Comments (6)
Should f[k] be interpreted as a vec(f)[k] when k is an integer for the time being?
from approxfun.jl.
I think that is cool. Out of curiosity, how expensive is vec(f) ?
from approxfun.jl.
It should be as expensive as making a copy of the coefficients, so not very. (It just deinterlaces the coefficients.)
On 2 Dec 2014, at 6:15 pm, Gustavo Goretkin [email protected] wrote:
I think that is cool. Out of curiosity, how expensive is vec(f) ?
—
Reply to this email directly or view it on GitHub #52 (comment).
from approxfun.jl.
Is there a reason not to store the coefficients in separate arrays in a vector-valued fun?
from approxfun.jl.
Right now the Fun type requires that the coefficients are a Vector{Float64} or Vector{Complex{Float64}}. This is consistent with the \ operator, which always works on infinite-length vectors. Doing special storage for different Fun types would complicate things
On 2 Dec 2014, at 6:26 pm, Gustavo Goretkin [email protected] wrote:
Is there a reason not to store the coefficients in separate arrays in a vector-valued fun?
—
Reply to this email directly or view it on GitHub #52 (comment).
from approxfun.jl.
Ah, that makes sense. An operator on a vector-valued fun also has its rows interlaced in the same way then, right?
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Related Issues (20)
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- Rounding errors lead to empty domain error when using abs
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- typos HOT 1
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