I want to remove all symmetric matrices from the list of all possible 0,1 matrices of (say) 2nd order. When I run this code, not all symmetric matrices are removed. Where am I wrong?`
T=Tuples((0,1),4) W=[matrix(2,2,v) for v in T] [W.remove(s) for s in W if s.is_symmetric() is true] show(W)
Single-line solution:
W = [M for v in Tuples((0,1),4) if not (M:=Matrix(2,2,v)).is_symmetric()]
It's slightly faster in this case to not construct the matrix unless necessary: W = [Matrix(2,2,v) for v in Tuples((0,1),4) if v[1] != v[2]]. This approach is of course more awkward when dealing with larger matrices.
And this is slightly faster: [Matrix(2,2,v) for v in [v for v in Tuples((0,1),4) if v[1] != v[2]]]: construct the appropriate list of tuples first and then convert to matrices. Maybe a bad idea if the list of tuples is going to be long. (I don't know why this should be any faster.)
@John: Then one can simply do [Matrix(2,2,v[:2]+[1-v[1],v[2]]) for v in Tuples((0,1),3)].
In your code W defines the range of the loop but changes inside.
It can be improved as follows:
T=Tuples((0,1),4)
W=[matrix(2,2,v) for v in T]
V=W.copy()
[V.remove(s) for s in W if s.is_symmetric()]
V
[
[0 1] [1 1] [0 0] [1 0] [0 1] [1 1] [0 0] [1 0]
[0 0], [0 0], [1 0], [1 0], [0 1], [0 1], [1 1], [1 1]
]
but it is better not to construct to many matrices:
import itertools
def f():
f1=lambda x:matrix(2,2,x)
f2=lambda v:v[1]!=v[2]
T=(item for item in itertools.product([0,1], repeat=4))
return list(map(f1,filter(f2,T)))
f()
[
[0 0] [0 0] [0 1] [0 1] [1 0] [1 0] [1 1] [1 1]
[1 0], [1 1], [0 0], [0 1], [1 0], [1 1], [0 0], [0 1]
]
timeit('f()')
625 loops, best of 3: 40.7 μs per loop
In intel python, replacing matrix by numpy.array one can obtain two times faster code
Why not use list comprehension right away?
[m for m in mm if not m.is_symmetric()]
Asked: 2023-03-11 13:11:21 +0100
Seen: 336 times
Last updated: Mar 12 '23
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