perfect, thank you!
On Tue, Mar 24, 2015 at 9:14 PM, Kiko wrote:
>
>
> 2015-03-24 11:02 GMT+01:00 questions anon :
>
>> I would like to find the nearest coord in a netcdf from a given latitude
>> and longitude.
>> I found some fantastic code that does this -
I would like to find the nearest coord in a netcdf from a given latitude
and longitude.
I found some fantastic code that does this -
http://nbviewer.ipython.org/github/Unidata/unidata-python-workshop/blob/master/netcdf-by-coordinates.ipynb
but I keep receiving this error - I am receiving a ValueErr
#x27;time.hour').max()
>
> It also might be worth looking at other more data analysis packages,
> either more generic (e.g., pandas, http://pandas.pydata.org/) or
> weather/climate data specific (e.g., Iris, http://scitools.org.uk/iris/and
> CDAT,
> http://www2-pcmdi.llnl.gov/cdat
I have hourly 2D temperature data in a monthly netcdf and I would like to
find the daily maximum temperature. The shape of the netcdf is (744, 106,
193)
I would like to use the year-month-day as a new list name (i.e. 2009-03-01,
2009-03-022009-03-31) and then add each of the hours worth of
tem
Hello all,
I have netcdf files that contain hourly rainfall data. Each netcdf file
includes one months worth of hours and I have 10 years worth of data. I
would like to calculate the sum of each month and then the mean of these
summed months across all of the years.
I have no problem firstly calc
Thank you Scott for your prompt response.
Your suggestion has fixed the problem and thank you for your clear
explanation of how it works.
thanks!!
On Thu, Nov 28, 2013 at 7:20 PM, Scott Sinclair wrote:
> On 28 November 2013 09:06, questions anon
> wrote:
> > I have a separate t
Hi All,
I just posted this on the SciPy forum but realised it might be more
appropriate here?
I have a separate text file for daily rainfall data that covers the whole
country. I would like to calculate the monthly mean, min, max and the mean
of the sum for one state.
The mean, max and min are jus
excellent thank you, that worked perfectly. I just need to remember this
feature next time I need it.
Thanks again
On Thu, Apr 12, 2012 at 11:41 PM, Tim Cera wrote:
> Use 'ma.max' instead of 'np.max'. This might be a bug OR an undocumented
> feature. :-)
>
> import numpy.ma as ma
> mar
I am trying to mask an array and then add the array to a list, so I can
then go on and calculate the max, min and mean of that list.
The mask seems to work when I check each array. I check each array by
finding the max, mean and mean and comparing with the unmasked array (they
are different).
Howev
sure I understand how to
write the second suggestion about the iterative approach but will have a go.
Thanks again
On Wed, Jan 25, 2012 at 1:26 PM, Brett Olsen wrote:
> On Tue, Jan 24, 2012 at 6:22 PM, questions anon
> wrote:
> > I need some help understanding how to loop through ma
I need some help understanding how to loop through many arrays to calculate
the 95th percentile.
I can easily do this by using numpy.concatenate to make one big array and
then finding the 95th percentile using numpy.percentile but this causes a
memory error when I want to run this on 100's of netcd
rint ncfiletime
On Tue, Jan 10, 2012 at 3:28 PM, Aronne Merrelli
wrote:
>
>
> On Mon, Jan 9, 2012 at 7:59 PM, questions anon
> wrote:
>
>> thank you, I seem to have made some progress (with lots of help)!!
>> I still seem to be having trouble with the time. Because it is
print ncfiletime
On Tue, Jan 10, 2012 at 10:22 AM, Benjamin Root wrote:
>
>
> On Monday, January 9, 2012, questions anon
> wrote:
> > thanks for the responses.
> > Unfortunately they are not matching shapes
> >>>> print TSFC.shape, TIME.shape, LAT.shape,
hanks
On Wed, Jan 4, 2012 at 10:29 PM, Derek Homeier <
de...@astro.physik.uni-goettingen.de> wrote:
> On 04.01.2012, at 5:10AM, questions anon wrote:
>
> > Thanks for your responses but I am still having difficuties with this
> problem. Using argmax gives me one very large value
(TSFC)
data are hourly for a whole month.
Are there any other ideas for finding the location and time of the maximum
value in an array?
Thanks
On Wed, Dec 21, 2011 at 3:38 PM, Benjamin Root wrote:
>
>
> On Tuesday, December 20, 2011, questions anon
> wrote:
> > ok thanks,
this may not
> be helpful, but just in case... I find it suspicious that you *seem* (by
> quickly glancing at the code) to be taking TIME[max(temperature)] instead
> of TIME[argmax(temperature)].
