Quoting Robert Kern robert.k...@gmail.com:
On Fri, Dec 9, 2011 at 11:00, Yang Zhang yanghates...@gmail.com wrote:
Thanks for the clarification. Alas. So is there no simple workaround
to making numpy work in environments such as Jepp?
I don't think so, no.
It is far from being an optimal
Hi,
I have never seen myself a NetCDF file but if your NetCDF file is
using HDF5 as format (possible since NetCDF 4 if I am not mistaken),
you should be able to use h5py or PyTables to access and or modify it.
Best regards,
Armando
Quoting Chao YUE chaoyue...@gmail.com:
Dear all,
I
Correct. I thought just multiplying by -1 and inverting the logical
condition would give me the same output.
This makes exactly what I want:
x= numpy.arange(10.)
delta=3
y=[x[0]]
for value in x:
... if (value-y[-1]) delta:
...y.append(value)
...
y
[0., 4., 8.]
Armando
? Well a loop or list comparison seems like a good choice to me. It is
much more obvious at the expense of two LOCs. Did you profile the two
possibilities and are they actually performance-critical?
cheers
The second is between 8 and ten times faster on my machine.
import numpy
import
.
Armando
Quoting Vicente Sole s...@esrf.fr:
? Well a loop or list comparison seems like a good choice to me. It is
much more obvious at the expense of two LOCs. Did you profile the two
possibilities and are they actually performance-critical?
cheers
The second is between 8 and ten times
Quoting josef.p...@gmail.com:
but the two options don't produce the same result in general, the
cumsum version doesn't restart from zero, I think
try
x0 = np.random.randint(5,size=30).cumsum()
with delta=3
I don't see a way around recursive looping
The x0 data are already sorted. It was
With A and X being arrays:
B=numpy.zeros(A.shape, A.dtype)
B[A0] = X
Armando
Quoting gerardob gberbeg...@gmail.com:
Let A be a square matrix of 0's and 1's, and let X be a one dimesional
vector.
The length of X is equal to the number of 1's that A has.
I would like to produce a new
Quoting Bruce Southey bsout...@gmail.com:
On 01/18/2010 12:47 PM, Vicente Sole wrote:
Quoting Bruce Southey bsout...@gmail.com:
If you obtain the code from any package then you are bound by the terms
of that code. So while a user might not be 'inconvenienced' by the LGPL,
they are required
