Hi Nathaniel,
The results of running memtest was a pass with no errors.
-Karl
Nathaniel Smith wrote:
>
> (You should still run memtest. It's very easy - just install it with your
> package manager, then reboot. Hold down the shift key while booting, and
> you'll get a boot menu. Choose memtest,
thanks for all your responses. I think I have FINALLY worked it out with
all of your help.
I just assigned one array from one ncfile to "a" at the beginning of my
code and then ran the loop and it worked!!
sorry for all the questions but I learn so much playing and getting ideas
from others.
Thanks
On 07.12.2011, at 5:54AM, questions anon wrote:
> sorry the 'all_TSFC' is for my other check of maximum using concatenate and
> N.max, I know that works so I am comparing it to this method. The only reason
> I need another method is for memory error issues.
> I like the code I have written so f
sorry the 'all_TSFC' is for my other check of maximum using concatenate and
N.max, I know that works so I am comparing it to this method. The only
reason I need another method is for memory error issues.
I like the code I have written so far as it makes sense to me. I can't get
the extra examples I
On 07/12/2011, at 1:49 PM, questions anon wrote:
> fillvalue=ncfile.variables['T_SFC']._FillValue
> TSFC=MA.masked_values(TSFC, fillvalue)
You can probably also eliminate the above two lines from your code.
> TSFC=ncfile.variables['T_SFC'][:]
If your NetCDF
On 07.12.2011, at 5:07AM, Olivier Delalleau wrote:
> I *think* it may work better if you replace the last 3 lines in your loop by:
>
> a=all_TSFC[0]
> if len(all_TSFC) > 1:
> N.maximum(a, TSFC, out=a)
>
> Not 100% sure that would work though, as I'm not en
I *think* it may work better if you replace the last 3 lines in your loop
by:
a=all_TSFC[0]
if len(all_TSFC) > 1:
N.maximum(a, TSFC, out=a)
Not 100% sure that would work though, as I'm not entirely confident I
understand your code.
-=- Olivier
2011/12/6 q
Something fancier I think,
I am able to compare the result with my previous method so I can easily see
I am doing something wrong.
see code below:
all_TSFC=[]
for (path, dirs, files) in os.walk(MainFolder):
for dir in dirs:
print dir
path=path+'/'
for ncfile in files:
Is 'a' a regular numpy array or something fancier?
-=- Olivier
2011/12/6 questions anon
> thanks again my only problem though is that the out=a in the loop does not
> seem to replace my a= outside the loop so my final a is whatever I started
> with for a.
> Not sure what I am doing wrong whethe
thanks again my only problem though is that the out=a in the loop does not
seem to replace my a= outside the loop so my final a is whatever I started
with for a.
Not sure what I am doing wrong whether it is something with the loop or
with the command.
On Wed, Dec 7, 2011 at 1:44 PM, wrote:
> On
On Tue, Dec 6, 2011 at 9:36 PM, Olivier Delalleau wrote:
> The "out=a" keyword will ensure your first array will keep being updated. So
> you can do something like:
>
> a = my_list_of_arrays[0]
> for b in my_list_of_arrays[1:]:
> numpy.maximum(a, b, out=a)
I didn't think of the out argument whi
The "out=a" keyword will ensure your first array will keep being updated.
So you can do something like:
a = my_list_of_arrays[0]
for b in my_list_of_arrays[1:]:
numpy.maximum(a, b, out=a)
-=- Olivier
2011/12/6 questions anon
> thanks for all of your help, that does look appropriate but I am
thanks for all of your help, that does look appropriate but I am not sure
how to loop it over thousands of files.
I need to keep the first array to compare with but replace any greater
values as I loop through each array comparing back to the same array. does
that make sense?
On Wed, Dec 7, 2011
Thanks, I didn't know you could specify the out array :)
(to the OP: my initial suggestion, although probably not very efficient,
seems to work with 2D arrays too, so I have no idea why it didn't work for
you -- but Nathaniel's one seems to be the ideal one anyway).
-=- Olivier
2011/12/6 Nathani
I think you want
np.maximum(a, b, out=a)
- Nathaniel
On Dec 6, 2011 9:04 PM, "questions anon" wrote:
> thanks for responding Josef but that is not really what I am looking for,
> I have a multidimensional array and if the next array has any values
> greater than what is in my first array I wan
If you need to do them one after the other, numpy.maximum(a, b) will do it
(it won't work in-place on 'a' though, it'll make a new copy).
