[SciPy-User] numpy mean

[questions anon]

perfect (using axis=0), thank you!


On Thu, Jul 28, 2011 at 3:06 PM, Oleksandr Huziy <guziy.sasha at gmail.com>wrote:

> Hi,
>
> you can do the following
> >>> arr = np.array([array1,array2,array3])
> >>> arr
> array([[ 2,  4,  8],
>        [ 4,  8, 12],
>        [ 9,  3, 15]])
> >>> np.mean(arr, axis = 1)
> array([ 4.66666667,  8.        ,  9.        ])
> >>> np.mean(arr, axis = 0)
> array([  5.        ,   5.        ,  11.66666667])
>
> cheers
> --
> Oleksandr
>
> 2011/7/28 questions anon <questions.anon at gmail.com>
>
>> Hi All,
>> I thought this would be a relatively easy thing to do but the more I look
>> the more confused I become!
>>
>> I have a netcdf file containing hourly temperature data for given region
>> for a month.
>> I would like to find the mean/average of particular periods (i.e. 3
>> hours).
>>
>> e.g. this is what I would like:
>>
>> array1=[2,4,8]
>> array2=[4,8,12]
>> array3=[9,3,15]
>>
>> meanofarrays=np.mean(array1,array2,array3)
>> print meanofarrays
>> >>>[5,5,11]
>>
>> Is there a routine that will do what I am after?
>> If not I seem to be able to sum the arrays together and then divide by
>> another array, but I will need to produce an array to match the extent and
>> all values will need to be equal to the number of arrays I have summed. Can
>> anyone help with producing this array?
>>
>> i.e.
>> sumofarray=[15,15,35]
>> numberofarrays=[3,3,3]
>> meanofarrays=np.divide[sumofarrays,numberofarrays]
>> print meanofarrays
>> >>>[5,5,11]
>>
>> Any feedback will be greatly appreciated!!!
>>
>>
>>
>>
>>
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>>
>
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