Yo,
> m.row() returns a new (copy) of the row of the matrix, so making that
> immutable would be inconsistent with how copy works for matrices.
>
> sage: m = matrix.ones(10)
> sage: m.set_immutable()
> sage: copy(m).is_immutable()
> False
I do not see why matrix.row(0) should return a *copy* of the row when
the matrix is immutable.
Depending on how matrices are implemented matrix.row(0) could very
well be a O(1) operation giving me a pointer toward an internal data
structure. Why not after all, since I can't modify what I have?
Plus there is nothing of 'copy' in the row(i) semantic. Ideally, here,
I would get a O(1) pointer toward internal data. As efficient as it
gets.
> At least you can do "v = m.row(0); v.set_immutable()".
I take the rows of a matrix, and multiply each of them by a set of values.
At first I wrote:
values = [...]
M.set_immutable()
M = [set([r*x for x in values]) for r in M]
But that does not work, because r is not immutable. Then I tried:
values = [...]
M = M.rows()
for r in M:
r.set_immutable()
M = [set([r*x for x in values]) for r in M]
But that does not work because multiplying an immutable row by
anything is not immutable anymore. So now I write
values = [...]
M = [[r*x for x in values] for r in M]
for r in M:
for x in r:
x.set_immutable()
M = map(set,M)
Easy, isn't it?
Nathann
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