[SciPy-User] 3D array problem in Python

[Thøger Rivera-Thorsen]

By the way;

I don't know if they are useful to you, but here are some introductory 
slides from a workshop I gave for our undergraduate studemts at 
Stockholm University. The Python part starts at slide 18.

Emil


On 12/30/2012 05:16 PM, Happyman wrote:
> Hi
>
> Thanks for your answer... I also tried to do so unfortunately I could 
> not manage that because of no very good experience in Python...
> The way you propose is really good indeed but how can i do it???
>
> Two huge problems I always encounter in Python are:
> 1) Array with scalars...always!!!!!
> 2) for loop      I can understand Python is script language but how to 
> deal with it???
>
> Especially, when I create some function in which arguments can take 
> one value are messed up with negotiation wit outer coming variables 
> into my function!!
> for example,
> arg1, arg2 are arrays!
> def f( arg1, arg2 ):
>        some process, loops everthing
>    return val
>
>
>
>
>
> Воскресенье, 30 декабря 2012, 16:47 +01:00 от Thøger Rivera-Thorsen 
> <thoger.emil at gmail.com>:
>
>     Use np.where() or logical indexing (same thing, really) to mask
>     your array, then perform the operations. Let's say your array is
>     called A:
>
>     A[float(A) == 0.0] = 0.0
>
>     A[float(A) != 0.0] = [...etc.]
>
>
>     This, of course, only works if the operation for an entry doesn't
>     depend on other entries in the array; but it should give you a
>     great speed gain.
>
>     Cheers;
>
>     Emil
>
>
>
>     On 12/30/2012 04:32 AM, Happyman wrote:
>>     Hello
>>
>>     I have 3 dimensional array  which I want  to calculate in a huge
>>     process. Everything is working well if I use ordinary way which
>>     is unsuitable in Python like the following:
>>
>>     nums=32
>>     rows=120
>>     cols=150
>>
>>     for k in range(0,nums):
>>               for i in range(0,rows):
>>      for j in range(0,cols):
>>                     if float ( R[ k ] [ i ] [ j ] ) == 0.0:
>>                                val11 [ i ] =0.0
>>                     else:
>>                                val11[ i ] [ j ], val22[ i ][ j ] =
>>     integrate.quad( lambda x :  F1(x)*F2(x) , 0 , pi)
>>
>>     But, this calculation takes so long time, let's say about  1 hour
>>     (theoretically)... Is there any better way to easily and fast
>>     calculate the process such as [ F( i ) for i in xlist ] or
>>     something like that rather than using for loop?
>>
>>
>>
>>
>>     _______________________________________________
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>
>

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