>
> -=- Olivier
>
> 2011/12/20 questions anon
>
>> I have a netcdf file
I have a netcdf file that contains hourly temperature data for a whole
month. I would like to find the maximum temperature within that file and
also the corresponding Latitude and Longitude and Time and then plot this.
Below is the code I have so far. I think everything is working except for
identi
.close()
for b in TSFC[:]:
N.maximum(a,b, out=a)
print a
On Wed, Dec 7, 2011 at 4:11 PM, Derek Homeier <
de...@astro.physik.uni-goettingen.de> wrote:
> On 07.12.2011, at 5:54AM, questions anon wrote:
>
> > sorry the 'all_TSFC' is for m
a)
>
> Not 100% sure that would work though, as I'm not entirely confident I
> understand your code.
>
>
> -=- Olivier
>
> 2011/12/6 questions anon
>
>> Something fancier I think,
>> I am able to compare the result with my previous method so I can easily
>
max(axis=0)
print "max is", Max,"a is", a
On Wed, Dec 7, 2011 at 2:34 PM, Olivier Delalleau wrote:
> Is 'a' a regular numpy array or something fancier?
>
>
> -=- Olivier
>
> 2011/12/6 questions anon
>
>> thanks again my only problem thou
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 que
iel's one seems to be the ideal one anyway).
>
> -=- Olivier
>
>
> 2011/12/6 Nathaniel Smith
>
>> I think you want
>> np.maximum(a, b, out=a)
>>
>> - Nathaniel
>> On Dec 6, 2011 9:04 PM, "questions anon"
>> wrote:
>>
>>
t; -=- Olivier
>
> 2011/12/6 questions anon
>
>> 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 Delalle
eau 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 maxim
sk] = 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 is
>> larger (in the same place as the
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
running_sum=N.array(TSFC[i])
>>
>> TSFC_avg=N.true_divide(running_sum, slice_counter)
>> N.set_printoptions(threshold='nan')
>> print "the TSFC_avg is:", TSFC_avg
>>
>>
>>
>>
>> On Tue, Dec 6, 2011 at 9:50
de:
>
> for each slice:
> indexnonNaN=np.isfinite(SliceOf Toto)
> SliceOf TotoWithoutNan= SliceOf Toto [indexnonNaN]
>
> and then perform all operation I want o on the last array.
>
> i hope it does answer your question
>
> Xavier
>
>
> 2011/12/6 questions anon
ce_counter)
N.set_printoptions(threshold='nan')
print "the TSFC_avg is:", TSFC_avg
On Tue, Dec 6, 2011 at 9:45 AM, David Cournapeau wrote:
> On Mon, Dec 5, 2011 at 5:29 PM, questions anon
> wrote:
> > Maybe I am asking the wrong question or could go about this anot
Dec 1, 2011 at 12:16 PM, questions anon wrote:
> I am trying to calculate the mean across many netcdf files. I cannot use
> numpy.mean because there are too many files to concatenate and I end up
> with a memory error. I have enabled the below code to do what I need but I
> have a few n
I would like to calculate the max and min of many netcdf files.
I know how to create one big array and then concatenate and find the
numpy.max but when I run this on 1000's of arrays I have a memory error.
What I would prefer is to loop through the arrays and produce the maximum
without having the
I am trying to calculate the mean across many netcdf files. I cannot use
numpy.mean because there are too many files to concatenate and I end up
with a memory error. I have enabled the below code to do what I need but I
have a few nan values in some of my arrays. Is there a way to ignore these
some
> numpy.tile(mask.mask, [number_of_repeats] + [1] * len(mask.mask.shape))
>
> (not sure that's the most elegant way to do it, but it should work)
>
>
> -=- Olivier
>
> 2011/11/21 questions anon
>
>> Excellent, thank you.
>> I just realised this does not wor
new array is x, you can use:
>
> numpy.ma.masked_array(x, mask=mask.mask)
>
> -=- Olivier
>
> 2011/11/21 questions anon
>
>> I am trying to mask one array using another array.
>>
>> I have created a masked array using
>> mask=MA.masked_equal(myarray,
>> 0),
&g
I am trying to mask one array using another array.
I have created a masked array using
mask=MA.masked_equal(myarray,
0),
that looks something like:
[1 - - 1,
1 1 - 1,
1 1 1 1,
- 1 - 1]
I have an array of values that I want to mask whereever my mask has a a '-'.
how do I do this?
I
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