dmitrey kirjoitti:
What's wrong?
start python shell;
from numpy import sin = all ok
from numpy import repmat =
Traceback (most recent call last):
File stdin, line 1, in module
ImportError: cannot import name repmat
In numpy, the equivalent function is called tile:
from numpy
if it was excluded for any reasons, corresponding changes in
http://www.scipy.org/NumPy_for_Matlab_Users should be done
D.
lorenzo bolla wrote:
it looks like repmat is not there anymore... why?
use numpy.repeat and numpy.tile, instead!
hth,
lorenzo.
On 4/30/07, *dmitrey* [EMAIL PROTECTED]
It's not on the matlab page simpy because numpy.tile didn't exist when
the page was created. It should be fixed. But repmat is still there
in numpy.matlib (I think that was what it was called.)
--bb
On 4/30/07, dmitrey [EMAIL PROTECTED] wrote:
if it was excluded for any reasons, corresponding
lorenzo bolla wrote:
me!
I have two cases.
1. I need that arctan2(1+0.0001j,1-0.01j) gives something
close to arctan2(1,1): any decent analytic prolungation will do!
This is the foreseeable use case described by Anne.
In any event, I stand not only corrected, but
On 4/30/07, David Goldsmith [EMAIL PROTECTED] wrote:
(hint what is arctan(0+1j)?)
Well, at the risk of embarrassing myself, using arctan(x+iy) = I get:
arctan(0+1i) = -i*log((0+i*1)/sqrt(0^2 + 1^2)) = -i*log(i/1) = -i*log(i)
= -i*log(exp(i*pi/2)) = -i*i*pi/2 = pi/2...
Is there some reason
hold on, david. the formula I posted previously from wolfram is ArcTan[x,y]
with x or y complex: its the same of arctan2(x,y). arctan is another
function (even though arctan2(y,x) should be a better arctan(y/x)).
the correct formula for y = arctan(x), with any x (real or complex), should
be (if
lorenzo bolla wrote:
hold on, david. the formula I posted previously from wolfram is
ArcTan[x,y] with x or y complex: its the same of arctan2(x,y). arctan
is another function (even though arctan2(y,x) should be a better
arctan(y/x)).
the correct formula for y = arctan(x), with any x (real
Timothy Hochberg wrote:
On 4/30/07, *David Goldsmith* [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED] wrote:
(hint what is arctan(0+1j)?)
Well, at the risk of embarrassing myself, using arctan(x+iy) = I get:
arctan(0+1i) = -i*log((0+i*1)/sqrt(0^2 + 1^2)) = -i*log(i/1) =
dmitrey wrote:
hi all,
please inform me what is the simplest way to check, does the vector x
that came to my func is float or integer. I.e. someone can pass to my
func for example x0 = numpy.array([1, 0, 0]) and it can yield wrong
unexpected results vs numpy.array([1.0, 0, 0]) .
Usually,