-=- Olivier
2011/12/6 questions anon
> thanks for responding Josef but that is not really what I am looking for,
> I have a multidimensional array and if the
I have 2d numpy arrays
On Wed, Dec 7, 2011 at 1:05 PM, Olivier Delalleau wrote:
> Weird, it worked for me (with a and b two 1d numpy arrays). Anyway,
> Josef's solution is probably much more efficient (especially if you can put
> all your arrays into a single tensor).
>
>
> -=- Olivier
>
> 2011/
Weird, it worked for me (with a and b two 1d numpy arrays). Anyway, Josef's
solution is probably much more efficient (especially if you can put all
your arrays into a single tensor).
-=- Olivier
2011/12/6 questions anon
> Hi Olivier,
> No that does not seem to do anything
> am I missing another
thanks for responding Josef but that is not really what I am looking for, I
have a multidimensional array and if the next array has any values greater
than what is in my first array I want to replace them. The data are
contained in netcdf files.
I can achieve what I want if I combine all of my arra
Hi Olivier,
No that does not seem to do anything
am I missing another step whereever b is greater than a replace b with a?
thanks
On Wed, Dec 7, 2011 at 11:55 AM, Olivier Delalleau wrote:
> It may not be the most efficient way to do this, but you can do:
> mask = b > a
> a[mask] = b[mask]
>
> -=
On Mon, Dec 5, 2011 at 12:43 PM, Ralf Gommers
wrote:
> Hi all,
>
> It's been a little over 6 months since the release of 1.6.0 and the NA
> debate has quieted down, so I'd like to ask your opinion on the timing of
> 1.7.0. It looks to me like we have a healthy amount of bug fixes and small
> impro
On Tue, Dec 6, 2011 at 7:55 PM, Olivier Delalleau wrote:
> It may not be the most efficient way to do this, but you can do:
> mask = b > a
> a[mask] = b[mask]
>
> -=- Olivier
>
> 2011/12/6 questions anon
>>
>> I would like to produce an array with the maximum values out of many
>> (1s) of arr
It may not be the most efficient way to do this, but you can do:
mask = b > a
a[mask] = b[mask]
-=- Olivier
2011/12/6 questions anon
> I would like to produce an array with the maximum values out of many
> (1s) of arrays.
> I need to loop through many multidimentional arrays and if a value
I would like to produce an array with the maximum values out of many
(1s) of arrays.
I need to loop through many multidimentional arrays and if a value is
larger (in the same place as the previous array) then I would like that
value to replace it.
e.g.
a=[1,1,2,2
11,2,2
1,1,2,2]
b=[1,1,3,2
2,1
On 06.12.2011, at 11:13PM, Wes McKinney wrote:
> This isn't the place for this discussion but we should start talking
> about building a *high performance* flat file loading solution with
> good column type inference and sensible defaults, etc. It's clear that
> loadtable is aiming for highest com
Hi,
How to make Numpy to match Matlab in term of performance ? I have tryied
with different options, using different MKL libraries and ICC versions,
still Numpy is below Matalb for certain basic tasks by ~2x. About 5 years
ago I was able to get about same speed, not anymore. Matlab suppose to use
On Tue, Dec 6, 2011 at 4:11 PM, Ralf Gommers
wrote:
>
>
> On Mon, Dec 5, 2011 at 8:43 PM, Ralf Gommers
> wrote:
>>
>> Hi all,
>>
>> It's been a little over 6 months since the release of 1.6.0 and the NA
>> debate has quieted down, so I'd like to ask your opinion on the timing of
>> 1.7.0. It look
On Mon, Dec 5, 2011 at 8:43 PM, Ralf Gommers wrote:
> Hi all,
>
> It's been a little over 6 months since the release of 1.6.0 and the NA
> debate has quieted down, so I'd like to ask your opinion on the timing of
> 1.7.0. It looks to me like we have a healthy amount of bug fixes and small
> improv
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 06/12/11 10:45, Paul Anton Letnes wrote:
> As a side note: since the built-in search isn't really all that good,
> would it be possible to put a customized google search box there
> instead?
