Andriy Basilisk wrote:
> Hello all,
>
> My challenge is this:
> I'm working on an application that parses numerical data from a text
> report using regular expressions, and then places the results in Numpy
> matrices for processing. The data contains integers, floats, and
> boolean values. The b
Hello all,
My challenge is this:
I'm working on an application that parses numerical data from a text
report using regular expressions, and then places the results in Numpy
matrices for processing. The data contains integers, floats, and
boolean values. The boolean values are represented in the
Hello,
I have just tried using alter_code1 on a code tree, and it appears to ignore
"from Numeric import *"
statements (does not change them to "from numpy.oldnumeric import *"). I
looked at the alter_code1.py and it says that the script should warn about
importing *. Although the script may no
OK, generally in code like this I leave the outer loops alone and try to
vectorize just the inner loop.I have some ideas in this direction, but
first, there seems to be some problems with the code at well. The code looks
like it is written to take non-square 'data' arrays. However,
for i in
Tim,
this is the best I could come up with until now:
import numpy as N
def naive_variogram(data, binsize=100., stepsize=5.):
"""calculates variograms along the rows and columns of the given
array which is supposed to contain equally spaced data with
stepsize stepsize"""
# how
On 6/22/07, Hanno Klemm <[EMAIL PROTECTED]> wrote:
Hi,
I have an array which represents regularly spaced spatial data. I now
would like to compute the (semi-)variogram, i.e.
gamma(h) = 1/N(h) \sum_{i,j\in N(h)} (z_i - z_j)**2,
where h is the (approximate) spatial difference between the
measu
Hi,
I have an array which represents regularly spaced spatial data. I now
would like to compute the (semi-)variogram, i.e.
gamma(h) = 1/N(h) \sum_{i,j\in N(h)} (z_i - z_j)**2,
where h is the (approximate) spatial difference between the
measurements z_i, and z_j, and N(h) is the number of measur
Hi,
I was wondering if the numpy function 'isinf' should return True for
complex infinity.
I encountered the following behavior that could be considered a bug:
>>> import numpy as N
>>> N.isinf(1j*N.inf)
True
>>> 1j/(N.array(1)-1)
(nannanj)
>>> N.isinf(1j/(N.array(1)-1))
False
>>> 1j/(N.
