Thank you, thank you, thank you!
I needed to find the max value (and corresponding TIME and LAT, LON) for
the entire month but I shouldn't have been using the tmax, instead I needed
to use the entire array. Below code works for those needing to do something
similar.
Thanks for all your help everyon
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 hourly data
> for a whole month I assume that is where my problem lies.
> When I run the following code I alwa
Do you mean that listval[0] is systematically equal to 0, or is it
something else?
-=- Olivier
2012/1/9 questions anon
> 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 hourly data
> for a whole month I assume t
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 hourly data
for a whole month I assume that is where my problem lies.
When I run the following code I alwayes receive the first timestamp of the
file. Not sure how to ge
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, LON.shape
> (721, 106, 193) (721,) (106,) (193,)
>
> So I still receive index out of bounds error:
tmax=TSFC.max(axis=0)
> nu
thanks for the responses.
Unfortunately they are not matching shapes
>>> print TSFC.shape, TIME.shape, LAT.shape, LON.shape
(721, 106, 193) (721,) (106,) (193,)
So I still receive index out of bounds error:
>>>tmax=TSFC.max(axis=0)
numpy array of max values for the month
>>>maxindex=tmax.argmax()
Oh, sorry, I hadn't paid enough attention to the way you are indexing A: if
you are using an array to index, it creates a copy, so using ".fill" will
fill the copy and you won't see the result.
Instead, use A[0:3], A[3:6], etc.
-=- Olivier
2012/1/9 "David Köpfer"
> Hi Oliver,
>
> thank you ver
Hi Oliver,
thank you very much for your reply, sadly it is not working as you and I hoped.
The array still stays at None even after the code.
I've also tried A[X] = [MyObject(...)]*len(X) but that just results in a Memory
error.
So is there really no way to avoid this broadcasting?
David
-
