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: if ncfile[-3:]=='.nc': print "dealing with ncfiles:", ncfile ncfile=os.path.join(path,ncfile) ncfile=Dataset(ncfile, 'r+', 'NETCDF4') TSFC=ncfile.variables['T_SFC'][:] fillvalue=ncfile.variables['T_SFC']._FillValue TSFC=MA.masked_values(TSFC, fillvalue) ncfile.close() all_TSFC.append(TSFC) a=TSFC[0] for b in TSFC[1:]: N.maximum(a,b,out=a) big_array=N.ma.concatenate(all_TSFC) Max=big_array.max(axis=0) print "max is", Max,"a is", a On Wed, Dec 7, 2011 at 2:34 PM, Olivier Delalleau <shish@keba.be> wrote:
Is 'a' a regular numpy array or something fancier?
-=- Olivier
2011/12/6 questions anon <questions.anon@gmail.com>
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, <josef.pktd@gmail.com> wrote:
On Tue, Dec 6, 2011 at 9:36 PM, Olivier Delalleau <shish@keba.be> 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 which makes it more efficient, but in my example I used Python's reduce which takes an iterable and not one huge array.
Josef
-=- Olivier
2011/12/6 questions anon <questions.anon@gmail.com>
thanks for all of your help, that does look appropriate but I am not
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 at 1:12 PM, Olivier Delalleau <shish@keba.be> wrote:
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 Nathaniel Smith <njs@pobox.com> > > I think you want > np.maximum(a, b, out=a) > > - Nathaniel > > On Dec 6, 2011 9:04 PM, "questions anon" <questions.anon@gmail.com> > 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 want to replace them. The data are >> contained in netcdf files. >> I can achieve what I want if I combine all of my arrays using numpy >> concatenate and then using the command numpy.max(myarray, axis=0) but >> because I have so many arrays I end up with a memory error so I need to find >> a way to get the maximum while looping. >> >> >> >> On Wed, Dec 7, 2011 at 12:36 PM, <josef.pktd@gmail.com> wrote: >>> >>> On Tue, Dec 6, 2011 at 7:55 PM, Olivier Delalleau <shish@keba.be> >>> 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 <questions.anon@gmail.com> >>> >> >>> >> I would like to produce an array with the maximum values out of >>> >> many >>> >> (10000s) 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
sure 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 >>> >> 2,1,0,0 >>> >> 1,1,2,0] >>> >> >>> >> where b>a replace with value in b, so the new a should be : >>> >> >>> >> a=[1,1,3,2] >>> >> 2,1,2,2 >>> >> 1,1,2,2] >>> >> >>> >> and then keep looping through many arrays and replace whenever >>> >> value is >>> >> larger. >>> >> >>> >> I have tried numpy.putmask but that results in >>> >> TypeError: putmask() argument 1 must be numpy.ndarray, not list >>> >> Any other ideas? Thanks >>> >>> if I understand correctly it's a minimum.reduce >>> >>> numpy >>> >>> >>> a = np.concatenate((np.arange(5)[::-1], >>> >>> np.arange(5)))*np.ones((4,3,1)) >>> >>> np.minimum.reduce(a, axis=2) >>> array([[ 0., 0., 0.], >>> [ 0., 0., 0.], >>> [ 0., 0., 0.], >>> [ 0., 0., 0.]]) >>> >>> a.T.shape >>> (10, 3, 4) >>> >>> python with iterable >>> >>> >>> reduce(np.maximum, a.T) >>> array([[ 4., 4., 4., 4.], >>> [ 4., 4., 4., 4.], >>> [ 4., 4., 4., 4.]]) >>> >>> reduce(np.minimum, a.T) >>> array([[ 0., 0., 0., 0.], >>> [ 0., 0., 0., 0.], >>> [ 0., 0., 0., 0.]]) >>> >>> Josef >>> >>> >> >>> >> _______________________________________________ >>> >> NumPy-Discussion mailing list >>> >> NumPy-Discussion@scipy.org >>> >> http://mail.scipy.org/mailman/listinfo/numpy-discussion >>> >> >>> > >>> > >>> > _______________________________________________ >>> > NumPy-Discussion mailing list >>> > NumPy-Discussion@scipy.org >>> > http://mail.scipy.org/mailman/listinfo/numpy-discussion >>> > >>> _______________________________________________ >>> NumPy-Discussion mailing list >>> NumPy-Discussion@scipy.org >>> http://mail.scipy.org/mailman/listinfo/numpy-discussion >> >> >> >> _______________________________________________ >> NumPy-Discussion mailing list >> NumPy-Discussion@scipy.org >> http://mail.scipy.org/mailman/listinfo/numpy-discussion >> > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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