project the lower triangle into the upper once I'm done?
In other words, how can I reverse the np.tril
method?
print DF_var.as_matrix()
# [[1 1 0 1 1 1 0 1 0 0 0]
# [1 1 1 1 1 0 1 0 1 1 1]
# [0 1 1 0 0 0 0 0 0 0 0]
# [1 1 0 1 0 0 0 0 0 0 0]
# [1 1 0 0 1 0 0 0 0 0 0]
# [1 0 0 0 0 1 1 0 0 0 0]
# [0 1 0 0 0 1 1 0 0 0 0]
# [1 0 0 0 0 0 0 1 1 0 0]
# [0 1 0 0 0 0 0 1 1 0 0]
# [0 1 0 0 0 0 0 0 0 1 0]
# [0 1 0 0 0 0 0 0 0 0 1]]
print np.tril(DF_var.as_matrix())
# [[1 0 0 0 0 0 0 0 0 0 0]
# [1 1 0 0 0 0 0 0 0 0 0]
# [0 1 1 0 0 0 0 0 0 0 0]
# [1 1 0 1 0 0 0 0 0 0 0]
# [1 1 0 0 1 0 0 0 0 0 0]
# [1 0 0 0 0 1 0 0 0 0 0]
# [0 1 0 0 0 1 1 0 0 0 0]
# [1 0 0 0 0 0 0 1 0 0 0]
# [0 1 0 0 0 0 0 1 1 0 0]
# [0 1 0 0 0 0 0 0 0 1 0]
# [0 1 0 0 0 0 0 0 0 0 1]]
How to convert it back to a full matrix?
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!
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
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