It is easy since Sphinx is being used. Copy searchbox.ht
On 6. des. 2011, at 17:32, Roger Binns wrote:
> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA1
>
> On 06/12/11 01:58, Pauli Virtanen wrote:
>> I think this cannot be helped --- it does not make sense to explain
>> basic Numpy concepts in every docstring, especially `axis` and `shape`
>> are v
Travis Oliphant wrote:
> My initial thoughts:
I don't have a horse in this race, but I do suggest people read Karl
Fogel's book before too much designing of governance structure:
http://producingoss.com/
(alas, it's not short, but it's a fairly easy read and you can get
convenient dead-tree or e
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 06/12/11 09:59, Chris Barker wrote:
> That would be a lot of links
Huh? The current page defines axis in terms of axis so if you don't know
what it is you have to look it up. The search is completely useless as I
showed. You want using numpy to
On 12/6/2011 9:54 AM, K.-Michael Aye wrote:
> I have a function f(x,y).
>
> I would like to calculate it at x = arange(20,101,20) and y = arange(2,30,2)
>
> How do I store that in a multi-dimensional array and preserve the grid
> points where I did the calculation
In [5]: X, Y = np.meshgrid(range
On 12/6/2011 8:32 AM, Roger Binns wrote:
>> I think this cannot be helped --- it does not make sense to explain
>> basic Numpy concepts in every docstring, especially `axis` and `shape`
>> are very common.
>
> They don't need to be explained on the page, but instead link to a page
> that does expla
Dear all,
I can't wrap my head around this. Mathematically it's not hard, I just
don't know how to store and access it without many loops.
I have a function f(x,y).
I would like to calculate it at x = arange(20,101,20) and y = arange(2,30,2)
How do I store that in a multi-dimensional array and
On Tue, Dec 6, 2011 at 2:51 AM, Xavier Barthelemy wrote:
> ok let me be more precise
>
> I have an Z array which is the elevation
> from this I extract a discrete array of Zero Crossing, and another
> discrete array of Crests.
> len(crest) is different than len(Xzeros). I have a threshold method
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 06/12/11 01:58, Pauli Virtanen wrote:
> I think this cannot be helped --- it does not make sense to explain
> basic Numpy concepts in every docstring, especially `axis` and `shape`
> are very common.
They don't need to be explained on the page, b
On Tue, Dec 6, 2011 at 7:53 AM, Matthew Brett wrote:
> Hi,
>
> On Tue, Dec 6, 2011 at 4:45 AM, Skipper Seabold wrote:
>> Hi,
>>
>> Is this intended?
>>
>> [~/]
>> [1]: np.result_type(np.uint, np.int)
>> [1]: dtype('float64')
>
> I would guess so - if your system ints are 64 bit. int64 can't
> co
Hi,
On Tue, Dec 6, 2011 at 4:45 AM, Skipper Seabold wrote:
> Hi,
>
> Is this intended?
>
> [~/]
> [1]: np.result_type(np.uint, np.int)
> [1]: dtype('float64')
I would guess so - if your system ints are 64 bit. int64 can't
contain the range for uint64, nor can uint64 contain all int64, If
there
Hi,
Is this intended?
[~/]
[1]: np.result_type(np.uint, np.int)
[1]: dtype('float64')
[~/]
[2]: np.version.version
[2]: '2.0.0.dev-aded70c'
Skipper
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy.org
http://mail.scipy.org/mailman/listinfo/nu
Yes I understood what you said. I know these tools, and I am using them.
I was just wandering if someone has a more one-liner-pythonic way to do it.
I don't think it's worth importing a new fortran module.
Thanks anyway :)
Xavier
2011/12/6 David Froger
> Excerpts from Xavier Barthelemy's message
06.12.2011 08:49, Roger Binns kirjoitti:
> Note that I am using regular Python lists (they were JSON at one point)
> and the fft documentation is incomprehensible to someone who hasn't used
> numpy before and only cares about fft (there are a lot of matches for
> Google searches about fft and pytho
Excerpts from Xavier Barthelemy's message of mar. déc. 06 08:51:22 +0100 2011:
> ok let me be more precise
>
> I have an Z array which is the elevation
> from this I extract a discrete array of Zero Crossing, and another discrete
> array of Crests.
> len(crest) is different than len(Xzeros). I hav
43 matches
Mail list logo
